Chapter test · nothing is marked until you submit

Zeros of Polynomials: Chapter Test

20 multiple-choice questions and 10 free-response questions, drawn from across the chapter and mixed together.

Free response

10 questions in parts, 144 points in total. Work them out on paper. There are no hints here: reveal each question's answer, worked solution, and rubric when you are ready to mark that one.

Free response · work it on paper

Question 1 of 10

1. Nothing left to break down . 13 points. Question 1 of 10.

A polynomial handed to you in factored form has already answered most counting questions, provided you notice that the word "roots" is asking more than one of them. Work throughout with

$p(x) = 3(x - 5)^3(x + 1)(x^2 + 16).$
1. Part A.
  Write the complete factorization of $p$ over the complex numbers. Then give the degree of $p$ , the number of its roots counted with multiplicity, and the number of its distinct roots.
  
  Write the expression An equation or an expression is enough here. Show how you built it. 4 points
2. Part B.
  Of the roots counted with multiplicity, state how many are real and how many are not, naming them. Then state the largest number of distinct roots any polynomial of this degree could have, and say what stops $p$ from reaching it.
  
  Carry your own answer forward Continue from the factorization and the counts you reached in Part A, whatever they came out to be. The credit here is for sorting your own root list and comparing it against the ceiling, not for repeating Part A.
  
  Solve and show your work Write each step out, and end with the value and its units. 4 points
3. Part C.
  Decide whether there is a polynomial of degree $6$ with complex coefficients whose distinct roots are exactly the four distinct roots of $p$ and none of whose roots is repeated. Justify your decision, and state the identity that settles it.
  
  Justify your claim State the claim, then give the reason it has to be true. 5 points
The answer
Part A

$p(x) = 3(x - 5)^3(x + 1)(x - 4i)(x + 4i)$ , of degree $6$ , with $6$ roots counted with multiplicity and $4$ distinct roots.
- The two nonreal factors may be written in either order; what is not the same is leaving $x^2 + 16$ standing, which is a factorization that has stopped early
Part B

Four are real, $5$ three times and $-1$ once; two are nonreal, $4i$ and $-4i$ . At most $6$ distinct roots are possible, and $p$ falls two short because $5$ takes three places.

Part C

There is none. Four distinct roots with every one simple supply exactly four linear factors, so the degree would be $4$ . The identity that settles it is that the multiplicities of the distinct roots add to the degree, and four multiplicities all equal to $1$ can only add to $4$ .
Worked solution

Part A

A factorization is complete only when every factor is linear, so the quadratic has to be split. Solving $x^2 + 16 = 0$ gives $x^2 = -16$ , and $(4i)^2 = 16i^2 = -16$ , as does $(-4i)^2$ .

$x^2 + 16 = (x - 4i)(x + 4i)$

Substituting that in leaves every factor linear.

$p(x) = 3(x - 5)^3(x + 1)(x - 4i)(x + 4i)$

Count the linear factors: $3 + 1 + 1 + 1 = 6$ , so the degree is $6$ and there are six roots counted with multiplicity, namely $5$ three times and then $-1$ , $4i$ and $-4i$ once each. Four different numbers appear on that list, so $p$ has four distinct roots, and the multiplicities $3, 1, 1, 1$ add to $6$ , which is the check worth running every time.

Part B

Sort the list of six roots by whether the imaginary part is zero. The numbers $5$ , $5$ , $5$ and $-1$ are real, and $4i$ and $-4i$ are not, so four of the six places are real and two are not.

$\underbrace{5,\ 5,\ 5,\ -1}_{\text{real}}, \qquad \underbrace{4i,\ -4i}_{\text{nonreal}}$

Notice the nonreal roots arrived in a matched pair, which the real coefficients of $p$ had already guaranteed.

The ceiling on distinct roots is the degree itself, because six linear factors can name at most six different numbers, and equality happens exactly when every root is simple. Here the root $5$ is not simple: it occupies three of the six places while contributing one value.

$6 - 4 = 2 = (3 - 1) + (1 - 1) + (1 - 1) + (1 - 1)$

The shortfall of $2$ is exactly the surplus copies that the repeated root spends, which is the whole reason the two counts differ.

Part C

Let the distinct roots be $s_1, s_2, s_3, s_4$ with multiplicities $m_1, m_2, m_3, m_4$ . Every one of the degree's linear factors carries exactly one of those values, so

$m_1 + m_2 + m_3 + m_4 = n.$

Suppose no root is repeated. Then every $m_i$ equals $1$ , and the left-hand side is $1$ added to itself four times, so $n = 4$ . A polynomial meeting the description would therefore have degree $4$ , not $6$ , and no such degree $6$ polynomial exists.

Read the same identity the other way and it tells you what a degree $6$ polynomial with these four values as its distinct roots must look like: the four multiplicities are positive integers adding to $6$ , so the surplus of $2$ has to be distributed among them. Either one value takes three places, as it does in $p$ , or two values take two places each. Both are genuinely available, so the repetition is forced but its shape is not.

It is worth naming what the argument never used. It did not use which numbers the four roots are, or whether any of them is real, or what the leading coefficient is. Only the count of distinct roots and the count of factors did any work, which is why the conclusion holds for every polynomial of degree $6$ with four distinct roots and not merely for this one.

In one line

The complete factorization is $p(x) = 3(x - 5)^3(x + 1)(x - 4i)(x + 4i)$ , of degree $6$ , with $6$ roots counted with multiplicity and $4$ distinct roots. Four of those six places are real, $5$ three times and $-1$ once, and two are nonreal, $4i$ and $-4i$ . A degree $6$ polynomial has at most $6$ distinct roots, and $p$ falls two short because $5$ takes three places while contributing one value. No polynomial of degree $6$ can have those four values as its distinct roots with none repeated, since four multiplicities all equal to $1$ add to $4$ , and the multiplicities of the distinct roots always add to the degree.

Rubric

Mark your own paper against this. Give yourself the points for each element you actually wrote down, not the ones you meant to.

Part A 4 points

Splits the quadratic factor into two linear factors over the complex numbers rather than leaving the factorization part-finished. . Worth 2 points.

Reports the degree and the two counts as separate numbers, and shows the multiplicities adding to the degree. . Worth 2 points.

Part B 4 points

Splits the root list by real and nonreal, counting with multiplicity rather than by distinct value. . Worth 2 points.

Names the ceiling as the degree and attributes the shortfall to the surplus copies a repeated root spends. . Worth 2 points.

Part C 5 points

Rules the polynomial out and derives the degree $4$ from every multiplicity being $1$ , rather than asserting that four roots cannot fill six places. . Worth 3 points. needs an explanation, not just an answer

States the identity that the multiplicities of the distinct roots add to the degree, and uses it as the ground of the argument. . Worth 2 points.
2. Counting under a real hypothesis . 15 points. Question 2 of 10.

A polynomial $p$ has real coefficients, degree $6$ , and leading coefficient $2$ . The number $2 - 7i$ is a root of $p$ of multiplicity $2$ , and $-4$ is the only real root of $p$ .
1. Part A.
  Give the multiplicity of the root $-4$ , and write out the complete root list of $p$ counted with multiplicity.
  
  Solve and show your work Write each step out, and end with the value and its units. 5 points
2. Part B.
  Write $p$ as a product of a constant, real linear factors, and real quadratic factors, with no factor breaking down further over the real numbers. Show the step that turns the conjugate pair into that quadratic.
  
  Carry your own answer forward Build the factorization from the root list you produced in Part A, even if that list was not the expected one. Credit here follows the conversion of a conjugate pair into a real quadratic and the assembly of the factors.
  
  Write the expression An equation or an expression is enough here. Show how you built it. 5 points
3. Part C.
  Suppose the degree had been $7$ instead of $6$ , with everything else in the description unchanged. Say what is then forced about the multiplicity of $-4$ and what is not, and explain why a polynomial of odd degree with real coefficients can never avoid a real root, however its nonreal roots are arranged.
  
  Justify your claim State the claim, then give the reason it has to be true. 5 points
The answer

Part A

The root $-4$ has multiplicity $2$ , and the complete list is $2 - 7i$ , $2 - 7i$ , $2 + 7i$ , $2 + 7i$ , $-4$ , $-4$ .

Part B

$p(x) = 2(x + 4)^2\bigl(x^2 - 4x + 53\bigr)^2$ .

Part C

The multiplicity of $-4$ is forced to be odd and at least $1$ , but not to any one value: it is $3$ if no further nonreal roots appear and $1$ if one more conjugate pair does. No odd degree can avoid a real root, since the nonreal roots occupy an even number of places and an even number taken from an odd one leaves at least one behind.

Worked solution

Part A

The coefficients are real, so a nonreal root drags its conjugate along, and it does so with the same multiplicity. That fixes four of the six places at once.

$2 - 7i,\quad 2 - 7i,\quad 2 + 7i,\quad 2 + 7i$

Two places are left. Neither can hold a nonreal root: a nonreal root would need its own conjugate as well, the two would use both remaining places, and then $-4$ would not be on the list at all, contradicting that it is a root.

So both remaining roots are real. The problem says $-4$ is the only real root of $p$ , so neither of them can be any other number, and both are $-4$ .

$6 = \underbrace{2 + 2}_{\text{nonreal}} + \underbrace{2}_{\text{real}}$

The multiplicity of $-4$ is therefore $2$ , and the complete list is $2 - 7i$ twice, $2 + 7i$ twice, and $-4$ twice.

Part B

The two real roots give the factor $(x + 4)$ twice. The conjugate pair is what needs work, and multiplying the pair together is the step engineered to clear the $i$ .

$\bigl(x - (2 - 7i)\bigr)\bigl(x - (2 + 7i)\bigr) = x^2 - \bigl[(2 - 7i) + (2 + 7i)\bigr]x + (2 - 7i)(2 + 7i)$

The sum is $4$ and the product is $2^2 + 7^2 = 53$ , both real, so the quadratic factor is $x^2 - 4x + 53$ . Its discriminant is $16 - 212 = -196$ , which is negative, so it has no real root and cannot be broken down any further over the reals.

The pair occurs twice, so the quadratic occurs twice as well, and the leading coefficient goes out front.

$p(x) = 2(x + 4)^2\bigl(x^2 - 4x + 53\bigr)^2$

Count the degrees as a check: $2 + 2 \cdot 2 = 6$ , which is $n = k + 2m$ with $k = 2$ real linear factors and $m = 2$ real quadratic ones.

Part C

Take the count first. Four places are still spent on $2 \pm 7i$ , so at degree $7$ there are three left rather than two. Every real place among them is $-4$ , since $-4$ is the only real root, and any nonreal place needs a partner, so the nonreal ones come two at a time. That leaves exactly two arrangements.

$7 = \underbrace{2 + 2}_{2 \pm 7i} + \underbrace{3}_{-4} \qquad\text{or}\qquad 7 = \underbrace{2 + 2}_{2 \pm 7i} + \underbrace{2}_{\text{a second pair}} + \underbrace{1}_{-4}$

So the multiplicity of $-4$ is $3$ or $1$ , and nothing in the description chooses between them: the stem fixes the multiplicity of $2 - 7i$ but never says it is the only nonreal root. What is forced is the parity, since three places minus an even number of nonreal ones is odd, and in particular $-4$ really is a root.

Now the general statement. Let $p$ be real of odd degree $n$ , and factor it over the reals as $k$ linear factors and $m$ quadratics of negative discriminant, so that

$n = k + 2m.$

The nonreal roots all come from the quadratics, two apiece, so they number $2m$ , an even number, always. Then $k = n - 2m$ is odd minus even, which is odd, and an odd number is at least $1$ . So at least one real linear factor $x - r$ survives, and its $r$ is a real root.

The argument never drew a curve. It is the pairing, not the picture, that forces the real root, and the picture of an odd-degree graph running off in opposite directions is a second, independent route to the same conclusion.

In one line

The root $-4$ has multiplicity $2$ , so the complete list is $2 - 7i$ , $2 - 7i$ , $2 + 7i$ , $2 + 7i$ , $-4$ , $-4$ , and the real factorization is $p(x) = 2(x + 4)^2(x^2 - 4x + 53)^2$ , whose quadratic has discriminant $-196$ and so is unbreakable over the reals. At degree $7$ the leftover count rises to three, and the multiplicity of $-4$ is then forced only to be odd: it is $3$ if no further nonreal roots appear and $1$ if a second conjugate pair does. No real polynomial of odd degree can avoid a real root, because its nonreal roots occupy an even number of places, so $k = n - 2m$ is odd and therefore at least $1$ .

Rubric

Mark your own paper against this. Give yourself the points for each element you actually wrote down, not the ones you meant to.

Part A 5 points

Pairs the nonreal root with its conjugate at the SAME multiplicity, not merely as one extra root. . Worth 2 points.

Rules out a nonreal occupant of the remaining places, then uses the only-real-root condition to force both of them to be $-4$ . . Worth 2 points.

Reports a list of six entries whose multiplicities add to the degree. . Worth 1 point.

Part B 5 points

Multiplies the conjugate pair into a quadratic and shows the sum and the product coming out real. . Worth 3 points.

Repeats the quadratic factor as often as the pair occurs and keeps the leading coefficient. . Worth 1 point.

Checks that the factor degrees add to $6$ , or that the quadratic's discriminant is negative so it is genuinely unbreakable over the reals. . Worth 1 point.

Part C 5 points

Says the multiplicity of $-4$ is forced only to be odd, and gives both arrangements that realise it, one with a second conjugate pair and one without. . Worth 2 points.

Argues the odd-degree claim from the nonreal roots occupying an even count, rather than from the shape of a graph. . Worth 3 points. needs an explanation, not just an answer
3. Which side, and where it turns . 15 points. Question 3 of 10.

A factored form is a blueprint of a picture. Everything asked below about

$P(x) = 2(x + 6)(x - 2)^2(x - 7)$

can be read from that form, and nothing below needs the expansion.
1. Part A.
  Give $P$ 's degree and its leading coefficient, and the direction of each arm. Then name every real zero with its multiplicity, and say for each whether the curve passes through the axis or turns back.
  
  Solve and show your work Write each step out, and end with the value and its units. 5 points
2. Part B.
  Take each interval that the real zeros leave behind, the unbounded ones included. Choose a value inside it, never a zero itself, evaluate $P$ there, and give the sign that value reveals. Add $P(0)$ as an anchor.
  
  Solve and show your work Write each step out, and end with the value and its units. 5 points
3. Part C.
  Name every set of $x$ for which $P(x) \ge 0$ . Then decide whether the point $(2, 0)$ is the highest point of the graph, and justify the decision.
  
  Carry your own answer forward Answer from your own sign row in Part B, even if it did not come out as expected. The credit here is for turning a sign row into a solution set and for the reasoning about the point on the axis, not for reproducing one particular set of test values.
  
  Justify your claim State the claim, then give the reason it has to be true. 5 points
The answer

Part A

Degree $4$ , leading coefficient $2$ , so both arms rise. The zeros are $-6$ and $7$ , each simple and each crossed, and $2$ , of multiplicity $2$ , where the curve touches and turns back.

Part B

$P(-7) = 2268$ , $P(0) = -336$ , $P(3) = -72$ , $P(8) = 1008$ , so the signs run positive, negative, negative, positive. The $y$ -intercept is $-336$ .

Part C

$P(x) \ge 0$ exactly for $x \le -6$ , at the single point $x = 2$ , and for $x \ge 7$ . The point $(2, 0)$ is the highest point of the curve on an interval around it, since $P$ is negative on both sides and zero there, but it is not the highest point overall, because both arms rise without bound.

Worked solution

Part A

Multiply the leading terms of the factors to get the leading term, since nothing else can contribute to the highest power.

$2 \cdot x \cdot x^2 \cdot x = 2x^4$

The degree is $4$ , which is even, and the leading coefficient $2$ is positive, so both arms point upward.

The zeros are the numbers that kill a factor, read with the sign flipped from what appears inside each bracket: $(x + 6)$ carries $-6$ , $(x - 2)^2$ carries $2$ , and $(x - 7)$ carries $7$ . Their exponents are the multiplicities, $1$ , $2$ and $1$ , and they add to $4$ , matching the degree.

Parity decides the behaviour at each: odd multiplicity crosses, even multiplicity touches and turns back. So the curve crosses at $-6$ , touches at $2$ , and crosses at $7$ .

Part B

Choose easy numbers strictly inside each interval and evaluate the factored form directly, since a product of four numbers is quicker than an expansion.

$P(-7) = 2(-1)(81)(-14) = 2268, \qquad P(0) = 2(6)(4)(-7) = -336$

$P(3) = 2(9)(1)(-4) = -72, \qquad P(8) = 2(14)(36)(1) = 1008$

One value settles an entire interval, because an interval containing no real zero cannot contain a sign change: a change would force a zero strictly between the two points, and there is none. So the signs run positive, negative, negative, positive.

The test value at $x = 0$ doubles as the anchor, so the curve passes through $(0, -336)$ .

Run the two checks the work owes. The outer signs are both positive, which is what two upward arms demanded. And the sign flips at $-6$ and at $7$ , the two odd zeros, and does not flip at $2$ , the even one, exactly as the multiplicities said.

Part C

Read the sign row and include the zeros, since the question allows equality.

$P(x) \ge 0 \quad\text{for}\quad x \le -6, \quad x = 2, \quad x \ge 7$

The isolated point is the interesting entry. At $x = 2$ the value is zero, but on either side of it the value is negative, so $2$ belongs to the solution set entirely on its own, with no interval of its own around it.

That same observation answers the second question in one direction. Since $P(2) = 0$ and $P$ is negative at every nearby $x$ on both sides, no point of the curve near $2$ is higher than $(2, 0)$ , so it is a genuine high point locally. This is the turning point the even multiplicity guarantees, and it is one of the very few turning points whose coordinates algebra can name exactly.

It is not the highest point of the whole graph. Both arms rise, and Part B already produced a value above it: $P(8) = 1008 > 0$ . One evaluated point is enough to refute a claim about the maximum, and it is the only kind of evidence worth offering, since a hump drawn without being evaluated proves nothing about how high the curve climbs.

In one line

The degree is $4$ with leading coefficient $2$ , so both arms rise; the zeros are $-6$ (multiplicity $1$ , crosses), $2$ (multiplicity $2$ , touches and turns back) and $7$ (multiplicity $1$ , crosses). Testing gives $P(-7) = 2268$ , $P(0) = -336$ , $P(3) = -72$ and $P(8) = 1008$ , so the signs run positive, negative, negative, positive, the flips fall at exactly the odd zeros, and the outer signs match the arms. Hence $P(x) \ge 0$ for $x \le -6$ , at the single point $x = 2$ , and for $x \ge 7$ . The point $(2, 0)$ is the highest point of the curve near $2$ , since $P$ is negative on both sides, but not the highest overall, as $P(8) = 1008$ shows.

Rubric

Mark your own paper against this. Give yourself the points for each element you actually wrote down, not the ones you meant to.

Part A 5 points

Gets the degree and leading coefficient by multiplying leading terms, without expanding, and turns them into the direction of both arms. . Worth 2 points.

Reads each zero with the correct sign and attaches its multiplicity, then converts the parity of each multiplicity into crossing or touching. . Worth 3 points.

Part B 5 points

Uses one test value strictly inside each of the four intervals, none of them a zero, and evaluates correctly. . Worth 2 points.

Reports a sign for every interval and gives the $y$ -intercept as a coordinate or a value of $P$ . . Worth 1 point.

Reconciles the outer signs with the arms and the pattern of flips with the multiplicities from Part A. . Worth 2 points.

Part C 5 points

Gives the solution set of $P(x) \ge 0$ including the isolated point at the even zero, not just the two unbounded pieces. . Worth 2 points.

Distinguishes a highest point near the zero from a highest point overall, and supports the second half with an evaluated value or with the direction of the arms. . Worth 3 points. needs an explanation, not just an answer
4. Totals from a cubic nobody can solve . 15 points. Question 4 of 10.

Not one of the candidates the rational root theorem permits for

$p(x) = 5x^3 - 15x^2 + 2x + 4$

returns the value zero, so its three roots $r_1$ , $r_2$ , $r_3$ are out of reach by hand. Every quantity below is still available.
1. Part A.
  Report $e_1$ , $e_2$ and $e_3$ for $p$ , and say in words what each of the three collects from the root list.
  
  Solve and show your work Write each step out, and end with the value and its units. 5 points
2. Part B.
  Find $\dfrac{1}{r_1} + \dfrac{1}{r_2} + \dfrac{1}{r_3}$ . State the condition this quantity needs in order to exist at all, and say which coefficient of $p$ certifies that the condition holds.
  
  Carry your own answer forward Use the symmetric sums you produced in Part A, even if they were not the expected values. Credit here follows building the reciprocal sum out of them and stating the condition it needs.
  
  Solve and show your work Write each step out, and end with the value and its units. 5 points
3. Part C.
  Both quantities above were computed without a single root being found. State the property an expression in $r_1$ , $r_2$ , $r_3$ must have for that to be possible, then give one expression in these roots that has the property and one that does not, saying what goes wrong for the second.
  
  Explain why it works A sentence or two. Reasons, not steps. 5 points
The answer

Part A

$e_1 = 3$ , $e_2 = \frac{2}{5}$ , $e_3 = -\frac{4}{5}$ : the sum of the roots, the sum of the three pairwise products, and the product of all three.

Part B

The sum is $-\frac{1}{2}$ . It needs every root to be nonzero, and the constant term certifies it: $p(0) = 4 \neq 0$ , so $0$ is not a root.

Part C

The expression must be symmetric: relabelling the roots must leave it unchanged. The product of the pairwise sums qualifies, since $(r_1 + r_2)(r_1 + r_3)(r_2 + r_3) = e_1e_2 - e_3 = 2$ here. The single product $r_1r_2$ does not, since swapping $r_2$ with $r_3$ turns it into $r_1r_3$ , so the coefficients cannot say which pair was meant.

Worked solution

Part A

Write out the coefficients before touching anything: $a_3 = 5$ , $a_2 = -15$ , $a_1 = 2$ , $a_0 = 4$ . Then apply $e_k = (-1)^k \frac{a_{n-k}}{a_n}$ three times, reading the parity of $k$ each time and dividing by $a_3$ every time.

$e_1 = -\frac{-15}{5} = 3, \qquad e_2 = +\frac{2}{5}, \qquad e_3 = -\frac{4}{5}$

The signs alternate because a term that uses $k$ of the roots drags $k$ minus signs out of the factors $(x - r_i)$ with it, and those multiply to $(-1)^k$ . It is nothing to do with sums against products, which is why $e_2$ is positive and $e_3$ is not.

In words, $e_1 = r_1 + r_2 + r_3$ , $e_2 = r_1r_2 + r_1r_3 + r_2r_3$ , and $e_3 = r_1r_2r_3$ .

Part B

Deal with the condition first, because a value computed for an expression that does not exist is worth nothing. A reciprocal $\frac{1}{r_i}$ needs $r_i \neq 0$ , and $0$ is a root of $p$ exactly when $p(0) = 0$ , that is exactly when the constant term is zero. Here the constant term is $4$ , so no root is zero and every reciprocal exists.

Now put the three fractions over a common denominator. The numerator collects every product of two of the roots and the denominator is the product of all three.

$\frac{1}{r_1} + \frac{1}{r_2} + \frac{1}{r_3} = \frac{r_2r_3 + r_1r_3 + r_1r_2}{r_1r_2r_3} = \frac{e_2}{e_3}$

Substituting the two values from Part A,

$\frac{e_2}{e_3} = \frac{\frac{2}{5}}{-\frac{4}{5}} = -\frac{1}{2}.$

The fifths cancel, which is worth noticing: dividing one symmetric sum by another cancels the leading coefficient, so this particular quantity would have come out the same for any nonzero multiple of $p$ .

Part C

The property is symmetry: an expression qualifies when relabelling the roots leaves its value unchanged. The reason that is the right condition is that the coefficients know the roots only as an unordered list. Vieta's formulas are read off the expansion of $(x - r_1)(x - r_2)(x - r_3)$ , and that product is unchanged by any reordering of its factors, so nothing recoverable from the coefficients can depend on which root was called $r_1$ .

One that qualifies. Take the product of the three pairwise sums. Swapping any two roots merely permutes the three factors, so the value cannot move. To reach it, notice that $r_1 + r_2 = e_1 - r_3$ and likewise for the others, so the product is the monic cubic $x^3 - e_1x^2 + e_2x - e_3$ evaluated at $x = e_1$ :

$(r_1 + r_2)(r_1 + r_3)(r_2 + r_3) = e_1^3 - e_1\cdot e_1^2 + e_2e_1 - e_3 = e_1e_2 - e_3 = 3\cdot\tfrac{2}{5} + \tfrac{4}{5} = 2.$

That is a definite number attached to three roots nobody has written down.

One that does not. The expression $r_1r_2$ is a product of two of the roots, but of two particular ones. Swapping the labels on $r_2$ and $r_3$ turns it into $r_1r_3$ , and the same polynomial would then be asked to supply a different answer to the same question. Since $p$ does not order its roots, the question has no answer to give, and indeed only the total $e_2 = r_1r_2 + r_1r_3 + r_2r_3$ is available. The distinction is exactly the one the chapter keeps making: $e_2$ is a sum of pairwise products and never a single one.

In one line

For $p(x) = 5x^3 - 15x^2 + 2x + 4$ the symmetric sums are $e_1 = 3$ , $e_2 = \frac{2}{5}$ and $e_3 = -\frac{4}{5}$ . The reciprocals of the roots sum to $\frac{e_2}{e_3} = -\frac{1}{2}$ , which exists because the constant term $4$ is nonzero and so no root is zero. What makes such a computation possible is that the expression be symmetric, unchanged by relabelling the roots: the product of the pairwise sums qualifies, being $e_1e_2 - e_3 = 2$ , while a single product $r_1r_2$ does not, since swapping two labels turns it into $r_1r_3$ and the coefficients hold no information about which pair was meant.

Rubric

Mark your own paper against this. Give yourself the points for each element you actually wrote down, not the ones you meant to.

Part A 5 points

Divides every value by the leading coefficient $5$ rather than reading the coefficients off directly. . Worth 2 points.

Attaches the sign factor $(-1)^k$ correctly, putting a leading minus on the readings for $e_1$ and $e_3$ and none on the reading for $e_2$ , and reports the resulting values with their own signs. . Worth 2 points.

Says what each symmetric sum collects, distinguishing the sum of pairwise products from a product of everything. . Worth 1 point.

Part B 5 points

Rewrites the sum of reciprocals as a ratio of symmetric sums before substituting any number. . Worth 2 points.

Evaluates the ratio correctly, keeping the sign of $e_3$ . . Worth 1 point.

States that no root may be zero and names the constant term, or $p(0)$ , as what certifies it. . Worth 2 points.

Part C 5 points

Names symmetry, that relabelling the roots does not change the value, as the property, rather than describing the method again. . Worth 2 points. needs an explanation, not just an answer

Supplies one qualifying expression with its value and one failing expression, and locates the failure in the dependence on which root got which label. . Worth 3 points.
5. Dividing until it refuses . 14 points. Question 5 of 10.

The polynomial

$P(x) = x^4 + x^3 - 18x^2 - 52x - 40$

arrives with nothing factored, but $-2$ is known to be one of its roots. How many copies of the corresponding factor it contains is a separate question, and the whole of Part A is about answering it rather than guessing it.
1. Part A.
  Confirm that $-2$ is a root, then find its multiplicity. Show the quotient produced at every stage, and name the value that ends the process.
  
  Solve and show your work Write each step out, and end with the value and its units. 5 points
2. Part B.
  Assemble the factorization of $P$ from your divisions. Name each root together with how many times it occurs, and verify that the total is the degree.
  
  Carry your own answer forward Assemble the factorization from the quotients your divisions produced in Part A, even if they were not the expected ones. Credit here follows the assembly of the factors and the check against the degree.
  
  Write the expression An equation or an expression is enough here. Show how you built it. 4 points
3. Part C.
  The definition of multiplicity demands a leftover factor that does not vanish at the root. Say which of the numbers you computed in Part A is that leftover factor's value, and explain what it would have meant about $P$ if a fourth division had come out even as well.
  
  Justify your claim State the claim, then give the reason it has to be true. 5 points
The answer

Part A

$P(-2) = 0$ and the multiplicity of $-2$ is $3$ . The quotients are $x^3 - x^2 - 16x - 20$ , then $x^2 - 3x - 10$ , then $x - 5$ , which is $-7$ at $-2$ , so the divisions stop.

Part B

$P(x) = (x + 2)^3(x - 5)$ , with $-2$ of multiplicity $3$ and $5$ simple, and $3 + 1 = 4$ , the degree.

Part C

The value is $-7$ , which is the last quotient $x - 5$ evaluated at $-2$ . Had a fourth division also come out even, $P$ would have been $(x + 2)^4$ times a constant, and since $P$ is monic of degree $4$ that forces $P(x) = (x + 2)^4$ , whose constant term is $16$ rather than $-40$ .

Worked solution

Part A

Confirm the given root first, since every later step inherits the error if it is wrong.

$P(-2) = 16 - 8 - 72 + 104 - 40 = 0$

So $-2$ is a root and $(x + 2)$ divides $P$ . That alone says the multiplicity is at least $1$ and nothing more.

Synthetic division of $1,\, 1,\, -18,\, -52,\, -40$ by $-2$ gives $1 - 2 = -1$ , then $-18 + 2 = -16$ , then $-52 + 32 = -20$ , then $-40 + 40 = 0$ . Remainder $0$ , quotient $x^3 - x^2 - 16x - 20$ .

Now test the quotient at $-2$ , not the original polynomial, because the live question is whether the factor goes into what is left: $-8 - 4 + 32 - 20 = 0$ . Dividing $1,\, -1,\, -16,\, -20$ by $-2$ leaves quotient $x^2 - 3x - 10$ and remainder $0$ .

Test again: $4 + 6 - 10 = 0$ . Dividing $1,\, -3,\, -10$ by $-2$ leaves quotient $x - 5$ and remainder $0$ .

One more test, and this is the one that decides the answer.

$-2 - 5 = -7 \neq 0$

The factor does not come out a fourth time, so the multiplicity of $-2$ is exactly $3$ .

Part B

Each clean division peeled off one copy of $(x + 2)$ , and the last quotient is what remains after all of them.

$P(x) = (x + 2)^3(x - 5)$

The leftover factor $x - 5$ is linear, so nothing is left to break down, and its root is $5$ .

The roots are therefore $-2$ with multiplicity $3$ and $5$ with multiplicity $1$ , and the multiplicities add to the degree.

$3 + 1 = 4 = \deg P$

The factorization also closes the search. A product is zero only when one of its factors is zero, so any number killing $P$ must kill $(x + 2)$ or $(x - 5)$ , which makes $-2$ and $5$ the only roots $P$ has, real or otherwise.

Part C

The definition asks for a shape: $r$ has multiplicity $m$ in $P$ when $P(x) = (x - r)^m\,q(x)$ for some polynomial $q$ with $q(r) \neq 0$ . Three clean divisions produce exactly that shape with $m = 3$ , and the $q$ it produces is the final quotient.

$P(x) = (x + 2)^3\,q(x), \qquad q(x) = x - 5, \qquad q(-2) = -7$

So the stopping value is not an incidental number that happened to be nonzero. It is the one quantity the definition insists on, and computing it is what turns "at least three" into "exactly three".

Now suppose the fourth test had returned zero instead. Then $(x + 2)$ would divide $q$ , and since $q$ has degree $1$ that forces $q$ to be a constant multiple of $(x + 2)$ , so $P$ would be a constant times $(x + 2)^4$ . The leading coefficient of $P$ is $1$ , so the constant is $1$ and $P$ would be $(x + 2)^4$ exactly, whose constant term is $2^4 = 16$ . The constant term of $P$ is $-40$ , so that outcome was never available.

That is worth noticing as a check on the arithmetic: the answer $m = 3$ was forced by the polynomial before any division was performed, and the divisions merely found it.

In one line

Since $P(-2) = 0$ the factor is present, and three clean divisions leave $x^3 - x^2 - 16x - 20$ , then $x^2 - 3x - 10$ , then $x - 5$ , which takes the value $-7$ at $-2$ , so the multiplicity of $-2$ is exactly $3$ and $P(x) = (x + 2)^3(x - 5)$ , with roots $-2$ of multiplicity $3$ and $5$ simple, adding to the degree $4$ . That value $-7$ is the leftover factor evaluated at the root, which is the quantity the definition of multiplicity requires to be nonzero. Had a fourth division come out even, $P$ would have been $(x + 2)^4$ , whose constant term is $16$ rather than $-40$ .

Rubric

Mark your own paper against this. Give yourself the points for each element you actually wrote down, not the ones you meant to.

Part A 5 points

Evaluates $P$ at $-2$ before dividing rather than taking the given root on trust. . Worth 1 point.

Performs each further division on the quotient the previous one produced, not on $P$ again, and reports the quotients correctly. . Worth 3 points.

Stops on a value that fails to be zero and names that value, rather than stopping after a fixed number of divisions. . Worth 1 point.

Part B 4 points

Assembles all the peeled factors with the correct exponent and keeps the final quotient as a factor. . Worth 2 points.

Lists both roots with multiplicities and checks that they add to $4$ . . Worth 2 points.

Part C 5 points

Identifies the stopping value as the final quotient evaluated at the root, and ties it to the $q(r) \neq 0$ clause of the definition. . Worth 3 points. needs an explanation, not just an answer

Says what a fourth clean division would have forced $P$ to be, and gives a feature of $P$ that rules it out. . Worth 2 points.
6. What a sign can and cannot count . 15 points. Question 6 of 10.

A polynomial $P$ with real coefficients has degree $6$ and can be written $P(x) = (x - 2)^m\,G(x)$ for some polynomial $G$ with $G(2) \neq 0$ . Three facts are known about it: $P(1) = 5$ , $P(3) = 40$ , and $2$ is the only real zero of $P$ anywhere in the interval $1 \le x \le 3$ .
1. Part A.
  Decide whether $m$ is odd or even, and justify the decision from the three facts.
  
  Justify your claim State the claim, then give the reason it has to be true. 5 points
2. Part B.
  List every value $m$ could take, and rule out each value your answer to Part A did not already exclude.
  
  Carry your own answer forward Start from the parity you settled in Part A, whatever you concluded there, and eliminate from the values that parity allows. The marks are for the elimination, not for revisiting Part A.
  
  Solve and show your work Write each step out, and end with the value and its units. 5 points
3. Part C.
  Explain why the sign of $P$ on either side of $x = 2$ records only the parity of $m$ and never its value, so that this evidence cannot separate the surviving values. Then name a computation that does settle which one $m$ is.
  
  Explain why it works A sentence or two. Reasons, not steps. 5 points
The answer

Part A

$m$ is even. Both $P(1)$ and $P(3)$ are positive and the only real zero between them is $2$ , so $P$ holds one sign on each side of $2$ and does not change sign across it, which happens at a real zero exactly when its multiplicity is even.

Part B

$m$ is $2$ or $4$ . Part A leaves the even values $2$ , $4$ and $6$ , and $m = 6$ would make $G$ a constant $a$ and $P(x) = a(x - 2)^6$ , which gives $P(1) = a = P(3)$ , contradicting $5 \neq 40$ .

Part C

Near $2$ the sign of $P$ is the sign of $G(2)$ times the sign of $(x - 2)^m$ , and every even $m$ makes that factor positive on both sides, so $m = 2$ and $m = 4$ are indistinguishable there and only the parity is recorded. Dividing $P$ by $(x - 2)$ repeatedly, stopping at the first nonzero remainder, returns $m$ itself.

Worked solution

Part A

A polynomial's graph is unbroken, so $P$ can change sign only where it has a real zero. On the interval $1 \le x \le 3$ the only real zero is $2$ , so on $1 \le x < 2$ the sign is constant and equal to the sign of $P(1)$ , and on $2 < x \le 3$ it is constant and equal to the sign of $P(3)$ .

$P(1) = 5 > 0, \qquad P(3) = 40 > 0$

Both are positive, so $P$ is positive immediately to the left of $2$ and positive immediately to the right of it. The sign does not change across the zero.

Now use the parity rule in the direction that fits. Writing $P(x) = (x - 2)^m G(x)$ with $G(2) \neq 0$ , the factor $G$ holds the fixed sign of $G(2)$ on a small interval around $2$ , while $(x - 2)^m$ is positive to the right of $2$ and, to the left, positive when $m$ is even and negative when $m$ is odd. A sign change across $2$ therefore happens exactly when $m$ is odd. It did not happen, so $m$ is even.

Part B

The multiplicity is at least $1$ and at most the degree, so $1 \le m \le 6$ , and Part A cut that down to the even values.

$m \in \{2,\ 4,\ 6\}$

Only $m = 6$ can still be excluded, and the two given values do it. If $m = 6$ , then $P(x) = (x - 2)^6 G(x)$ with $\deg G = 0$ , so $G$ is a nonzero constant $a$ , and

$P(1) = a(-1)^6 = a, \qquad P(3) = a(1)^6 = a.$

Those two are equal, but $P(1) = 5$ and $P(3) = 40$ are not, so $m = 6$ is impossible.

Nothing rules out $2$ or $4$ , and both genuinely occur. In each case the leftover factor $G$ still has degree to spare, four or two, so it can take the values $5$ at $x = 1$ and $40$ at $x = 3$ while keeping every real zero of its own outside the interval and staying nonzero at $2$ . The given data therefore cannot separate the two cases.

Part C

Look at what the sign near the zero actually depends on. On a small interval around $2$ the factor $G$ holds the fixed sign of $G(2)$ , so the sign of $P$ there is that fixed sign multiplied by the sign of $(x - 2)^m$ . For any even $m$ the second factor is positive on both sides.

$(x - 2)^2 > 0 \quad\text{and}\quad (x - 2)^4 > 0 \qquad \text{for every } x \neq 2$

So $m = 2$ and $m = 4$ give exactly the same local picture, and no test value taken on either side of $2$ can tell them apart. What that picture records is the parity of $m$ and nothing more. The difference between the two cases is how fast the curve leaves the axis, and speed is not a sign.

Be careful how far that reaches. It is a statement about the signs immediately around $2$ , not a claim that $m$ is beyond all reach. Signs elsewhere on the line are governed by $G$ , and a particular $G$ carrying real zeros of its own can leave room for only one of the two values, since the larger $m$ leaves $G$ less degree to spend. Without knowing $G$ , the picture at the zero fixes the parity alone.

What settles $m$ outright, and needs nothing about $G$ , is division. Divide $P$ by $(x - 2)$ , then divide each quotient by $(x - 2)$ again, and count the divisions that leave remainder zero; the first nonzero remainder stops the count, and the count is $m$ . That is the definition being carried out rather than inferred.

In one line

Since $P(1)$ and $P(3)$ are both positive and $2$ is the only real zero between them, $P$ does not change sign across $2$ , so $m$ is even. That leaves $2$ , $4$ and $6$ , and $m = 6$ would force $G$ to be a constant $a$ and give $P(1) = a = P(3)$ , contradicting $5 \neq 40$ , so $m$ is $2$ or $4$ . The signs on either side of $2$ can go no further, because $(x - 2)^m$ is positive on both sides for both surviving values, so the local picture records only the parity. Dividing $P$ by $(x - 2)$ repeatedly and stopping at the first nonzero remainder returns $m$ itself.

Rubric

Mark your own paper against this. Give yourself the points for each element you actually wrote down, not the ones you meant to.

Part A 5 points

Establishes that $P$ holds one sign on each side of $2$ , using that the interval contains no other real zero. . Worth 2 points.

Concludes even multiplicity from the absence of a sign change, invoking the parity rule rather than asserting it from the size of the values. . Worth 3 points. needs an explanation, not just an answer

Part B 5 points

Bounds $m$ by the degree and keeps only the values the parity from Part A allows. . Worth 2 points.

Excludes the largest value by showing that it forces $G$ to be constant and $P(1) = P(3)$ . . Worth 2 points.

Reports the surviving values as a list rather than picking one of them. . Worth 1 point.

Part C 5 points

Locates the limitation in the sign of $(x - 2)^m$ being the same on both sides of $2$ for every even exponent, rather than saying only that signs are not precise enough. . Worth 3 points. needs an explanation, not just an answer

Names repeated division by $(x - 2)$ , stopping at the first nonzero remainder, as the computation that returns $m$ . . Worth 2 points.
7. What the coefficient system buys . 15 points. Question 7 of 10.

Every coefficient of

$q(x) = x^4 - 2x^3 - 42x^2 - 146x + 325$

is rational, and $5 - 2\sqrt{3}$ is one of its roots.
1. Part A.
  Name the second root that the coefficients force, and give the monic quadratic, with rational coefficients throughout, that must therefore be a factor of $q$ . State also which condition on $\sqrt{3}$ the theorem needs before it can supply that partner at all.
  
  Write the expression An equation or an expression is enough here. Show how you built it. 5 points
2. Part B.
  Carry out the division of $q$ by that quadratic, then give all four roots of $q$ and say how many of them are real.
  
  Carry your own answer forward Divide by the quadratic you produced in Part A, even if it was not the expected one. Credit here follows the division itself and the finishing of the quotient over the complex numbers.
  
  Solve and show your work Write each step out, and end with the value and its units. 5 points
3. Part C.
  Two different pairings acted inside this one polynomial. For each, state the swap it performs on a number, the numbers that swap leaves untouched, and hence the coefficient system its theorem has to demand. Then suppose a single coefficient of $q$ were replaced by an irrational real number: say which of the two theorems would still be available for the result, and which would not.
  
  Compare the two methods Say what each one costs you, and when you would reach for it. 5 points
The answer

Part A

The second root is $5 + 2\sqrt{3}$ and the forced factor is $x^2 - 10x + 13$ . The theorem needs $\sqrt{3}$ to be irrational, with $5$ , $-2$ and $3$ all rational and the coefficient $-2$ nonzero.

Part B

$q(x) = (x^2 - 10x + 13)(x^2 + 8x + 25)$ , so the four roots are $5 - 2\sqrt{3}$ , $5 + 2\sqrt{3}$ , $-4 + 3i$ and $-4 - 3i$ . Two of them are real.

Part C

The surd swap sends $a + b\sqrt{3}$ to $a - b\sqrt{3}$ and fixes exactly the rationals, so its theorem demands rational coefficients. Conjugation sends $a + bi$ to $a - bi$ and fixes exactly the reals, so it demands only real ones. An irrational coefficient keeps the conjugate root theorem available and takes the surd theorem away.

Worked solution

Part A

Check the hypotheses before invoking the theorem, because there are four of them and each does work. Every coefficient of $q$ is rational. The root is written $a + b\sqrt{d}$ with $a = 5$ , $b = -2$ and $d = 3$ , all three rational. The coefficient $b$ is not zero. And $\sqrt{3}$ is irrational, which is what makes that writing unique and the swap well defined at all. With the four in place the partner $5 + 2\sqrt{3}$ is a root.

Because $b \neq 0$ the two partners are different numbers, so their linear factors both divide $q$ and so does their product. Multiply the pair, which is the product engineered to clear the radical.

$\bigl(x - (5 - 2\sqrt{3})\bigr)\bigl(x - (5 + 2\sqrt{3})\bigr) = x^2 - 10x + \bigl(5^2 - 2^2\cdot 3\bigr) = x^2 - 10x + 13$

The radical cancels twice over: in the sum, because the swap moved only a sign, and in the product, because that product is a difference of squares. Note that the $2$ in front of the radical is squared along with the radical, so the constant is $25 - 12$ and not $25 - 3$ . Both coefficients came out rational, as they had to.

Part B

Long division, staying with rational numbers throughout. Dividing $x^4$ by $x^2$ gives $x^2$ , and $x^2(x^2 - 10x + 13) = x^4 - 10x^3 + 13x^2$ ; subtracting leaves $8x^3 - 55x^2 - 146x + 325$ . Dividing $8x^3$ by $x^2$ gives $8x$ , and $8x(x^2 - 10x + 13) = 8x^3 - 80x^2 + 104x$ ; subtracting leaves $25x^2 - 250x + 325$ . Dividing $25x^2$ by $x^2$ gives $25$ , and $25(x^2 - 10x + 13) = 25x^2 - 250x + 325$ ; subtracting leaves nothing.

$x^4 - 2x^3 - 42x^2 - 146x + 325 = (x^2 - 10x + 13)(x^2 + 8x + 25)$

The zero remainder confirms the factor. Finish the quotient with the quadratic formula, since its discriminant $64 - 100 = -36$ is negative.

$x = \frac{-8 \pm \sqrt{-36}}{2} = \frac{-8 \pm 6i}{2} = -4 \pm 3i$

The four roots are $5 - 2\sqrt{3}$ , $5 + 2\sqrt{3}$ , $-4 + 3i$ and $-4 - 3i$ . Two are real, and their being irrational is no obstacle: an irrational number is a perfectly good real root. The other two contribute no point on the real line at all, and they arrived as a conjugate pair, which the real coefficients had already guaranteed.

Part C

Set the two side by side, because they look like twins and cost different amounts.

$\underbrace{a + b\sqrt{3} \ \mapsto\ a - b\sqrt{3}}_{\text{fixes exactly the rationals}}, \qquad \underbrace{a + bi \ \mapsto\ a - bi}_{\text{fixes exactly the reals}}$

Each theorem is proved the same way. The swap respects addition and multiplication, so it passes cleanly through the whole polynomial; and it leaves every coefficient alone, so each coefficient emerges unchanged and what is left is the polynomial evaluated at the partner. That second step is the only place a hypothesis is spent, and it is spent on exactly the set the swap fixes. The surd swap fixes the rationals, so it needs rational coefficients. Conjugation fixes the reals, so it needs real ones. Nothing else separates the two arguments.

Now make one coefficient of $q$ irrational, say by replacing $325$ with $325 + \sqrt{2}$ . Every coefficient of the result is still a real number, so the conjugate root theorem still applies: any nonreal root it has still drags its conjugate along. The surd theorem does not apply, because the hypothesis it spends has failed.

Say carefully what that does and does not mean. A theorem that does not apply is not a theorem that has been contradicted, and the altered polynomial might still happen to carry a matched surd pair. What is gone is the guarantee, and the guarantee was the whole reason for checking the coefficient system first.

In one line

The rational coefficients force the partner $5 + 2\sqrt{3}$ and hence the factor $x^2 - 10x + 13$ , the theorem needing $\sqrt{3}$ irrational with $5$ , $-2$ and $3$ rational and $-2$ nonzero. Dividing gives $q(x) = (x^2 - 10x + 13)(x^2 + 8x + 25)$ , so the four roots are $5 \pm 2\sqrt{3}$ and $-4 \pm 3i$ , of which two are real. The surd swap fixes exactly the rationals and conjugation fixes exactly the reals, which is why one theorem demands rational coefficients and the other only real ones; replacing a single coefficient of $q$ by an irrational real number would keep the conjugate root theorem available and take the surd theorem away.

Rubric

Mark your own paper against this. Give yourself the points for each element you actually wrote down, not the ones you meant to.

Part A 5 points

Checks the theorem's hypotheses on $q$ and on the root before invoking it, including that $\sqrt{3}$ is irrational. . Worth 2 points.

Builds the quadratic from the sum and the product of the pair, squaring the coefficient of the radical along with the radical. . Worth 3 points.

Part B 5 points

Carries out the division and reports a zero remainder with the correct quotient. . Worth 2 points.

Finishes the quotient over the complex numbers, keeping both nonreal roots. . Worth 2 points.

States how many roots are real, counting the irrational ones as real. . Worth 1 point.

Part C 5 points

Gives all three items for each pairing, the swap, the set it fixes and the coefficient system demanded, keeping rational and real distinct. . Worth 3 points. needs an explanation, not just an answer

Rules correctly on both theorems under the altered coefficient, and treats an unavailable theorem as a lost guarantee rather than as a false conclusion. . Worth 2 points.
8. The same quartic in two languages . 12 points. Question 8 of 10.

A polynomial can be written to display its roots or to display its coefficients, and the two readings have to agree. Work throughout with

$p(x) = 5(x - 2)^2(x + 3)(x - 4).$
1. Part A.
  List the roots of $p$ with their multiplicities, check the multiplicities against the degree, and state how many distinct roots $p$ has.
  
  Solve and show your work Write each step out, and end with the value and its units. 3 points
2. Part B.
  Compute the sum and the product of the roots directly from the list. Then, given that $p$ expands to $5x^4 - 25x^3 - 20x^2 + 220x - 240$ , obtain the same two values from the coefficients instead, and confirm that the two routes agree.
  
  Carry your own answer forward Use for the first route the root list you wrote in Part A, even if it was not the expected one. Credit here follows computing each total along both routes and comparing them.
  
  Solve and show your work Write each step out, and end with the value and its units. 5 points
3. Part C.
  Two counts have been in play: roots taken with multiplicity, and distinct roots. Say which of them the degree is always equal to and why, then give a monic quartic for which the two counts are the same number.
  
  Explain why it works A sentence or two. Reasons, not steps. 4 points
The answer

Part A

The roots are $2$ with multiplicity $2$ , $-3$ simple and $4$ simple, and $2 + 1 + 1 = 4$ is the degree. There are $3$ distinct roots.

Part B

From the list, $2 + 2 - 3 + 4 = 5$ and $(2)(2)(-3)(4) = -48$ . From the coefficients, $e_1 = -\frac{-25}{5} = 5$ and $e_4 = +\frac{-240}{5} = -48$ , so the two routes agree.

Part C

The degree always equals the count taken with multiplicity, since each of the $n$ linear factors of the complete factorization carries exactly one entry of the list. The two counts agree when every root is simple, as for $x(x - 1)(x - 3)(x - 7)$ .

Worked solution

Part A

The constant $5$ carries no root at all. Each remaining factor vanishes at the number that makes its bracket zero, with the sign flipped from what is written: $(x - 2)$ carries $2$ , and the exponent on it is that root's multiplicity, while $(x + 3)$ carries $-3$ and $(x - 4)$ carries $4$ , once each.

$2,\quad 2,\quad -3,\quad 4$

That is four entries, so the degree is $4$ and there are four roots counted with multiplicity, and the multiplicities of the distinct roots add to the degree.

$2 + 1 + 1 = 4$

Three different numbers appear on the list, so $p$ has three distinct roots.

Part B

From the list, add all four entries and then multiply all four, remembering that the repeated root appears twice in each.

$e_1 = 2 + 2 + (-3) + 4 = 5, \qquad e_4 = (2)(2)(-3)(4) = -48$

From the coefficients, use $e_k = (-1)^k \frac{a_{n-k}}{a_n}$ with $a_4 = 5$ . For $k = 1$ the sign is negative, and for $k = 4$ it is positive, since $(-1)^4 = 1$ . Every value is divided by the leading coefficient, which is the step most often dropped.

$e_1 = -\frac{a_3}{a_4} = -\frac{-25}{5} = 5, \qquad e_4 = +\frac{a_0}{a_4} = \frac{-240}{5} = -48$

Both totals match. Watch the sign on the product in particular: it came out with a plus in front because the degree is even, and the same reading on a cubic would have carried a minus.

Part C

The complete factorization of a polynomial of degree $n$ has exactly $n$ linear factors, and each one carries exactly one entry of the root list. So the count taken with multiplicity is the degree, always, with no conditions attached.

$m_1 + m_2 + \cdots + m_k = n$

The count of distinct roots is the number $k$ of different values on that list. It is at most $n$ , and usually less. Here the identity reads $2 + 1 + 1 = 4$ with $k = 3$ : the repeated root spends two places while contributing one value.

The two counts agree exactly when every multiplicity is $1$ , because $k$ ones add to $k$ , which then has to equal $n$ . Any quartic with four different roots does it, for instance

$x(x - 1)(x - 3)(x - 7),$

whose roots are $0$ , $1$ , $3$ and $7$ : four counted with multiplicity, and four distinct.

In one line

The roots are $2$ with multiplicity $2$ , $-3$ simple and $4$ simple, so $2 + 1 + 1 = 4$ matches the degree and there are $3$ distinct roots. The sum of the roots is $5$ and their product is $-48$ by both routes, from the list and from $e_1 = -\frac{-25}{5}$ and $e_4 = +\frac{-240}{5}$ . The degree is always the count taken with multiplicity, since each of the $n$ linear factors carries one entry of the list, and the two counts agree exactly when every root is simple, as for $x(x - 1)(x - 3)(x - 7)$ .

Rubric

Mark your own paper against this. Give yourself the points for each element you actually wrote down, not the ones you meant to.

Part A 3 points

Reads each root with the sign flipped from its bracket and attaches the exponent as the multiplicity. . Worth 2 points.

Checks the multiplicities against the degree and reports the count of distinct roots separately. . Worth 1 point.

Part B 5 points

Computes both totals from the root list with the repeated root counted twice. . Worth 2 points.

Computes both totals from the coefficients with the sign $(-1)^k$ read from the parity of $k$ . . Worth 2 points.

States that the two routes agree, rather than leaving four numbers side by side. . Worth 1 point.

Part C 4 points

Identifies the degree with the count taken with multiplicity and grounds it in one root per linear factor. . Worth 2 points. needs an explanation, not just an answer

Supplies a monic quartic whose roots are four different numbers, and says that agreement is exactly the case where every root is simple. . Worth 2 points.
9. One intercept and one extra root . 15 points. Question 9 of 10.

The graph of a monic quartic $P$ with real coefficients meets the $x$ -axis at $x = -2$ and nowhere else, and at that point it touches the axis and turns back. It is also known that $P(0) = 52$ and that $-3 + 2i$ is a root of $P$ .
1. Part A.
  Determine the multiplicity of the zero $-2$ , ruling out the other value the picture leaves open, and say how many of $P$ 's roots are nonreal.
  
  Justify your claim State the claim, then give the reason it has to be true. 5 points
2. Part B.
  Write $P$ in factored form over the real numbers and then in standard form.
  
  Carry your own answer forward Build on the multiplicity and the count of nonreal roots you settled in Part A, even if they were not the expected ones. Credit here follows turning the given nonreal root into a real quadratic factor and assembling $P$ .
  
  Write the expression An equation or an expression is enough here. Show how you built it. 5 points
3. Part C.
  Suppose the root $-3 + 2i$ had not been given, and you knew only that $P$ is a monic quartic with real coefficients whose graph meets the axis only at $x = -2$ , touching it there, with $P(0) = 52$ . Say what would still be determined and what would not, and support the second half with a different quartic meeting every one of those conditions.
  
  Explain why it works A sentence or two. Reasons, not steps. 5 points
The answer

Part A

The multiplicity of $-2$ is $2$ , and two roots are nonreal. A touch means even multiplicity, so $2$ or $4$ ; multiplicity $4$ would force $P(x) = (x + 2)^4$ , whose value at $0$ is $16$ rather than $52$ . The two remaining roots cannot be real, since $-2$ is the only intercept.

Part B

$P(x) = (x + 2)^2\bigl(x^2 + 6x + 13\bigr) = x^4 + 10x^3 + 41x^2 + 76x + 52$ .

Part C

Still determined: the multiplicity $2$ at $-2$ , that the other two roots are one nonreal conjugate pair, and that the pair's product is $13$ , since $4 \cdot 13 = 52$ . Not determined: the pair's sum. For instance $(x + 2)^2(x^2 + 13)$ meets every stated condition and is a different polynomial.

Worked solution

Part A

The curve touches the axis at $-2$ and turns back, and that happens at a real zero exactly when its multiplicity is even. The degree is $4$ , so the multiplicity is $2$ or $4$ .

Suppose it were $4$ . Then all four linear factors carry $-2$ , and since $P$ is monic there is nothing else left to choose:

$P(x) = (x + 2)^4, \qquad P(0) = 2^4 = 16.$

But $P(0) = 52$ , so that case is impossible and the multiplicity is $2$ .

That leaves two of the four places on the root list. Neither can be real, because every real root is an $x$ -intercept and $-2$ is the only one, and neither can be a further copy of $-2$ , since the multiplicity has just been fixed at $2$ . So exactly two roots are nonreal, a single conjugate pair, and the count is consistent, because the nonreal roots of a real polynomial always come to an even number.

Part B

The coefficients are real, so the given root drags its conjugate $-3 - 2i$ along, and that pair is the conjugate pair Part A predicted. Multiplying the two linear factors clears the $i$ :

$\bigl(x - (-3 + 2i)\bigr)\bigl(x - (-3 - 2i)\bigr) = x^2 + 6x + \bigl((-3)^2 + 2^2\bigr) = x^2 + 6x + 13$

Its discriminant is $36 - 52 = -16$ , negative, so it has no real root and contributes no intercept, exactly as required.

Combining with the double root at $-2$ , and with leading coefficient $1$ because $P$ is monic,

$P(x) = (x + 2)^2\bigl(x^2 + 6x + 13\bigr).$

Expanding, $(x^2 + 4x + 4)(x^2 + 6x + 13) = x^4 + 10x^3 + 41x^2 + 76x + 52$ , whose constant term is $52$ . That is the given value of $P(0)$ , so the whole construction checks.

Part C

Everything Part A established used only the stated conditions, so the multiplicity $2$ and the single nonreal pair survive the loss of the extra root. Write what is left as

$P(x) = (x + 2)^2\bigl(x^2 + bx + c\bigr), \qquad b^2 - 4c < 0,$

with the inequality saying that the quadratic contributes no intercept. The value at $0$ pins one of the two unknowns, and only one:

$P(0) = 4c = 52 \quad\Longrightarrow\quad c = 13.$

So the product of the nonreal pair is $13$ , while their sum, which is $-b$ , is untouched by any of the data. The only surviving restriction is $b^2 < 52$ , so every $b$ with $\lvert b \rvert < 2\sqrt{13}$ gives a polynomial meeting every stated condition, and there are infinitely many of them.

One is enough to make the point. Taking $b = 0$ ,

$(x + 2)^2\bigl(x^2 + 13\bigr) = x^4 + 4x^3 + 17x^2 + 52x + 52,$

which is monic, has real coefficients, takes the value $52$ at $0$ , touches the axis at $-2$ and meets it nowhere else. It is not the $P$ of Part B, and no drawing could be used to tell the two apart at the axis, because they differ only in a pair of roots that leaves no mark there.

In one line

The zero $-2$ has multiplicity $2$ , since a touch forces an even multiplicity and multiplicity $4$ would make $P(x) = (x + 2)^4$ with $P(0) = 16$ rather than $52$ ; the two remaining roots are therefore a nonreal conjugate pair. With $-3 + 2i$ given, that pair is $-3 \pm 2i$ and $P(x) = (x + 2)^2(x^2 + 6x + 13) = x^4 + 10x^3 + 41x^2 + 76x + 52$ . Without it, the multiplicity, the single nonreal pair and the pair's product $13$ are still fixed by $P(0) = 4c = 52$ , but the pair's sum is not: $(x + 2)^2(x^2 + 13)$ meets every stated condition and is a different polynomial.

Rubric

Mark your own paper against this. Give yourself the points for each element you actually wrote down, not the ones you meant to.

Part A 5 points

Converts the touch into even multiplicity and narrows the options to $2$ and $4$ using the degree. . Worth 2 points.

Eliminates multiplicity $4$ by computing the value at $0$ that it would force, rather than by asserting it looks wrong. . Worth 2 points. needs an explanation, not just an answer

Concludes that the two remaining roots are nonreal, using that every real root would show as an intercept. . Worth 1 point.

Part B 5 points

Supplies the conjugate of the given root and multiplies the pair into a real quadratic. . Worth 2 points.

Assembles $P$ with the repeated real factor and leading coefficient $1$ , then expands correctly. . Worth 2 points.

Checks the constant term against the given value of $P(0)$ , or checks that the quadratic has negative discriminant. . Worth 1 point.

Part C 5 points

Separates what the conditions fix, the multiplicity and the product of the nonreal pair, from what they leave free, the pair's sum. . Worth 3 points. needs an explanation, not just an answer

Exhibits a second quartic satisfying every stated condition, and checks at least its value at $0$ and its lack of a second intercept. . Worth 2 points.
10. Two exponents and one condition . 15 points. Question 10 of 10.

A polynomial of degree $7$ is built as

$P(x) = (x + 1)^a(x - 3)^b\bigl(x^2 + 5\bigr),$

where $a$ and $b$ are positive integers. It is known that $P(x) > 0$ at every $x$ strictly between $-1$ and $3$ .
1. Part A.
  Find every pair $(a, b)$ consistent with the degree and with the sign condition.
  
  Solve and show your work Write each step out, and end with the value and its units. 5 points
2. Part B.
  For each surviving pair, state what the graph does at $x = -1$ and at $x = 3$ , give the sign of $P$ on each of the three intervals the real zeros create, and say how many $x$ -intercepts the graph has.
  
  Carry your own answer forward Describe the pairs you found in Part A, even if they were not the expected ones. Credit here follows turning each pair into behaviour at the two zeros and into a sign row.
  
  Explain what it means Words, not just symbols. Say what the number is telling you about the situation. 5 points
3. Part C.
  Explain why no sign chart, however many test values it uses, could decide which of the two pairs is the right one. Then name a single computation that does decide it, and give the value it returns in each case.
  
  Justify your claim State the claim, then give the reason it has to be true. 5 points
The answer

Part A

The pairs are $(a, b) = (3, 2)$ and $(a, b) = (1, 4)$ . The degree forces $a + b = 5$ , and the sign condition forces $b$ to be even.

Part B

Both pairs give the same description: the curve crosses at $-1$ , touches at $3$ and turns back, and $P$ is negative for $x < -1$ and positive on $-1 < x < 3$ and on $x > 3$ . Each has exactly two $x$ -intercepts, since $x^2 + 5$ has no real zero.

Part C

A sign chart records only signs, and the sign of $P$ at any point depends on the exponents only through their parities, which the two pairs share, so both give identical signs everywhere. Evaluating $P(0)$ decides it: $(3, 2)$ returns $45$ and $(1, 4)$ returns $405$ .

Worked solution

Part A

The degree is the total of the exponents, and the quadratic factor contributes $2$ of them.

$a + b + 2 = 7 \quad\Longrightarrow\quad a + b = 5$

Now read the sign on $-1 < x < 3$ factor by factor. There $x + 1$ is positive, so $(x + 1)^a$ is positive whatever $a$ is; $x^2 + 5$ is positive for every real $x$ ; and $x - 3$ is negative, so $(x - 3)^b$ carries the sign $(-1)^b$ .

$\operatorname{sign} P(x) = (+1)\cdot(-1)^b\cdot(+1) = (-1)^b$

For $P$ to be positive there, $b$ must be even. With $a$ and $b$ positive integers adding to $5$ , the even values available to $b$ are $2$ and $4$ , giving $a = 3$ and $a = 1$ .

Part B

Take the pairs one at a time and read the parity of each exponent, since parity is what decides crossing against touching.

For $(a, b) = (3, 2)$ the exponent at $-1$ is odd and the one at $3$ is even, so the curve crosses at $-1$ and touches at $3$ . For $(a, b) = (1, 4)$ the exponent at $-1$ is again odd and the one at $3$ again even, so the description is word for word the same.

The sign row agrees too. To the left of $-1$ both $x + 1$ and $x - 3$ are negative and $x^2 + 5$ is positive, so

$\operatorname{sign} P = (-1)^a(-1)^b = (-1)^{a+b} = (-1)^5 = -1,$

which depends only on the total $a + b = 5$ and therefore cannot tell the pairs apart. Between $-1$ and $3$ the sign is positive, as the given condition already said, and to the right of $3$ every factor is positive.

The real zeros are $-1$ and $3$ in both cases. The factor $x^2 + 5$ is never zero for a real $x$ , so its two roots are nonreal and leave no mark on the axis, and the graph has exactly two $x$ -intercepts even though $P$ has seven roots counted with multiplicity.

Part C

The sign of a product is the product of the signs of its factors, and the sign of $(x - r)^m$ at a given $x$ depends on the exponent only through whether it is odd or even. Both surviving pairs have $a$ odd and $b$ even, so at every real $x$ apart from the two zeros the two candidates have matching signs.

$(3,2):\ (-1)^3,\ (-1)^2 \qquad\text{against}\qquad (1,4):\ (-1)^1,\ (-1)^4$

A sign chart is a record of nothing but those signs, on intervals cut by the real zeros, and the real zeros are the same two numbers in both cases. So however many test values are added, each one returns the same sign for both candidates and the chart has nothing left to separate them with. What genuinely differs is how fast the curve leaves the axis at each zero, and speed is not a sign. The flatness at $3$ would look different if the two were drawn accurately, but a drawing is a suggestion about a multiplicity, never a proof of one.

A value, rather than a sign, settles it in one step. Take $x = 0$ , which is easy in both cases:

$(3,2):\ (1)^3(-3)^2(5) = 45, \qquad (1,4):\ (1)^1(-3)^4(5) = 405.$

Those are different numbers, so a single evaluation of $P$ at $0$ identifies the pair. Repeated division by $(x - 3)$ , counting the clean divisions until a remainder fails to vanish, would return $b$ directly and would serve just as well.

In one line

The degree gives $a + b = 5$ and the sign condition on $-1 < x < 3$ forces $b$ even, so the pairs are $(3, 2)$ and $(1, 4)$ . Both describe the same picture: crossing at $-1$ , touching at $3$ , negative for $x < -1$ and positive on the other two intervals, with exactly two $x$ -intercepts because $x^2 + 5$ has no real zero. No sign chart can separate them, since the sign at any point depends on the exponents only through their parities and those agree; evaluating $P(0)$ does separate them, returning $45$ for $(3, 2)$ and $405$ for $(1, 4)$ .

Rubric

Mark your own paper against this. Give yourself the points for each element you actually wrote down, not the ones you meant to.

Part A 5 points

Uses the degree to get $a + b = 5$ , counting the quadratic factor's two degrees. . Worth 2 points.

Determines the sign on the middle interval factor by factor and concludes that $b$ is even. . Worth 2 points.

Reports both surviving pairs rather than settling on one of them. . Worth 1 point.

Part B 5 points

Reads crossing or touching from the parity of each exponent, for both pairs. . Worth 2 points.

Gives a sign for all three intervals and notes that the two pairs produce the same row. . Worth 2 points.

Counts the $x$ -intercepts as two and attributes the other roots to multiplicity and to the nonreal pair. . Worth 1 point.

Part C 5 points

Locates the obstruction in the sign depending on the exponents only through their parity, which the two pairs share. . Worth 3 points. needs an explanation, not just an answer

Names a computation that returns a value rather than a sign, and gives its two different outcomes. . Worth 2 points.

Marking rubrics

Work every question first. Then mark your own paper against these, giving yourself the points for each element you actually wrote down.

1. Nothing left to break down 13 points

Part A 4

Splits the quadratic factor into two linear factors over the complex numbers rather than leaving the factorization part-finished. 2
Reports the degree and the two counts as separate numbers, and shows the multiplicities adding to the degree. 2

Part B 4

Splits the root list by real and nonreal, counting with multiplicity rather than by distinct value. 2
Names the ceiling as the degree and attributes the shortfall to the surplus copies a repeated root spends. 2

Part C 5

Rules the polynomial out and derives the degree $4$ from every multiplicity being $1$ , rather than asserting that four roots cannot fill six places. 3
States the identity that the multiplicities of the distinct roots add to the degree, and uses it as the ground of the argument. 2

2. Counting under a real hypothesis 15 points

Part A 5

Pairs the nonreal root with its conjugate at the SAME multiplicity, not merely as one extra root. 2
Rules out a nonreal occupant of the remaining places, then uses the only-real-root condition to force both of them to be $-4$ . 2
Reports a list of six entries whose multiplicities add to the degree. 1

Part B 5

Multiplies the conjugate pair into a quadratic and shows the sum and the product coming out real. 3
Repeats the quadratic factor as often as the pair occurs and keeps the leading coefficient. 1
Checks that the factor degrees add to $6$ , or that the quadratic's discriminant is negative so it is genuinely unbreakable over the reals. 1

Part C 5

Says the multiplicity of $-4$ is forced only to be odd, and gives both arrangements that realise it, one with a second conjugate pair and one without. 2
Argues the odd-degree claim from the nonreal roots occupying an even count, rather than from the shape of a graph. 3

3. Which side, and where it turns 15 points

Part A 5

Gets the degree and leading coefficient by multiplying leading terms, without expanding, and turns them into the direction of both arms. 2
Reads each zero with the correct sign and attaches its multiplicity, then converts the parity of each multiplicity into crossing or touching. 3

Part B 5

Uses one test value strictly inside each of the four intervals, none of them a zero, and evaluates correctly. 2
Reports a sign for every interval and gives the $y$ -intercept as a coordinate or a value of $P$ . 1
Reconciles the outer signs with the arms and the pattern of flips with the multiplicities from Part A. 2

Part C 5

Gives the solution set of $P(x) \ge 0$ including the isolated point at the even zero, not just the two unbounded pieces. 2
Distinguishes a highest point near the zero from a highest point overall, and supports the second half with an evaluated value or with the direction of the arms. 3

4. Totals from a cubic nobody can solve 15 points

Part A 5

Divides every value by the leading coefficient $5$ rather than reading the coefficients off directly. 2
Attaches the sign factor $(-1)^k$ correctly, putting a leading minus on the readings for $e_1$ and $e_3$ and none on the reading for $e_2$ , and reports the resulting values with their own signs. 2
Says what each symmetric sum collects, distinguishing the sum of pairwise products from a product of everything. 1

Part B 5

Rewrites the sum of reciprocals as a ratio of symmetric sums before substituting any number. 2
Evaluates the ratio correctly, keeping the sign of $e_3$ . 1
States that no root may be zero and names the constant term, or $p(0)$ , as what certifies it. 2

Part C 5

Names symmetry, that relabelling the roots does not change the value, as the property, rather than describing the method again. 2
Supplies one qualifying expression with its value and one failing expression, and locates the failure in the dependence on which root got which label. 3

5. Dividing until it refuses 14 points

Part A 5

Evaluates $P$ at $-2$ before dividing rather than taking the given root on trust. 1
Performs each further division on the quotient the previous one produced, not on $P$ again, and reports the quotients correctly. 3
Stops on a value that fails to be zero and names that value, rather than stopping after a fixed number of divisions. 1

Part B 4

Assembles all the peeled factors with the correct exponent and keeps the final quotient as a factor. 2
Lists both roots with multiplicities and checks that they add to $4$ . 2

Part C 5

Identifies the stopping value as the final quotient evaluated at the root, and ties it to the $q(r) \neq 0$ clause of the definition. 3
Says what a fourth clean division would have forced $P$ to be, and gives a feature of $P$ that rules it out. 2

6. What a sign can and cannot count 15 points

Part A 5

Establishes that $P$ holds one sign on each side of $2$ , using that the interval contains no other real zero. 2
Concludes even multiplicity from the absence of a sign change, invoking the parity rule rather than asserting it from the size of the values. 3

Part B 5

Bounds $m$ by the degree and keeps only the values the parity from Part A allows. 2
Excludes the largest value by showing that it forces $G$ to be constant and $P(1) = P(3)$ . 2
Reports the surviving values as a list rather than picking one of them. 1

Part C 5

Locates the limitation in the sign of $(x - 2)^m$ being the same on both sides of $2$ for every even exponent, rather than saying only that signs are not precise enough. 3
Names repeated division by $(x - 2)$ , stopping at the first nonzero remainder, as the computation that returns $m$ . 2

7. What the coefficient system buys 15 points

Part A 5

Checks the theorem's hypotheses on $q$ and on the root before invoking it, including that $\sqrt{3}$ is irrational. 2
Builds the quadratic from the sum and the product of the pair, squaring the coefficient of the radical along with the radical. 3

Part B 5

Carries out the division and reports a zero remainder with the correct quotient. 2
Finishes the quotient over the complex numbers, keeping both nonreal roots. 2
States how many roots are real, counting the irrational ones as real. 1

Part C 5

Gives all three items for each pairing, the swap, the set it fixes and the coefficient system demanded, keeping rational and real distinct. 3
Rules correctly on both theorems under the altered coefficient, and treats an unavailable theorem as a lost guarantee rather than as a false conclusion. 2

8. The same quartic in two languages 12 points

Part A 3

Reads each root with the sign flipped from its bracket and attaches the exponent as the multiplicity. 2
Checks the multiplicities against the degree and reports the count of distinct roots separately. 1

Part B 5

Computes both totals from the root list with the repeated root counted twice. 2
Computes both totals from the coefficients with the sign $(-1)^k$ read from the parity of $k$ . 2
States that the two routes agree, rather than leaving four numbers side by side. 1

Part C 4

Identifies the degree with the count taken with multiplicity and grounds it in one root per linear factor. 2
Supplies a monic quartic whose roots are four different numbers, and says that agreement is exactly the case where every root is simple. 2

9. One intercept and one extra root 15 points

Part A 5

Converts the touch into even multiplicity and narrows the options to $2$ and $4$ using the degree. 2
Eliminates multiplicity $4$ by computing the value at $0$ that it would force, rather than by asserting it looks wrong. 2
Concludes that the two remaining roots are nonreal, using that every real root would show as an intercept. 1

Part B 5

Supplies the conjugate of the given root and multiplies the pair into a real quadratic. 2
Assembles $P$ with the repeated real factor and leading coefficient $1$ , then expands correctly. 2
Checks the constant term against the given value of $P(0)$ , or checks that the quadratic has negative discriminant. 1

Part C 5

Separates what the conditions fix, the multiplicity and the product of the nonreal pair, from what they leave free, the pair's sum. 3
Exhibits a second quartic satisfying every stated condition, and checks at least its value at $0$ and its lack of a second intercept. 2

10. Two exponents and one condition 15 points

Part A 5

Uses the degree to get $a + b = 5$ , counting the quadratic factor's two degrees. 2
Determines the sign on the middle interval factor by factor and concludes that $b$ is even. 2
Reports both surviving pairs rather than settling on one of them. 1

Part B 5

Reads crossing or touching from the parity of each exponent, for both pairs. 2
Gives a sign for all three intervals and notes that the two pairs produce the same row. 2
Counts the $x$ -intercepts as two and attributes the other roots to multiplicity and to the nonreal pair. 1

Part C 5

Locates the obstruction in the sign depending on the exponents only through their parity, which the two pairs share. 3
Names a computation that returns a value rather than a sign, and gives its two different outcomes. 2

Multiple choice

Free response