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Polynomial Division and Roots: 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, 179 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. Two polynomials neither expanded nor drawn . 17 points. Question 1 of 10.

Two polynomials are described rather than written out in full. The first, $F(x) = (x + 3)^2(2x - 7)\left(x^2 + 5\right)$ , arrives in factors. The second, $G$ , is described only by its degree and by how often its graph meets the horizontal axis.
1. Part A.
  State what $F$ has for a degree, for a leading term and for a constant term. None of the three calls for an expansion.
  
  Write the expression An equation or an expression is enough here. Show how you built it. 5 points
2. Part B.
  Describe each far end of the graph of $F$ , and say how many distinct real zeros $F$ has, giving the reason each factor does or does not contribute one.
  
  Carry your own answer forward Read the far ends off the leading term you reported in part A, whatever it was. The credit is for the reasoning you run on your own leading term, not for landing on a particular pair of ends.
  
  Explain what it means Words, not just symbols. Say what the number is telling you about the situation. 6 points
3. Part C.
  The second polynomial, $G$ , has degree $6$ and its graph crosses the horizontal axis at six different points. Decide how many turning points that graph must have, giving both the largest and the smallest number the description allows, and name the fact that supplies each bound.
  
  Justify your claim State the claim, then give the reason it has to be true. 6 points
The answer
Part A

$F$ has degree $5$ , leading term $2x^5$ and constant term $-315$ .
- the leading term may be written $2x^5$ or as the coefficient $2$ on $x^5$ ; $18x^5$ is not the same, since the repeated factor contributes $x^2$ and not $3^2x^2$
Part B

The graph falls on the far left and rises on the far right. $F$ has two distinct real zeros, $-3$ from the repeated factor and $\tfrac{7}{2}$ from $2x - 7$ ; the factor $x^2 + 5$ is never zero, so it contributes none.

Part C

Exactly five. The degree caps the turning points at $6 - 1 = 5$ , and six distinct real zeros force at least $6 - 1 = 5$ of them, so the two bounds meet and no other count is possible.
Worked solution

Part A

Degrees add across a product and leading terms multiply, so the factored form gives up all three answers with no expansion.

$\deg F = 2 + 1 + 2 = 5, \qquad \left(x^2\right)(2x)\left(x^2\right) = 2x^5$

The constant term is the value at $x = 0$ , which is the product of the factors' own constants:

$F(0) = (3)^2(-7)(5) = 9 \times (-35) = -315.$

Part B

The far ends belong to the leading term alone, which part A found to be $2x^5$ : an odd degree with a positive leading coefficient, so the graph falls on the far left and rises on the far right.

A product is zero exactly when one of its factors is, so take the factors one at a time:

$(x + 3)^2 = 0 \ \Rightarrow\ x = -3, \qquad 2x - 7 = 0 \ \Rightarrow\ x = \tfrac{7}{2}, \qquad x^2 + 5 \ge 5 > 0.$

The squared factor supplies the single number $-3$ however many times it appears, and $x^2 + 5$ never reaches zero for a real input. So there are two distinct real zeros, comfortably inside the ceiling of $5$ that degree $5$ allows.

Part C

Two separate facts press on the count from opposite sides.

From above. A polynomial of degree $n$ has a graph with at most $n - 1$ turning points, so

$\#\{\text{turning points}\} \le 6 - 1 = 5.$

From below. Between two neighbouring zeros the graph leaves the axis and comes back to it, so it must stop rising or stop falling somewhere in between. Six distinct zeros have five neighbouring pairs, each forcing at least one turning point:

$\#\{\text{turning points}\} \ge 6 - 1 = 5.$

The two bounds are the same number, so the count is pinned: exactly five. Note how differently the two arguments work. The ceiling is a fact about degree and holds whatever the zeros do; the floor is a fact about the zeros and would collapse if any two of the six coincided, since only distinct zeros have a stretch of graph between them.

In one line

$F(x) = (x + 3)^2(2x - 7)\left(x^2 + 5\right)$ has degree $5$ , leading term $2x^5$ and constant term $-315$ , so its graph falls on the far left and rises on the far right. It has two distinct real zeros, $-3$ and $\tfrac{7}{2}$ , since $x^2 + 5$ never vanishes and a repeated factor still names one number. The graph of $G$ must have exactly five turning points: degree $6$ allows at most $5$ , and six distinct real zeros force at least $5$ , so the ceiling and the floor meet.

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

Adds the factors' degrees, counting the squared factor twice, to reach degree $5$ . . Worth 2 points.

Multiplies the factors' leading terms rather than their constants, reaching $2x^5$ . . Worth 2 points.

Obtains the constant term as the product of the factors at $x = 0$ , keeping the sign. . Worth 1 point.

Part B 6 points

Uses the parity of the degree and the sign of the leading coefficient to name both far ends. . Worth 2 points.

Sets each factor to zero in turn and solves, including the fractional zero from $2x - 7$ . . Worth 2 points.

Says why $x^2 + 5$ contributes no real zero, and counts $-3$ once rather than twice. . Worth 2 points.

Part C 6 points

States the ceiling $n - 1$ and applies it to degree $6$ . . Worth 2 points.

Argues the floor from the six distinct zeros, explaining why each neighbouring pair forces a turn. . Worth 3 points. needs an explanation, not just an answer

Concludes that the two bounds coincide, so the count is exactly five rather than at most five. . Worth 1 point.
2. The working was lost, the answer was not . 17 points. Question 2 of 10.

A division was carried out and then the working was thrown away. What survives is the divisor $D(x) = x^2 - 3x + 1$ , the quotient $Q(x) = 2x^2 - x - 5$ and the remainder $R(x) = -5x + 1$ .
1. Part A.
  Rebuild the dividend $P(x)$ in standard form, and say why no second dividend could have produced this same set of three.
  
  Write the expression An equation or an expression is enough here. Show how you built it. 5 points
2. Part B.
  Decide which of $7$ , $x^2 + 4$ , $6x - 11$ and $0$ could be the remainder of some division by this same $D(x)$ , giving for each one the reason that settles it.
  
  Justify your claim State the claim, then give the reason it has to be true. 6 points
3. Part C.
  Now suppose the dividend had been $4x - 9$ instead of the polynomial from part A, with the same divisor $D(x)$ . Give the quotient and the remainder, and explain how the algorithm's two promises settle the answer without a single pass of the loop.
  
  Explain why it works A sentence or two. Reasons, not steps. 6 points
The answer
Part A

$P(x) = 2x^4 - 7x^3 + 9x - 4$ . It is the only possible dividend because $P$ is computed from $D$ , $Q$ and $R$ rather than chosen.
- $2x^4 - 7x^3 + 0x^2 + 9x - 4$ is the same polynomial with the empty column shown; $2x^4 - 7x^3 + 14x - 5$ is not, since it drops the remainder
Part B

All but $x^2 + 4$ . A remainder must be the zero polynomial or have degree below the divisor's degree $2$ , so $7$ (degree $0$ ), $6x - 11$ (degree $1$ ) and $0$ all qualify, while $x^2 + 4$ has degree $2$ and still contains a whole copy of $D$ .

Part C

The quotient is $0$ and the remainder is $4x - 9$ . That pair keeps the identity and keeps the degree condition, since $\deg(4x - 9) = 1 < 2$ , and only one pair can keep both, so it is the answer.
Worked solution

Part A

The division algorithm's identity runs in both directions, so multiply back.

$\left(x^2 - 3x + 1\right)\left(2x^2 - x - 5\right) = 2x^4 - 7x^3 + 0x^2 + 14x - 5$

Collecting that product: $2x^4$ from the leading pair; $-x^3 - 6x^3 = -7x^3$ ; $-5x^2 + 3x^2 + 2x^2 = 0$ ; $15x - x = 14x$ ; and $-5$ . Now add the remainder:

$P(x) = 2x^4 - 7x^3 + 0x^2 + 14x - 5 + (-5x + 1) = 2x^4 - 7x^3 + 9x - 4.$

Uniqueness is not even needed here. The right-hand side $DQ + R$ is a single polynomial once the three are fixed, so there is nothing left to choose.

Part B

The divisor has degree $2$ , so the stopping condition reads: the remainder is the zero polynomial, or its degree is at most $1$ .

$R = 0 \quad \text{or} \quad \deg R < \deg D = 2$

Taking them in turn: $7$ has degree $0$ , so it passes; $6x - 11$ has degree $1$ , so it passes; $0$ is named separately by the condition, since the zero polynomial has no degree at all, and it passes. That leaves $x^2 + 4$ , whose degree is $2$ , exactly the divisor's. It is disqualified, and the reason is visible rather than formal: one more copy of $D$ can still be taken out of it, since

$x^2 + 4 = \left(x^2 - 3x + 1\right) + (3x + 3),$

which moves a $1$ into the quotient and leaves $3x + 3$ , a genuine remainder. A leftover that still contains a copy of the divisor is an unfinished division, not the answer to one.

Part C

Test the candidate pair $Q = 0$ and $R = 4x - 9$ against the two promises.

$4x - 9 = \left(x^2 - 3x + 1\right)\cdot 0 + (4x - 9), \qquad \deg(4x - 9) = 1 < 2 = \deg D$

The identity holds trivially, and the degree condition holds because a linear leftover is already smaller, in degree, than a quadratic divisor. Uniqueness closes the argument: the division algorithm promises exactly one pair keeping both conditions, so any pair that keeps both is that pair. The loop agrees without being run, because its stopping test is already satisfied before the first pass, and the corresponding integer statement is the familiar one that $3$ divided by $5$ gives quotient $0$ and remainder $3$ .

Watch out. A quotient of $0$ is a finished answer, not a failure to divide. Nothing in the algorithm promises the quotient is nonzero.

In one line

Multiplying back gives $P(x) = \left(x^2 - 3x + 1\right)\left(2x^2 - x - 5\right) + (-5x + 1) = 2x^4 - 7x^3 + 9x - 4$ , and no other dividend is possible, since $DQ + R$ is computed rather than chosen. Of the four proposed leftovers, $7$ , $6x - 11$ and $0$ are legal remainders for this divisor while $x^2 + 4$ is not, because it still contains a whole copy of $D$ . Dividing $4x - 9$ by the same divisor gives quotient $0$ and remainder $4x - 9$ : that pair keeps the identity and the degree condition, and uniqueness makes anything keeping both the answer.

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 identity $P = DQ + R$ rather than $P = DQ$ , so the remainder is added in. . Worth 2 points.

Expands the product correctly, including the $x^2$ column that cancels to zero. . Worth 2 points.

Says that $DQ + R$ is determined once $D$ , $Q$ and $R$ are given, so the dividend is not a choice. . Worth 1 point.

Part B 6 points

States the stopping condition in terms of degree, naming the zero polynomial as its own case. . Worth 2 points.

Gives a verdict for all four, with the degree that settles each. . Worth 2 points.

Explains why matching the divisor's degree disqualifies a leftover, rather than only asserting the rule. . Worth 2 points. needs an explanation, not just an answer

Part C 6 points

Reports the pair $Q = 0$ , $R = 4x - 9$ . . Worth 1 point.

Checks the pair against both promises, the identity and the degree condition, rather than one of them. . Worth 2 points.

Appeals to uniqueness to explain why keeping both promises is enough to be the answer. . Worth 3 points. needs an explanation, not just an answer
3. One unknown coefficient, then one tableau . 17 points. Question 3 of 10.

The cubic $P(x) = 3x^3 + kx^2 - 8x + 1$ carries a single unknown coefficient. One stated remainder fixes it, and a second division then does the rest of the work.
1. Part A.
  Dividing $P(x)$ by $x - 2$ leaves remainder $17$ . Find $k$ .
  
  Solve and show your work Write each step out, and end with the value and its units. 5 points
2. Part B.
  Using your value of $k$ , run a synthetic tableau that divides $P(x)$ by $x + 1$ . Give the corner number and the finished bottom row, say what each stretch of that row means, and multiply back as a check.
  
  Carry your own answer forward Use the value of $k$ you found in part A, whatever it was, and say which value you are working with. The credit here is for running the tableau on your own coefficients.
  
  Write the expression An equation or an expression is enough here. Show how you built it. 6 points
3. Part C.
  Suppose the condition in part A had been that $x - 2$ divides $P(x)$ exactly, instead of leaving remainder $17$ . Decide whether some value of $k$ makes that true, and say what changes and what does not in the equation you solved.
  
  Justify your claim State the claim, then give the reason it has to be true. 6 points
The answer
Part A

$k = 2$ .
- $k = 2$ may be reached by a tableau with corner $2$ , whose last cell is $9 + 4k$ ; a value of $k = 6$ answers a different question, about division by $x + 2$
Part B

With $k = 2$ : corner $-1$ , quotient $3x^2 - x - 7$ , remainder $8$ .
- the bottom row $3,\; -1,\; -7,\; 8$ says the same thing; a quotient written $3x^3 - x^2 - 7x$ has been read one degree too high
Part C

Yes, at $k = -\tfrac{9}{4}$ . Only the right-hand side changes: the evaluation at $2$ is set equal to $0$ rather than to $17$ , and nothing in the theorem requires the solution to be an integer.
Worked solution

Part A

The remainder on dividing by $x - 2$ is the value at $2$ , so set that value equal to the stated remainder.

$P(2) = 3(8) + 4k - 16 + 1 = 9 + 4k = 17 \quad\Longrightarrow\quad k = 2$

Checking, $3x^3 + 2x^2 - 8x + 1$ at $x = 2$ gives $24 + 8 - 16 + 1 = 17$ .

Part B

With $k = 2$ the coefficients are $3,\; 2,\; -8,\; 1$ , and $x + 1 = x - (-1)$ , so the corner number is $-1$ . Bring down $3$ ; then $3 \times (-1) = -3$ and $2 - 3 = -1$ ; then $-1 \times (-1) = 1$ and $-8 + 1 = -7$ ; then $-7 \times (-1) = 7$ and $1 + 7 = 8$ .

$\begin{array}{c|rrrr} -1 & 3 & 2 & -8 & 1 \\ & & -3 & 1 & 7 \\ \hline & 3 & -1 & -7 & 8 \end{array}$

So the quotient is $3x^2 - x - 7$ and the remainder is $8$ . Expanding as a check, $(x + 1)\left(3x^2 - x - 7\right) = 3x^3 + 2x^2 - 8x - 7$ , and adding $8$ restores $3x^3 + 2x^2 - 8x + 1$ .

Part C

Exact division is not a different kind of condition. It is the same condition with a different number on the right, because dividing exactly means leaving remainder $0$ , and the remainder is still the value at $2$ .

$P(2) = 9 + 4k = 0 \quad\Longrightarrow\quad k = -\tfrac{9}{4}$

So such a $k$ exists. The left-hand side is untouched: the same evaluation, at the same input, built from the same coefficients. What changes is only the target value.

One thing worth saying plainly is that a fractional answer is not a symptom of an error here. The problem never insisted on integer coefficients, and $3x^3 - \tfrac{9}{4}x^2 - 8x + 1$ is a perfectly good polynomial with $x - 2$ as a factor. It is the Rational Root Theorem, not the Factor Theorem, that asks for integer coefficients, and nothing in this part invoked it.

In one line

The stated remainder gives $P(2) = 9 + 4k = 17$ , so $k = 2$ and $P(x) = 3x^3 + 2x^2 - 8x + 1$ . Dividing that by $x + 1$ with corner $-1$ produces the bottom row $3,\; -1,\; -7,\; 8$ , so the quotient is $3x^2 - x - 7$ and the remainder is $8$ , which expanding confirms. Had the condition been exact division by $x - 2$ , the same evaluation would have been set equal to $0$ instead of $17$ , giving $k = -\tfrac{9}{4}$ : a legitimate answer, since nothing in the Factor Theorem asks for an integer.

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 at $2$ , the input that makes the divisor zero, rather than at $-2$ . . Worth 2 points.

Sets the evaluation equal to $17$ and solves the resulting linear equation. . Worth 2 points.

Reports $k$ as a single number and checks it against the stated remainder. . Worth 1 point.

Part B 6 points

Puts $-1$ in the corner, flipping the sign of the divisor's constant. . Worth 2 points.

Runs bring down, multiply, add across all four columns with correct signs. . Worth 2 points.

Reads the bottom row as a quadratic quotient plus a separate remainder, and confirms by expanding. . Worth 2 points.

Part C 6 points

Recognises exact division as remainder $0$ and sets the same evaluation equal to $0$ . . Worth 2 points.

Solves for $k$ and answers the decision asked for rather than only computing. . Worth 2 points.

Explains that only the right-hand side changed, and that a non-integer $k$ is legitimate here. . Worth 2 points. needs an explanation, not just an answer
4. Built to a specification, then read back . 18 points. Question 4 of 10.

A polynomial is wanted with three prescribed roots, $-4$ , $\tfrac{1}{3}$ and $\tfrac{3}{2}$ , of the least degree those roots allow, and with leading coefficient $6$ .
1. Part A.
  Write down a polynomial meeting the whole specification, in factored form and expanded.
  
  Write the expression An equation or an expression is enough here. Show how you built it. 6 points
2. Part B.
  Describe the two far ends of the graph of your polynomial, give its $y$ -intercept, and say exactly how many turning points that graph must have.
  
  Carry your own answer forward Read all three answers off the polynomial you produced in part A, whatever it was, and say which polynomial you are reading. The credit is for the reasoning, not for matching a particular intercept.
  
  Explain what it means Words, not just symbols. Say what the number is telling you about the situation. 6 points
3. Part C.
  A classmate offers $Q(x) = 18(x + 4)\left(x - \tfrac{1}{3}\right)\left(x - \tfrac{3}{2}\right)$ and claims it meets the same specification. Rule on the claim, say what a list of roots does and does not fix, and say what a specification would have had to ask for in order to select $Q$ .
  
  Justify your claim State the claim, then give the reason it has to be true. 6 points
The answer
Part A

$P(x) = (x + 4)(3x - 1)(2x - 3) = 6x^3 + 13x^2 - 41x + 12$ .
- $6(x + 4)\left(x - \tfrac{1}{3}\right)\left(x - \tfrac{3}{2}\right)$ is the same polynomial with the $6$ still outside; $(x + 4)\left(x - \tfrac{1}{3}\right)\left(x - \tfrac{3}{2}\right)$ is not, since its leading coefficient is $1$
Part B

The graph falls on the far left and rises on the far right, and it meets the $y$ -axis at $12$ . It must have exactly two turning points: three distinct real zeros force at least two, and degree $3$ allows at most two.

Part C

It fails. $Q$ has the three prescribed roots and the least degree, but its leading coefficient is $18$ rather than $6$ . A root list fixes a polynomial only up to a nonzero constant multiple, so a specification selecting $Q$ would have to name $18$ as the leading coefficient, or give a point on the graph that picks it out.
Worked solution

Part A

Each prescribed root $r$ contributes the factor $x - r$ , and three roots are the fewest factors that can carry them, so the least degree is $3$ .

$P(x) = 6(x + 4)\left(x - \tfrac{1}{3}\right)\left(x - \tfrac{3}{2}\right)$

The $6$ can be shared out to clear both fractions, since $6 = 3 \times 2$ : giving the $3$ to the second factor and the $2$ to the third turns them into $3x - 1$ and $2x - 3$ , which is tidier to expand.

$(3x - 1)(2x - 3) = 6x^2 - 11x + 3$

$(x + 4)\left(6x^2 - 11x + 3\right) = 6x^3 - 11x^2 + 3x + 24x^2 - 44x + 12 = 6x^3 + 13x^2 - 41x + 12$

The leading coefficient is $6$ , as required, and $P(-4) = -384 + 208 + 164 + 12 = 0$ checks one root.

Part B

The degree is $3$ , odd, and the leading coefficient $6$ is positive, so the two ends disagree and the right one points up: the graph falls on the far left and rises on the far right.

The $y$ -intercept is the value at $0$ , which is the constant term:

$P(0) = 12.$

For the turns, press from both sides. The ceiling gives at most $3 - 1 = 2$ . The floor gives at least one turn between each neighbouring pair of the three distinct zeros, which is $3 - 1 = 2$ again. The bounds coincide, so the count is exactly $2$ .

Part C

Check the claim clause by clause. The three roots are right, since $Q$ vanishes wherever a factor does, and multiplying by $18$ cannot move a zero of a product. The degree is right, at $3$ . What fails is the last clause:

$Q(x) = 18(x + 4)\left(x - \tfrac{1}{3}\right)\left(x - \tfrac{3}{2}\right) = 3\,P(x) = 18x^3 + 39x^2 - 123x + 36,$

whose leading coefficient is $18$ , not $6$ .

That is the general situation rather than an accident of this example. Every nonzero multiple of a polynomial has exactly the same roots, so a root list can never single out one polynomial; it produces a whole family $a(x - r_1)(x - r_2)(x - r_3)$ , one member for each nonzero $a$ . Exactly one further condition is needed to choose the member, and the specification chose it by naming the leading coefficient. A point on the graph would have done the same job: asking instead for the graph to pass through $(1, -30)$ would have selected $Q$ , since $Q(1) = 18(5)\left(\tfrac{2}{3}\right)\left(-\tfrac{1}{2}\right) = -30$ .

In one line

The specification is met by $P(x) = (x + 4)(3x - 1)(2x - 3) = 6x^3 + 13x^2 - 41x + 12$ . Its graph falls on the far left and rises on the far right, meets the $y$ -axis at $12$ , and has exactly two turning points, since degree $3$ allows at most two and three distinct real zeros force at least two. The classmate's $Q$ has the right roots and the right degree but leading coefficient $18$ , so it fails: a root list fixes a polynomial only up to a nonzero constant multiple, and the specification would have had to name $18$ , or a point such as $(1, -30)$ , to select it.

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 6 points

Turns each root into the factor $x - r$ with the correct sign, including the negative root. . Worth 2 points.

Attaches the scalar $6$ so that the leading coefficient is $6$ , rather than leaving a monic product. . Worth 2 points.

Expands correctly to $6x^3 + 13x^2 - 41x + 12$ . . Worth 2 points.

Part B 6 points

Names both far ends from the parity of the degree and the sign of the leading coefficient. . Worth 2 points.

Identifies the $y$ -intercept as the constant term, or as the value at $0$ . . Worth 1 point.

Pins the number of turning points between the degree ceiling and the floor the distinct zeros force. . Worth 3 points.

Part C 6 points

Checks each clause of the specification separately and identifies the leading coefficient as the one that fails. . Worth 2 points.

States that roots fix a polynomial only up to a nonzero constant multiple, with a reason rather than an assertion. . Worth 3 points. needs an explanation, not just an answer

Names a condition that would have selected $Q$ : the leading coefficient $18$ , or a specific point on its graph. . Worth 1 point.
5. What the list reaches, and what it cannot . 17 points. Question 5 of 10.

The polynomial $P(x) = 2x^3 + x^2 - 7x + 3$ has integer coefficients, so its rational roots are confined to a short list. What the list cannot reach has to be located some other way.
1. Part A.
  Write out the finite set of fractions inside which any rational root of $P$ has to lie, in lowest terms and without repeats, then say which member of that set is a root.
  
  Write the expression An equation or an expression is enough here. Show how you built it. 5 points
2. Part B.
  Take the corresponding linear factor out of $P$ and report the polynomial left behind. Then show that neither of the two roots still outstanding is rational.
  
  Carry your own answer forward Peel off the factor belonging to the root you found in part A, whatever it was, and work inside the quotient it leaves rather than returning to $P$ .
  
  Write the expression An equation or an expression is enough here. Show how you built it. 6 points
3. Part C.
  Evaluate $x^2 + x - 3$ at $-3$ , $-2$ , $1$ and $2$ , then use the four values to trap each remaining root between consecutive integers. Say what the trapping establishes and what it leaves unnamed.
  
  Carry your own answer forward Evaluate the quadratic you produced in part B, whatever it was, and trap its roots using your own four values.
  
  Explain what it means Words, not just symbols. Say what the number is telling you about the situation. 6 points
The answer
Part A

The permitted numbers are $\pm 1$ , $\pm 3$ , $\pm\tfrac{1}{2}$ and $\pm\tfrac{3}{2}$ , eight in all, and $\tfrac{1}{2}$ is a root.
- the list may be written as $\pm\tfrac{p}{q}$ with $p \in \{1, 3\}$ and $q \in \{1, 2\}$ ; a list containing $\pm\tfrac{1}{3}$ has taken the denominators from the constant term
Part B

$P(x) = (2x - 1)\left(x^2 + x - 3\right)$ . The quadratic's own candidates are $\pm 1$ and $\pm 3$ , whose values are $-1$ , $-3$ , $9$ and $3$ , none of them zero, so neither remaining root is rational.

Part C

The values are $3$ , $-1$ , $-1$ and $3$ , so one root lies between $-3$ and $-2$ and the other between $1$ and $2$ . Each sign change proves a root sits inside that interval; neither names the number, and the numbers are irrational.
Worked solution

Part A

The numerator of a rational root divides the constant term $3$ and the denominator divides the leading coefficient $2$ .

$p \in \{1, 3\}, \qquad q \in \{1, 2\} \quad\Longrightarrow\quad \pm 1, \ \pm 3, \ \pm\tfrac{1}{2}, \ \pm\tfrac{3}{2}$

Test the small ones first. $P(1) = 2 + 1 - 7 + 3 = -1$ and $P(-1) = -2 + 1 + 7 + 3 = 9$ , so neither works. Next,

$P\!\left(\tfrac{1}{2}\right) = \tfrac{1}{4} + \tfrac{1}{4} - \tfrac{7}{2} + 3 = 0,$

so $\tfrac{1}{2}$ is a root.

Part B

Run the tableau with corner $\tfrac{1}{2}$ on the coefficients $2,\; 1,\; -7,\; 3$ . The bottom row is $2,\; 2,\; -6,\; 0$ , so

$P(x) = \left(x - \tfrac{1}{2}\right)\left(2x^2 + 2x - 6\right) = (2x - 1)\left(x^2 + x - 3\right),$

where the $2$ has been moved from the quadratic into the linear factor to clear the fraction.

The remaining roots are the roots of $x^2 + x - 3$ , which is monic with integer coefficients, so any rational root of it would be an integer dividing $3$ . That is four numbers, and all four fail:

$1 + 1 - 3 = -1, \qquad 1 - 1 - 3 = -3, \qquad 9 + 3 - 3 = 9, \qquad 9 - 3 - 3 = 3.$

So neither of the two remaining roots is rational.

Part C

Evaluate at the four inputs.

$9 - 3 - 3 = 3, \qquad 4 - 2 - 3 = -1, \qquad 1 + 1 - 3 = -1, \qquad 4 + 2 - 3 = 3$

The sign changes from $+$ to $-$ across $[-3, -2]$ and from $-$ to $+$ across $[1, 2]$ . A polynomial graph is unbroken, so it cannot cross from one side of the axis to the other without meeting it, and each interval therefore holds a root. Two roots is all a quadratic has, so these are both of them, and each interval holds exactly one.

What the argument delivers is a location, and only a location. It proves the root is there, and it can be tightened as far as patience allows by halving the interval, but it never produces the number. Part B already showed neither root is rational, so no amount of tightening will ever land on a fraction.

In one line

The theorem permits $\pm 1$ , $\pm 3$ , $\pm\tfrac{1}{2}$ and $\pm\tfrac{3}{2}$ , and $\tfrac{1}{2}$ is a root, so $P(x) = (2x - 1)\left(x^2 + x - 3\right)$ . The quotient's own candidates $\pm 1$ and $\pm 3$ give $-1$ , $-3$ , $9$ and $3$ , so neither remaining root is rational. Evaluating the quotient at $-3$ , $-2$ , $1$ and $2$ gives $3$ , $-1$ , $-1$ and $3$ , trapping one root between $-3$ and $-2$ and the other between $1$ and $2$ : the sign changes prove where each root is without ever naming it.

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

Takes numerators from the constant term and denominators from the leading coefficient, not the reverse. . Worth 2 points.

Produces all eight numbers, both signs included, reduced and without duplicates. . Worth 2 points.

Tests candidates and reports $\tfrac{1}{2}$ as a root, with the evaluation that establishes it. . Worth 1 point.

Part B 6 points

Divides by the factor the root supplies and reports a quadratic quotient, one degree lower. . Worth 2 points.

Clears the fraction by moving the leading coefficient into the linear factor, giving $(2x - 1)$ . . Worth 1 point.

Builds the quotient's own candidate list and tests every entry, rather than reusing the list from part A. . Worth 3 points.

Part C 6 points

Computes all four values correctly. . Worth 2 points.

Identifies both sign changes and names the interval each root lies in. . Worth 2 points.

Says that the trap locates a root without naming it, and connects that to the failed candidate list. . Worth 2 points.
6. One value, three questions . 18 points. Question 6 of 10.

A single value of a polynomial settles more than it looks. Throughout this question $P$ is a polynomial with $P(4) = 6$ , and three separate questions are pressed on that one number.
1. Part A.
  Decide, for each of $P(x)$ , $P(x) - 6$ and $P(x) + 6$ , whether $x - 4$ is a factor of it. Give the evaluation that settles each case.
  
  Justify your claim State the claim, then give the reason it has to be true. 6 points
2. Part B.
  Suppose in addition that $P(x) = (x - 4)\,Q(x) + 6$ with $Q(x) = 2x^2 + 3x - 1$ . Find $P(7)$ , and state the remainder when $P(x)$ is divided by $x - 7$ .
  
  Solve and show your work Write each step out, and end with the value and its units. 6 points
3. Part C.
  Decide whether $2x - 8$ is a factor of $P(x) - 6$ , and then state what the two-way factor statement looks like for a divisor $ax - b$ with $a \ne 0$ .
  
  Justify your claim State the claim, then give the reason it has to be true. 6 points
The answer

Part A

It is a factor of $P(x) - 6$ only. The three values at $4$ are $6$ , $0$ and $12$ , and the Factor Theorem converts the middle one into a factor while ruling the other two out.
Part B

$P(7) = 360$ , and $360$ is also the remainder on dividing $P(x)$ by $x - 7$ .
- $360$ may be reached by expanding $P$ first, which gives $2x^3 - 5x^2 - 13x + 10$ ; an answer of $354$ has left the $+6$ off the identity
Part C

Yes. Since $2x - 8 = 2(x - 4)$ and multiplying a divisor by a nonzero constant cannot change whether a division comes out exact, the quotient simply picks up a factor of $\tfrac{1}{2}$ . In general, $ax - b$ is a factor of a polynomial exactly when that polynomial vanishes at $\tfrac{b}{a}$ .
Worked solution

Part A

The Factor Theorem is a two-way statement, so one evaluation settles each case in whichever direction it happens to fall. Evaluate all three at $4$ :

$P(4) = 6, \qquad P(4) - 6 = 0, \qquad P(4) + 6 = 12$

A value of $0$ means $x - 4$ is a factor, so $x - 4$ divides $P(x) - 6$ exactly. A value that is not $0$ means it is not a factor, which rules out both $P(x)$ itself and $P(x) + 6$ : were $x - 4$ a factor of either, evaluating the factorization at $4$ would force the value to be $0$ , and it is not.

Notice that the negative verdicts use the theorem in its contrapositive form. Nothing about $P$ beyond the single number $P(4)$ was needed for any of the three.

Part B

The identity holds at every input, so evaluate it at $7$ rather than expanding.

$Q(7) = 2(49) + 21 - 1 = 118, \qquad P(7) = (7 - 4)(118) + 6 = 354 + 6 = 360$

The second question is the same number wearing a different hat. The remainder on dividing $P(x)$ by $x - 7$ is $P(7)$ by the Remainder Theorem, so it is $360$ . No second division is needed, and no expansion of $P$ either.

Part C

Part A gave $P(x) - 6 = (x - 4)\,S(x)$ for some polynomial $S$ . Pull the constant across:

$P(x) - 6 = (x - 4)\,S(x) = (2x - 8)\cdot\tfrac{1}{2}S(x),$

and $\tfrac{1}{2}S(x)$ is still a polynomial, since halving every coefficient of a polynomial leaves a polynomial. So $2x - 8$ is a factor, with the quotient halved and nothing else disturbed.

The general statement follows the same way. For $a \ne 0$ ,

$ax - b = a\left(x - \tfrac{b}{a}\right),$

so $ax - b$ divides a polynomial exactly when $x - \tfrac{b}{a}$ does, and by the Factor Theorem that happens exactly when the polynomial vanishes at $\tfrac{b}{a}$ . The rule of thumb is unchanged: test the value that kills the candidate factor. For $2x - 8$ that value is $4$ , which is why part A already answered this part.

In one line

Of the three, $x - 4$ is a factor of $P(x) - 6$ alone, since the values at $4$ are $6$ , $0$ and $12$ and only a value of $0$ produces a factor. With $Q(x) = 2x^2 + 3x - 1$ , evaluating the identity at $7$ gives $P(7) = 3(118) + 6 = 360$ , which is also the remainder on dividing by $x - 7$ . Finally $2x - 8 = 2(x - 4)$ is a factor of $P(x) - 6$ too, the quotient merely picking up a factor of $\tfrac{1}{2}$ : in general $ax - b$ is a factor exactly when the polynomial vanishes at $\tfrac{b}{a}$ .

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 6 points

Evaluates all three polynomials at $4$ rather than attempting a division. . Worth 2 points.

Gives the right verdict for all three cases. . Worth 2 points.

Explains how a nonzero value rules a factor out, not only how a zero value establishes one. . Worth 2 points. needs an explanation, not just an answer

Part B 6 points

Evaluates the given identity at $7$ instead of expanding $P$ first. . Worth 2 points.

Computes $Q(7)$ and assembles $P(7)$ with the $+6$ retained. . Worth 2 points.

Names the remainder on division by $x - 7$ as that same value, citing the Remainder Theorem. . Worth 2 points.

Part C 6 points

Writes $2x - 8$ as a constant times $x - 4$ and rules that it is a factor. . Worth 2 points.

Explains why a nonzero constant multiple of a divisor divides exactly the same polynomials, naming what happens to the quotient. . Worth 3 points. needs an explanation, not just an answer

States the general condition for $ax - b$ as vanishing at $\tfrac{b}{a}$ . . Worth 1 point.
7. A quadratic divisor, and what one division certifies . 19 points. Question 7 of 10.

Take $P(x) = 2x^4 - 7x^3 - 7x^2 + 38x - 24$ and $D(x) = x^2 - 5x + 6$ . Whether $D$ is a factor of $P$ is a question one division answers, and the answer carries rather more than a quotient.
1. Part A.
  Divide $P(x)$ by $D(x)$ , reporting the quotient and the remainder.
  
  Write the expression An equation or an expression is enough here. Show how you built it. 6 points
2. Part B.
  Your part A result hands over two roots of $P$ with no further evaluation. Name them and say what entitles you to them. Then say how many further roots $P$ can have, and whether any of them is rational.
  
  Carry your own answer forward Argue from the quotient and remainder you reported in part A, whatever they were, and say which pair you are using.
  
  Justify your claim State the claim, then give the reason it has to be true. 7 points
3. Part C.
  Explain why the single division in part A proves that $D$ is a factor of $P$ . Say what a table of values of $P$ alone could never supply, however long it ran, and why a divisor with no rational root leaves no convenient input to test at all. Then say what the uniqueness half of the division algorithm contributes.
  
  Explain why it works A sentence or two. Reasons, not steps. 6 points
The answer
Part A

The quotient is $2x^2 + 3x - 4$ and the remainder is $0$ .
- the answer may be written as the identity $P(x) = \left(x^2 - 5x + 6\right)\left(2x^2 + 3x - 4\right)$ ; a nonzero remainder means an arithmetic slip, since this division comes out exact
Part B

$2$ and $3$ , because $D(x) = (x - 2)(x - 3)$ and an exact division makes $P = D\,Q$ , which vanishes wherever $D$ does. The quadratic quotient supplies at most two further roots, and neither is rational: the quotient's candidates $\pm 1, \pm 2, \pm 4, \pm\tfrac{1}{2}$ all fail.

Part C

A remainder of $0$ turns the identity into $P = D\,Q$ , true at every input, which is what being a factor means. A table of values never yields the cofactor $Q$ , and a divisor with no rational root offers no input worth testing. Uniqueness then makes one division decisive: an exact factorization by $D$ would pair with remainder $0$ .
Worked solution

Part A

Both polynomials are already in standard form with no gaps, so run the loop three times.

First pass: $2x^4 \div x^2 = 2x^2$ , and $2x^2\left(x^2 - 5x + 6\right) = 2x^4 - 10x^3 + 12x^2$ ; subtracting leaves $3x^3 - 19x^2 + 38x - 24$ . Second pass: $3x^3 \div x^2 = 3x$ , and $3x\left(x^2 - 5x + 6\right) = 3x^3 - 15x^2 + 18x$ ; subtracting leaves $-4x^2 + 20x - 24$ . Third pass: $-4x^2 \div x^2 = -4$ , and $-4\left(x^2 - 5x + 6\right) = -4x^2 + 20x - 24$ ; subtracting leaves $0$ .

$2x^4 - 7x^3 - 7x^2 + 38x - 24 = \left(x^2 - 5x + 6\right)\left(2x^2 + 3x - 4\right)$

The forecast checks out before the arithmetic does: $\deg Q = 4 - 2 = 2$ , and the quotient's leading coefficient is $2 \div 1 = 2$ .

Part B

The remainder was $0$ , so the identity collapses to a product: $P(x) = D(x)\,Q(x)$ . Splitting the divisor,

$D(x) = x^2 - 5x + 6 = (x - 2)(x - 3),$

and a product is zero exactly when one factor is, so $P(2) = 0$ and $P(3) = 0$ without either being computed.

How many more can there be? The quotient $2x^2 + 3x - 4$ has degree $2$ , so it contributes at most two roots, and $P$ has degree $4$ , which caps the total at four in any case. Testing the quotient's own candidate list, which is $\pm 1$ , $\pm 2$ , $\pm 4$ and $\pm\tfrac{1}{2}$ , gives $1$ , $-5$ , $10$ , $-2$ , $40$ , $16$ , $-2$ and $-5$ ; none is zero, so neither remaining root is rational.

Watch out. The two roots named here came from the DIVISOR, not from the quotient. It is the exactness of the division that transfers them to $P$ .

Part C

Why the division settles it. Being a factor is a statement about an identity: $D$ is a factor of $P$ when $P(x) = D(x)\,Q(x)$ for some polynomial $Q$ and for every value of $x$ . The division produced exactly such a $Q$ and certified that nothing was left over, so the statement is established as written.

What evaluations cannot supply. An evaluation tests one input. Here, as it happens, two of them would have settled the factor question on their own: $D(x) = (x - 2)(x - 3)$ , so $P(2) = P(3) = 0$ would show both linear factors divide $P$ , and since they share no root their product does too. What no table of values of $P$ ever supplies is the cofactor $Q$ , and that is half of what a factorization is for. Nor does the shortcut generalise: for a divisor with no rational root there is no convenient input to test in the first place. Divide by $x^2 + 1$ and no real input makes the divisor vanish at all, so the table has nowhere to look, while the division runs exactly as before.

What uniqueness contributes. Without it, an exact division might be one lucky route among several, and failing to find one would prove nothing. The division algorithm promises exactly one pair $(Q, R)$ keeping the identity and the degree condition together, so:

$P = D\,S \ \text{ for some polynomial } S \quad\Longrightarrow\quad (S, 0) \ \text{ keeps both promises} \quad\Longrightarrow\quad S = Q, \ 0 = R.$

So one division decides the question in both directions at once. A remainder of $0$ proves $D$ is a factor, and a nonzero remainder proves it is not.

In one line

The division is exact: $2x^4 - 7x^3 - 7x^2 + 38x - 24 = \left(x^2 - 5x + 6\right)\left(2x^2 + 3x - 4\right)$ , quotient $2x^2 + 3x - 4$ with remainder $0$ . Since $x^2 - 5x + 6 = (x - 2)(x - 3)$ , the numbers $2$ and $3$ are roots of $P$ with nothing further computed, and the quadratic quotient supplies at most two more, neither rational, since its candidates $\pm 1, \pm 2, \pm 4, \pm\tfrac{1}{2}$ all fail. The one division settles the factor question because a remainder of $0$ makes the identity a product holding at every input, and it hands over the cofactor, which no table of values of $P$ ever supplies and which a divisor with no rational root gives no input to hunt for; uniqueness then means any exact factorization by $D$ would have to be the pair the division already returned.

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 6 points

Forms each quotient term by dividing leading term by leading term, and subtracts the entire product each pass. . Worth 3 points.

Reaches quotient $2x^2 + 3x - 4$ with remainder $0$ . . Worth 2 points.

Checks the answer, by expanding $DQ$ or against the degree and leading-coefficient forecast. . Worth 1 point.

Part B 7 points

Factors the divisor and names $2$ and $3$ as roots of $P$ . . Worth 2 points.

Explains that a remainder of $0$ turns the identity into a product, which is what transfers the divisor's roots to $P$ . . Worth 3 points. needs an explanation, not just an answer

Bounds the further roots by the quotient's degree and tests the quotient's candidate list to rule out rational ones. . Worth 2 points.

Part C 6 points

States that a remainder of $0$ makes the identity a product valid at every input, which is the definition of a factor. . Worth 2 points. needs an explanation, not just an answer

Says what a table of values of $P$ cannot supply, naming the cofactor, and why a divisor with no rational root leaves nothing to test. . Worth 2 points. needs an explanation, not just an answer

Uses uniqueness to argue that a nonzero remainder rules a factorization out, not merely that a zero remainder establishes one. . Worth 2 points. needs an explanation, not just an answer
8. Where the filter has nothing to grip . 18 points. Question 8 of 10.

The polynomial $P(x) = 6x^4 - 5x^3 - 17x^2 + 6x$ has constant term $0$ , which is precisely the case in which the Rational Root Theorem, applied exactly as printed, gives nothing away.
1. Part A.
  Say what the theorem's numerator condition asserts about $P$ as printed, and why that assertion excludes nothing. Then rewrite $P$ in a form the theorem can grip, and give the candidate list for the polynomial that remains.
  
  Justify your claim State the claim, then give the reason it has to be true. 6 points
2. Part B.
  Find every rational root of $P$ .
  
  Carry your own answer forward Work from the factored form and the candidate list you produced in part A, whatever they were, rather than starting again from $P$ as printed.
  
  Solve and show your work Write each step out, and end with the value and its units. 6 points
3. Part C.
  A classmate summarises part B like this: "For a polynomial with zero constant term, the rational roots are $0$ together with every number on the cofactor's candidate list." Refute the summary with a specific polynomial, demonstrating the failure rather than asserting it, and write down a corrected sentence.
  
  Construct a counterexample Give one specific case, and show it breaks the claim. 6 points
The answer

Part A

The condition reads $p \mid 0$ , which every integer satisfies, so it excludes nothing. Factoring gives $P(x) = x\left(6x^3 - 5x^2 - 17x + 6\right)$ , whose cubic has candidates $\pm 1, \pm 2, \pm 3, \pm 6, \pm\tfrac{1}{2}, \pm\tfrac{3}{2}, \pm\tfrac{1}{3}, \pm\tfrac{2}{3}, \pm\tfrac{1}{6}$ .
Part B

$x = 0$ , $x = 2$ , $x = \tfrac{1}{3}$ and $x = -\tfrac{3}{2}$ .
- the same four roots are named by the factorization $P(x) = x(x - 2)(3x - 1)(2x + 3)$ ; a list omitting $0$ has forgotten the factor that was taken out
Part C

Take $R(x) = x\left(x^2 - 3\right)$ . Its cofactor's candidates are $\pm 1$ and $\pm 3$ , whose values are $-2$ , $-2$ , $6$ and $6$ , so none is a root and $0$ is the only rational root of $R$ . Corrected: the rational roots are $0$ together with whichever candidates testing confirms.
Worked solution

Part A

The theorem says the numerator of a rational root divides the constant term. Here the constant term is $0$ , and every integer divides $0$ , since $0 = p \times 0$ whatever $p$ is. The condition is therefore satisfied by every numerator and rules out nothing at all; the denominator condition still bites, but on its own it leaves infinitely many candidates.

The repair is to take the largest power of $x$ out first, which costs nothing and hands over one root immediately:

$P(x) = x\left(6x^3 - 5x^2 - 17x + 6\right).$

The cofactor has constant term $6$ and leading coefficient $6$ , so its numerators divide $6$ and so do its denominators:

$\pm 1, \ \pm 2, \ \pm 3, \ \pm 6, \ \pm\tfrac{1}{2}, \ \pm\tfrac{3}{2}, \ \pm\tfrac{1}{3}, \ \pm\tfrac{2}{3}, \ \pm\tfrac{1}{6}$

Eighteen numbers once duplicates such as $\tfrac{2}{2}$ and $\tfrac{6}{3}$ have been reduced away.

Part B

The factor $x$ gives the root $0$ at once. Inside the cubic, test candidates smallest first. $6 - 5 - 17 + 6 = -10$ rules out $1$ , and $-6 - 5 + 17 + 6 = 12$ rules out $-1$ . Next try $2$ :

$6(8) - 5(4) - 17(2) + 6 = 48 - 20 - 34 + 6 = 0$

so $2$ is a root. Peel it off with corner $2$ on $6,\; -5,\; -17,\; 6$ , whose bottom row is $6,\; 7,\; -3,\; 0$ :

$6x^3 - 5x^2 - 17x + 6 = (x - 2)\left(6x^2 + 7x - 3\right) = (x - 2)(3x - 1)(2x + 3)$

The quadratic quotient ends the hunt, and it splits into two linear factors, so the full factorization is $P(x) = x(x - 2)(3x - 1)(2x + 3)$ and the rational roots are $0$ , $2$ , $\tfrac{1}{3}$ and $-\tfrac{3}{2}$ . A quartic has room for at most four roots, so the list is complete.

Part C

The summary quietly converts the theorem into its converse, and one polynomial is enough to break it. Take

$R(x) = x\left(x^2 - 3\right) = x^3 - 3x.$

The cofactor $x^2 - 3$ is monic with constant term $-3$ , so its candidates are the integers dividing $3$ , namely $\pm 1$ and $\pm 3$ . Test all four:

$1 - 3 = -2, \qquad 1 - 3 = -2, \qquad 9 - 3 = 6, \qquad 9 - 3 = 6$

Not one of them is zero. So the classmate's recipe would report five rational roots, $0, 1, -1, 3, -3$ , where $R$ has exactly one, namely $0$ . Substituting $1$ into $R$ gives $1 - 3 = -2$ , not $0$ , which is the failure demonstrated rather than described.

The corrected sentence keeps the direction the theorem actually proves: the rational roots are $0$ together with those candidates that testing confirms. The list marks the only places a rational root is permitted to sit, and never says which of those places is occupied.

In one line

As printed, the numerator condition on $P(x) = 6x^4 - 5x^3 - 17x^2 + 6x$ reads $p \mid 0$ , which every integer satisfies, so it filters nothing; factoring gives $P(x) = x\left(6x^3 - 5x^2 - 17x + 6\right)$ , whose candidates are $\pm 1, \pm 2, \pm 3, \pm 6, \pm\tfrac{1}{2}, \pm\tfrac{3}{2}, \pm\tfrac{1}{3}, \pm\tfrac{2}{3}, \pm\tfrac{1}{6}$ . Testing finds $2$ , and peeling it off leaves $6x^2 + 7x - 3 = (3x - 1)(2x + 3)$ , so $P(x) = x(x - 2)(3x - 1)(2x + 3)$ and the rational roots are $0$ , $2$ , $\tfrac{1}{3}$ and $-\tfrac{3}{2}$ . The classmate's summary is the converse of the theorem and fails: $R(x) = x\left(x^2 - 3\right)$ has candidates $\pm 1$ and $\pm 3$ with values $-2$ , $-2$ , $6$ and $6$ , so its only rational root is $0$ .

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 6 points

Says that every integer divides $0$ , so the numerator condition is vacuous here. . Worth 2 points. needs an explanation, not just an answer

Factors out the largest power of $x$ to produce a cofactor with nonzero constant term. . Worth 2 points.

Builds the cofactor's list correctly, reduced and deduplicated, with both signs. . Worth 2 points.

Part B 6 points

Reports $0$ as a root, from the factor taken out in part A. . Worth 1 point.

Tests candidates against the cofactor and peels off a factor when one succeeds. . Worth 3 points.

Finishes the quadratic quotient and reports all four roots, noting that a quartic can have no more. . Worth 2 points.

Part C 6 points

Produces a specific polynomial with zero constant term whose cofactor has no rational root. . Worth 2 points.

Builds the cofactor's candidate list and evaluates every entry, showing the failure rather than asserting it. . Worth 2 points.

Writes a corrected sentence that keeps the one-way direction the theorem proves. . Worth 2 points.
9. Two ways to look for a root . 20 points. Question 9 of 10.

The polynomial $f(x) = x^3 - 6x^2 + 3x + 5$ has integer coefficients. This question puts two tools on it in turn, a candidate list and a table of values, and asks what each of them is in a position to settle.
1. Part A.
  Evaluate $f$ at $-1$ , $0$ , $1$ , $2$ , $5$ and $6$ using synthetic tableaux, and name the theorem that lets a tableau's last cell serve as a value of $f$ .
  
  Solve and show your work Write each step out, and end with the value and its units. 6 points
2. Part B.
  Use your six values to trap each real root of $f$ between two consecutive entries of your list, and justify the claim that you have found every real root of $f$ .
  
  Carry your own answer forward Trap the roots using the six values you computed in part A, whatever they were, and state which values you are reading the signs from.
  
  Justify your claim State the claim, then give the reason it has to be true. 7 points
3. Part C.
  Explain why no root of $f$ is rational, and say what the trapping argument does and does not deliver about the roots.
  
  Explain why it works A sentence or two. Reasons, not steps. 7 points
The answer
Part A

$f(-1) = -5$ , $f(0) = 5$ , $f(1) = 3$ , $f(2) = -5$ , $f(5) = -5$ and $f(6) = 23$ . The Remainder Theorem is what makes the last cell $f(c)$ .
- the same six values arise from direct substitution; a value of $f(-1) = -3$ has taken $(-1)^3$ as $+1$
Part B

The sign changes across $[-1, 0]$ , $[1, 2]$ and $[5, 6]$ trap one root in each. That is three roots, which is all a cubic can have, so there are no others and each interval holds exactly one.

Part C

$f$ is monic with constant term $5$ , so any rational root would be an integer dividing $5$ , and the four candidates give $3$ , $-5$ , $-5$ and $-285$ , none of them zero. So no root is rational. The trapping proves where each root is and how close two integers can get to it, and never produces the number itself.
Worked solution

Part A

Each tableau runs on the top row $1,\; -6,\; 3,\; 5$ with the input in the corner. For $c = 6$ , for instance, bring down $1$ ; then $6$ and $-6 + 6 = 0$ ; then $0$ and $3 + 0 = 3$ ; then $18$ and $5 + 18 = 23$ .

$\begin{array}{c|rrrr} 6 & 1 & -6 & 3 & 5 \\ & & 6 & 0 & 18 \\ \hline & 1 & 0 & 3 & 23 \end{array}$

Running the same three steps at each input gives

$f(-1) = -5, \quad f(0) = 5, \quad f(1) = 3, \quad f(2) = -5, \quad f(5) = -5, \quad f(6) = 23.$

The last cell is the remainder on dividing by $x - c$ , and the Remainder Theorem says that remainder is $f(c)$ . One multiply and one add per coefficient does the whole job, with no power of the input ever built.

Part B

Read the signs along the list: $f(-1) = -5$ , $f(0) = 5$ , $f(1) = 3$ , $f(2) = -5$ , $f(5) = -5$ , $f(6) = 23$ . The sign flips three times, and an unbroken curve cannot flip sign without meeting the axis, so

$-1 < r_1 < 0, \qquad 1 < r_2 < 2, \qquad 5 < r_3 < 6.$

Those are three different real roots, since the three intervals do not overlap. A nonzero polynomial of degree $3$ has at most three distinct roots, so the count is closed: there is no room for a fourth, and consequently none of these intervals can hold two, or the total would exceed three.

Watch out. No sign change between $2$ and $5$ does not prove there is no root there. It happens that there is none, but matching signs guarantee nothing; what closes that gap is the ceiling on the count, not the table.

Part C

Why no root is rational. The coefficients are integers and the leading coefficient is $1$ , so a rational root $\tfrac{p}{q}$ in lowest terms needs $q \mid 1$ , forcing $q = 1$ : every rational root would be an integer dividing $5$ . That is four numbers, and part A has already disposed of three of them.

$f(1) = 3, \qquad f(-1) = -5, \qquad f(5) = -5, \qquad f(-5) = -125 - 150 - 15 + 5 = -285$

None is zero, so $f$ has no rational root. Combined with part B, each of the three real roots is an irrational number.

What the trap delivers. It delivers location and certainty of existence: a root is inside $[5, 6]$ , and that is proved rather than guessed. It also delivers as much precision as patience allows, since the interval can be halved repeatedly, each halving keeping whichever side carries the sign change.

What it never delivers is the number. No finite sequence of these steps can end on the root, and part A's failed candidate list explains why no exact fraction will ever appear: there is no fraction to land on.

In one line

The six tableaux give $f(-1) = -5$ , $f(0) = 5$ , $f(1) = 3$ , $f(2) = -5$ , $f(5) = -5$ and $f(6) = 23$ , each last cell being $f(c)$ by the Remainder Theorem. The three sign changes trap one root in $(-1, 0)$ , one in $(1, 2)$ and one in $(5, 6)$ , and a cubic has room for no more, so those are all of them. None is rational: $f$ is monic with constant term $5$ , so a rational root would be an integer dividing $5$ , and $f(1) = 3$ , $f(-1) = -5$ , $f(5) = -5$ and $f(-5) = -285$ are all nonzero. The trap proves where each root lies and tightens as far as wanted, but it never names the number.

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 6 points

Sets up each tableau with the corner holding the input and the top row holding every coefficient. . Worth 2 points.

Produces all six values correctly, with signs handled at the negative input. . Worth 3 points.

Names the Remainder Theorem as the entitlement to read a value of $f$ from a division. . Worth 1 point.

Part B 7 points

Identifies all three sign changes and names the interval each root lies in. . Worth 3 points.

Argues from unbrokenness that a sign change forces a zero, rather than asserting it. . Worth 2 points. needs an explanation, not just an answer

Closes the count with the degree ceiling, noting that matching signs on their own prove nothing. . Worth 2 points.

Part C 7 points

Uses the monic case to reduce the candidates to integers dividing $5$ , and tests all four. . Worth 3 points.

Concludes that each trapped root is irrational, joining the failed list to part B's three roots. . Worth 2 points. needs an explanation, not just an answer

Separates what the trapping proves, existence and location, from what it cannot supply, the number itself. . Worth 2 points. needs an explanation, not just an answer
10. A crate that has to hold thirty cubic metres . 18 points. Question 10 of 10.

A rectangular crate is built to a single dial setting $x$ , measured in metres. Its height is $x$ metres, its base is $2x - 1$ metres wide and $x + 3$ metres long, and the crate is required to hold exactly $30$ cubic metres.
1. Part A.
  Express the crate's volume as a polynomial in $x$ in standard form. Then rearrange the thirty cubic metre requirement so that one side reads zero and every coefficient is a whole number, and say which values of $x$ describe a crate that could actually be built.
  
  Model the situation Name your unknown first, then write every other quantity in terms of that one letter. 6 points
2. Part B.
  Find the height, then give all three measurements in metres. Say also how large the permitted set of fractions is, and how much of it your part A restriction throws away.
  
  Carry your own answer forward Build the candidate list from the equation you wrote in part A, whatever it was, and prune it with your own restriction on $x$ .
  
  Solve and show your work Write each step out, and end with the value and its units. 6 points
3. Part C.
  Take the linear factor supplied by the height you found out of the equation, then use the polynomial that remains to decide whether a second value of $x$ could meet the requirement. Say what your decision settles about the crate.
  
  Carry your own answer forward Work from the height you reported in part B, whatever it was, and reason about the quotient that your own division leaves behind.
  
  Explain what it means Words, not just symbols. Say what the number is telling you about the situation. 6 points
The answer

Part A

$V(x) = 2x^3 + 5x^2 - 3x$ , so the requirement is $2x^3 + 5x^2 - 3x - 30 = 0$ . The crate exists only for $x > \tfrac{1}{2}$ , since the width $2x - 1$ must be positive.
Part B

The height is $x = 2$ metres, so the crate is $2$ m high on a base $3$ m by $5$ m. The theorem permits $24$ candidates, of which $11$ clear $x > \tfrac{1}{2}$ .
- the three measurements may be given in any order as $2$ m, $3$ m and $5$ m; a height of $\tfrac{5}{2}$ would give a volume of $55$ cubic metres, not $30$
Part C

Dividing leaves $2x^2 + 9x + 15$ , whose discriminant $81 - 120 = -39$ is negative, so it has no real roots. The equation therefore has $x = 2$ as its only real solution, and no second crate of these proportions holds $30$ cubic metres.
Worked solution

Part A

Volume is height times width times length, so multiply the three given measurements.

$V(x) = x(2x - 1)(x + 3) = x\left(2x^2 + 5x - 3\right) = 2x^3 + 5x^2 - 3x$

Setting that equal to $30$ and moving everything to one side gives the equation the theorem can read:

$2x^3 + 5x^2 - 3x - 30 = 0.$

Moving the $30$ across is what supplies a nonzero constant term. The volume polynomial itself has constant term $0$ , on which the numerator condition would say nothing.

For the restriction, all three measurements must be positive: $x > 0$ , $2x - 1 > 0$ and $x + 3 > 0$ . The middle one is the binding condition, so the situation allows exactly $x > \tfrac{1}{2}$ .

Part B

The constant term is $-30$ and the leading coefficient is $2$ , so numerators divide $30$ and denominators divide $2$ :

$\pm 1, \ \pm 2, \ \pm 3, \ \pm 5, \ \pm 6, \ \pm 10, \ \pm 15, \ \pm 30, \ \pm\tfrac{1}{2}, \ \pm\tfrac{3}{2}, \ \pm\tfrac{5}{2}, \ \pm\tfrac{15}{2}$

That is $12$ positive values and their negatives, so $24$ candidates. The restriction $x > \tfrac{1}{2}$ kills every negative one and $\tfrac{1}{2}$ itself, leaving $11$ .

Test the small survivors. At $x = 1$ the left side is $2 + 5 - 3 - 30 = -26$ , and at $x = 2$ :

$2(8) + 5(4) - 6 - 30 = 16 + 20 - 36 = 0.$

So the height is $2$ metres, the width is $2(2) - 1 = 3$ metres and the length is $2 + 3 = 5$ metres. The volume checks: $2 \times 3 \times 5 = 30$ cubic metres.

Part C

Run the tableau with corner $2$ on the coefficients $2,\; 5,\; -3,\; -30$ . The bottom row is $2,\; 9,\; 15,\; 0$ , the final $0$ re-confirming the root, so

$2x^3 + 5x^2 - 3x - 30 = (x - 2)\left(2x^2 + 9x + 15\right).$

A quadratic quotient ends the hunt. Its discriminant is

$9^2 - 4(2)(15) = 81 - 120 = -39 < 0,$

so it has no real roots at all, and in particular none above $\tfrac{1}{2}$ . The only real solution of the equation is $x = 2$ .

Read back into the situation, that is a uniqueness statement: the dial setting is forced. There is exactly one crate of these proportions with a volume of $30$ cubic metres, and no second setting, large or small, would have done. Note that the restriction $x > \tfrac{1}{2}$ was not even needed to reach it here, since the quotient offered no competition at all.

In one line

The volume is $V(x) = x(2x - 1)(x + 3) = 2x^3 + 5x^2 - 3x$ , so the requirement becomes $2x^3 + 5x^2 - 3x - 30 = 0$ with $x > \tfrac{1}{2}$ . The theorem permits $24$ candidates, $11$ of which clear the restriction, and $x = 2$ works: the crate is $2$ m high on a base $3$ m by $5$ m, a volume of $30$ cubic metres. Dividing out $x - 2$ leaves $2x^2 + 9x + 15$ , whose discriminant $81 - 120 = -39$ is negative, so there is no second real solution and the dial setting is unique.

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 6 points

Builds the volume as the product of the three measurements and expands it to standard form. . Worth 2 points.

Moves the $30$ across to produce a polynomial equation with integer coefficients and zero on one side. . Worth 2 points.

States $x > \tfrac{1}{2}$ and identifies the width as the measurement that binds. . Worth 2 points.

Part B 6 points

Builds the candidate list from the equation's two end coefficients, reduced and deduplicated. . Worth 2 points.

Prunes with the restriction and reports both counts, $24$ permitted and $11$ surviving. . Worth 2 points.

Reports all three measurements in metres and checks the volume against the requirement. . Worth 2 points.

Part C 6 points

Divides by the linear factor the root supplies and reports the quadratic quotient. . Worth 2 points.

Decides the quotient has no real roots, by the discriminant or an equivalent argument. . Worth 2 points.

Reads the result back as a statement about the crate, that the dial setting is unique. . 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. Two polynomials neither expanded nor drawn 17 points

Part A 5

Adds the factors' degrees, counting the squared factor twice, to reach degree $5$ . 2
Multiplies the factors' leading terms rather than their constants, reaching $2x^5$ . 2
Obtains the constant term as the product of the factors at $x = 0$ , keeping the sign. 1

Part B 6

Uses the parity of the degree and the sign of the leading coefficient to name both far ends. 2
Sets each factor to zero in turn and solves, including the fractional zero from $2x - 7$ . 2
Says why $x^2 + 5$ contributes no real zero, and counts $-3$ once rather than twice. 2

Part C 6

States the ceiling $n - 1$ and applies it to degree $6$ . 2
Argues the floor from the six distinct zeros, explaining why each neighbouring pair forces a turn. 3
Concludes that the two bounds coincide, so the count is exactly five rather than at most five. 1

2. The working was lost, the answer was not 17 points

Part A 5

Uses the identity $P = DQ + R$ rather than $P = DQ$ , so the remainder is added in. 2
Expands the product correctly, including the $x^2$ column that cancels to zero. 2
Says that $DQ + R$ is determined once $D$ , $Q$ and $R$ are given, so the dividend is not a choice. 1

Part B 6

States the stopping condition in terms of degree, naming the zero polynomial as its own case. 2
Gives a verdict for all four, with the degree that settles each. 2
Explains why matching the divisor's degree disqualifies a leftover, rather than only asserting the rule. 2

Part C 6

Reports the pair $Q = 0$ , $R = 4x - 9$ . 1
Checks the pair against both promises, the identity and the degree condition, rather than one of them. 2
Appeals to uniqueness to explain why keeping both promises is enough to be the answer. 3

3. One unknown coefficient, then one tableau 17 points

Part A 5

Evaluates at $2$ , the input that makes the divisor zero, rather than at $-2$ . 2
Sets the evaluation equal to $17$ and solves the resulting linear equation. 2
Reports $k$ as a single number and checks it against the stated remainder. 1

Part B 6

Puts $-1$ in the corner, flipping the sign of the divisor's constant. 2
Runs bring down, multiply, add across all four columns with correct signs. 2
Reads the bottom row as a quadratic quotient plus a separate remainder, and confirms by expanding. 2

Part C 6

Recognises exact division as remainder $0$ and sets the same evaluation equal to $0$ . 2
Solves for $k$ and answers the decision asked for rather than only computing. 2
Explains that only the right-hand side changed, and that a non-integer $k$ is legitimate here. 2

4. Built to a specification, then read back 18 points

Part A 6

Turns each root into the factor $x - r$ with the correct sign, including the negative root. 2
Attaches the scalar $6$ so that the leading coefficient is $6$ , rather than leaving a monic product. 2
Expands correctly to $6x^3 + 13x^2 - 41x + 12$ . 2

Part B 6

Names both far ends from the parity of the degree and the sign of the leading coefficient. 2
Identifies the $y$ -intercept as the constant term, or as the value at $0$ . 1
Pins the number of turning points between the degree ceiling and the floor the distinct zeros force. 3

Part C 6

Checks each clause of the specification separately and identifies the leading coefficient as the one that fails. 2
States that roots fix a polynomial only up to a nonzero constant multiple, with a reason rather than an assertion. 3
Names a condition that would have selected $Q$ : the leading coefficient $18$ , or a specific point on its graph. 1

5. What the list reaches, and what it cannot 17 points

Part A 5

Takes numerators from the constant term and denominators from the leading coefficient, not the reverse. 2
Produces all eight numbers, both signs included, reduced and without duplicates. 2
Tests candidates and reports $\tfrac{1}{2}$ as a root, with the evaluation that establishes it. 1

Part B 6

Divides by the factor the root supplies and reports a quadratic quotient, one degree lower. 2
Clears the fraction by moving the leading coefficient into the linear factor, giving $(2x - 1)$ . 1
Builds the quotient's own candidate list and tests every entry, rather than reusing the list from part A. 3

Part C 6

Computes all four values correctly. 2
Identifies both sign changes and names the interval each root lies in. 2
Says that the trap locates a root without naming it, and connects that to the failed candidate list. 2

6. One value, three questions 18 points

Part A 6

Evaluates all three polynomials at $4$ rather than attempting a division. 2
Gives the right verdict for all three cases. 2
Explains how a nonzero value rules a factor out, not only how a zero value establishes one. 2

Part B 6

Evaluates the given identity at $7$ instead of expanding $P$ first. 2
Computes $Q(7)$ and assembles $P(7)$ with the $+6$ retained. 2
Names the remainder on division by $x - 7$ as that same value, citing the Remainder Theorem. 2

Part C 6

Writes $2x - 8$ as a constant times $x - 4$ and rules that it is a factor. 2
Explains why a nonzero constant multiple of a divisor divides exactly the same polynomials, naming what happens to the quotient. 3
States the general condition for $ax - b$ as vanishing at $\tfrac{b}{a}$ . 1

7. A quadratic divisor, and what one division certifies 19 points

Part A 6

Forms each quotient term by dividing leading term by leading term, and subtracts the entire product each pass. 3
Reaches quotient $2x^2 + 3x - 4$ with remainder $0$ . 2
Checks the answer, by expanding $DQ$ or against the degree and leading-coefficient forecast. 1

Part B 7

Factors the divisor and names $2$ and $3$ as roots of $P$ . 2
Explains that a remainder of $0$ turns the identity into a product, which is what transfers the divisor's roots to $P$ . 3
Bounds the further roots by the quotient's degree and tests the quotient's candidate list to rule out rational ones. 2

Part C 6

States that a remainder of $0$ makes the identity a product valid at every input, which is the definition of a factor. 2
Says what a table of values of $P$ cannot supply, naming the cofactor, and why a divisor with no rational root leaves nothing to test. 2
Uses uniqueness to argue that a nonzero remainder rules a factorization out, not merely that a zero remainder establishes one. 2

8. Where the filter has nothing to grip 18 points

Part A 6

Says that every integer divides $0$ , so the numerator condition is vacuous here. 2
Factors out the largest power of $x$ to produce a cofactor with nonzero constant term. 2
Builds the cofactor's list correctly, reduced and deduplicated, with both signs. 2

Part B 6

Reports $0$ as a root, from the factor taken out in part A. 1
Tests candidates against the cofactor and peels off a factor when one succeeds. 3
Finishes the quadratic quotient and reports all four roots, noting that a quartic can have no more. 2

Part C 6

Produces a specific polynomial with zero constant term whose cofactor has no rational root. 2
Builds the cofactor's candidate list and evaluates every entry, showing the failure rather than asserting it. 2
Writes a corrected sentence that keeps the one-way direction the theorem proves. 2

9. Two ways to look for a root 20 points

Part A 6

Sets up each tableau with the corner holding the input and the top row holding every coefficient. 2
Produces all six values correctly, with signs handled at the negative input. 3
Names the Remainder Theorem as the entitlement to read a value of $f$ from a division. 1

Part B 7

Identifies all three sign changes and names the interval each root lies in. 3
Argues from unbrokenness that a sign change forces a zero, rather than asserting it. 2
Closes the count with the degree ceiling, noting that matching signs on their own prove nothing. 2

Part C 7

Uses the monic case to reduce the candidates to integers dividing $5$ , and tests all four. 3
Concludes that each trapped root is irrational, joining the failed list to part B's three roots. 2
Separates what the trapping proves, existence and location, from what it cannot supply, the number itself. 2

10. A crate that has to hold thirty cubic metres 18 points

Part A 6

Builds the volume as the product of the three measurements and expands it to standard form. 2
Moves the $30$ across to produce a polynomial equation with integer coefficients and zero on one side. 2
States $x > \tfrac{1}{2}$ and identifies the width as the measurement that binds. 2

Part B 6

Builds the candidate list from the equation's two end coefficients, reduced and deduplicated. 2
Prunes with the restriction and reports both counts, $24$ permitted and $11$ surviving. 2
Reports all three measurements in metres and checks the volume against the requirement. 2

Part C 6

Divides by the linear factor the root supplies and reports the quadratic quotient. 2
Decides the quotient has no real roots, by the discriminant or an equivalent argument. 2
Reads the result back as a statement about the crate, that the dial setting is unique. 2

Multiple choice

Free response