Graphing Polynomial Functions: Free Response

Read the degree off the exponents rather than by expanding: the three factors contribute $2 + 1 + 1$ powers of $x$, so the degree is $4$.

The leading coefficient is the product of every factor's leading term, and the middle factor contributes a $2$ that is easy to walk past:

$-(x)^2 \cdot (2x) \cdot (x) = -2x^4$

The degree is even and the leading coefficient is negative, so the two arms agree with each other and both point downward.

A zero comes from each factor set equal to zero, and the exponent on that factor is its multiplicity:

$x + 4 = 0, \qquad 2x - 3 = 0, \qquad x - 5 = 0$

So the zeros are $-4$, carrying the exponent $2$, and $\frac{3}{2}$ and $5$, each carrying the exponent $1$. The multiplicities add to $2 + 1 + 1 = 4$, which is the degree, as it must be: the factorization is complete, so there is nothing left over for a non-real root.

The zeros in increasing order are $-4$, $\frac{3}{2}$ and $5$, so the four intervals are $x < -4$, then $-4 < x < \frac{3}{2}$, then $\frac{3}{2} < x < 5$, then $x > 5$. Take $-5$, $0$, $2$ and $6$: each is strictly inside its interval and each keeps the arithmetic small.

$P(-5) = -(1)(-13)(-10) = -130, \qquad P(0) = -(16)(-3)(-5) = -240$

$P(2) = -(36)(1)(-3) = 108, \qquad P(6) = -(100)(9)(1) = -900$

Only the sign of each result is being asked for, so the sizes can be ignored entirely. Reading left to right the signs are negative, negative, positive, negative, and the one interval carrying $P(x) > 0$ is $\frac{3}{2} < x < 5$.

The value at $0$ is worth keeping: it is the y-intercept as well as a test value, so the curve passes through $(0, -240)$.

The graph of a polynomial is one unbroken curve. Suppose two points $u < v$ of the same interval gave values of opposite sign. The piece of curve joining them runs from below the axis to above it, and being unbroken it cannot leap the axis, so it meets the axis somewhere between:

$P(u)P(v) < 0 \ \Longrightarrow \ P(c) = 0 \ \text{ for some } c \text{ with } u < c < v$

But the interval was cut so that it contains no zero, so no such $c$ exists, and the supposition is impossible. Every point of the interval therefore carries the sign that one test value revealed. That is what makes a sign chart cheap: four evaluations settle the whole line.

A zero fails as a test value for a plainer reason. Feeding it in returns $0$, and $0$ is neither positive nor negative, so it reports no sign at all. Its position is no better than its value: it sits on the boundary between two intervals that may carry different signs, belonging to neither of them, so it could not speak for either one even if it had a sign to offer. A test value has to be strictly inside the interval it reports on.

Take each zero in turn and split $P$ into the factor that vanishes there and everything else. At $r = \frac{3}{2}$,

$P(x) = (2x-3) \cdot \big[ -(x+4)^2(x-5) \big]$

and the bracket is nonzero at $\frac{3}{2}$, so near that point it holds one sign while $2x - 3$ runs from negative to positive. The product therefore changes sign. The same argument at $5$ gives the same verdict: an odd multiplicity means the vanishing factor itself changes sign.

At $-4$ the vanishing factor is $(x+4)^2$, which is positive on both sides and zero only at $-4$ itself. The rest of the polynomial holds one sign nearby, so the product holds that same sign on both sides. The curve reaches the axis at $-4$ and returns the way it came.

So the sign changes at exactly the zeros of odd multiplicity, which here are $\frac{3}{2}$ and $5$.

The arms supply a check that used none of this. An even degree with a negative leading coefficient sends both arms downward, so $P$ has to be negative on the far left and on the far right:

$P(-5) = -130 < 0, \qquad P(6) = -900 < 0$

Both outer signs are negative, so the check passes. Had one of them come out positive, the two routes would be contradicting each other and one of them would carry an error to be found before anything is drawn.

There are two x-intercepts and the picture distinguishes them. At $x = -4$ the curve arrives from below the axis and continues above it, so it crosses. At $x = 1$ it arrives from above, reaches the axis, and returns above it, so it touches and turns back.

The arms are read separately from the intercepts. The left arm falls away downward and the right arm rises, so they point in opposite directions. A polynomial's two arms agree with each other exactly when the degree is even, so opposite arms mean the degree is odd.

The sign of the leading coefficient then comes from the right-hand arm alone, because far out on either side the polynomial follows its leading term:

$P(x) \ \text{follows} \ a_n x^n \ \text{when} \ |x| \ \text{is large}$

For large positive $x$ that term is positive exactly when $a_n > 0$. The right arm rises, so $a_n > 0$. Note that this reads only the arms: the intercepts played no part in it, which is what makes the two readings independent.

At a crossing the multiplicity is odd, so the smallest it can be is $1$. At a touch the multiplicity is even, so the smallest it can be is $2$. Those are the only two intercepts, and the multiplicities of the real zeros can never total more than the degree, so

$1 + 2 = 3$

is a floor for the degree, reached by $a(x+4)(x-1)^2$.

Now ask how a different polynomial could produce the identical picture. Only two things can change. A multiplicity could be larger, but it cannot change parity without changing the picture, since odd means crossing and even means touching, so it can only rise by $2$, or $4$, and so on. Or there could be roots that are not real, and those arrive in conjugate pairs because the coefficients are real, so they too add to the degree two at a time. Every route upward moves in twos, which leaves

$\deg P \in \{3, 5, 7, 9, \ldots\}$

and rules out $4$, $6$, $8$ and the rest of the even numbers.

That conclusion was reached without ever looking at the arms, so Part A is now an independent check on it: opposite arms mean an odd degree, and every degree on this list is odd. Two different readings of the same picture agree.

Write $P(x) = (x-r)^m G(x)$ with $G(r) \neq 0$. Near $r$ the factor $G$ barely moves, sitting close to the fixed number $G(r)$, so the height of the curve at a distance $d$ from the zero is about $|G(r)| \, d^m$. Take $d$ smaller than $1$ and each extra power of $m$ multiplies that height by another factor of $d$:

$d = 0.1: \qquad d^1 = 0.1, \quad d^2 = 0.01, \quad d^3 = 0.001$

So a larger multiplicity leaves the axis far more slowly, and the curve lies down against it before pulling away. The flatness at $-4$ is therefore evidence for a multiplicity of $3$ or more, and since the multiplicity at a crossing must be odd, $3$ is the next value up from $1$.

Evidence is not proof, and here is why. The height near the zero is $|G(r)| \, d^m$, a product of two things, and a picture reports only the product. A simple zero with a small $|G(r)|$ produces exactly the same heights as a triple zero with $|G(r)| = 1$:

$\tfrac{1}{100}(0.1) = 0.001 = (0.1)^3$

At this distance from the zero the two curves sit at the same height, and the drawing carries no scale on its vertical axis to separate them. Flatness measures the drawing as much as it measures the exponent.

If the suggestion is right and the multiplicity at $-4$ is $3$, the floor becomes $3 + 2 = 5$, still odd, so it remains consistent with the arms.

What no drawing can ever show is a non-real root. An x-intercept is a real number where the height is zero, and a root such as $2 + 5i$ is not a location on the x-axis at all. The picture is silent about how many such roots $P$ has, and they are precisely the reason the degree list in Part B runs on forever.

A product is zero exactly when one of its factors is zero, so set each factor to zero in turn. That gives $x = 1$, $x = 3$ and $x = 8$, and all three lie inside the range $0 \le x \le 9$. The exponent on each factor is that zero's multiplicity, so $3$ carries multiplicity $2$ while $1$ and $8$ carry multiplicity $1$.

In the language of the situation, $x$ counts hundreds of chairs, so the workshop breaks even at $100$, $300$ and $800$ chairs a week.

The report at $x = 0$ is the value of the model when nothing is made:

$P(0) = -(-1)(9)(-8) = -72$

Profit is measured in thousands of dollars, so that is a loss of $72$ thousand dollars, which is what the model reports for a week in which no chairs are made.

The three break-even levels cut the range into four stretches. Take one value from each, none of them a break-even level, and work it out from the factors:

$P(0) = -(-1)(9)(-8) = -72, \qquad P(2) = -(1)(1)(-6) = 6$

$P(5) = -(4)(4)(-3) = 48, \qquad P(9) = -(8)(36)(1) = -288$

So the model is negative on $0 \le x < 1$, positive on $1 < x < 3$, positive again on $3 < x < 8$, and negative on $8 < x \le 9$. Two of those four values, $x = 0$ and $x = 9$, are the ends of the range rather than points inside it, and that is legitimate here: what a test value has to avoid is a zero, and neither end is one.

One point needs care, because the question asks for a strict inequality. The profit at $x = 3$ is exactly $0$, which is breaking even and not making a profit, so $x = 3$ has to be left out even though the stretches on both sides of it are profitable. The profitable set is $1 < x < 8$ with the single level $x = 3$ removed.

The outer signs are worth one glance before moving on. The degree is $4$ and the leading coefficient is $-1$, so both arms point downward, and both outer stretches came out negative. The two readings agree.

The level $x = 3$ carries multiplicity $2$. Write the model as

$P(x) = (x-3)^2 \cdot \big[ -(x-1)(x-8) \big]$

The squared factor is positive on both sides of $3$ and the bracket is positive nearby, so the profit is positive on both sides and exactly zero at $3$ itself. In the workshop's terms: as production approaches $300$ chairs a week from either direction the weekly profit shrinks away to nothing, and at $300$ it is exactly zero. The business has a bad week at exactly that output and is in profit on either side of it. It never falls into loss there. (How the profit moves in between is a separate question, and not one this work answers: the sign is settled, the shape is not.)

If that level were instead a zero of multiplicity $1$, the odd exponent would change the sign of the model as $x$ passed $3$. The story would then be quite different: production of a little under $300$ chairs and production of a little over $300$ chairs would sit on opposite sides of break-even, one profitable and one loss-making, and $300$ would be the exact tipping point between them.

That contrast is the whole content of the sign rule. A change of sign always requires a break-even level, but a break-even level does not always deliver a change of sign, and the multiplicity is what decides which kind this one is.

The candidate list needs both end coefficients. Numerators divide the constant term $-18$ and denominators divide the leading coefficient $3$:

$p \in \{1, 2, 3, 6, 9, 18\}, \qquad q \in \{1, 3\}$

so the candidates are $\pm 1, \pm 2, \pm 3, \pm 6, \pm 9, \pm 18, \pm\frac{1}{3}, \pm\frac{2}{3}$. Spend the cheap ones first: $P(1) = 24$, $P(-1) = -64$, $P(2) = 20$, and then

$P(3) = 243 - 351 - 45 + 171 - 18 = 0$

Divide it out. Synthetic division by $3$ on the coefficients $3, -13, -5, 57, -18$ gives $3, -4, -17, 6$ with remainder $0$, so the quotient is $3x^3 - 4x^2 - 17x + 6$.

Now the step that measures multiplicity: divide the QUOTIENT by $3$ again, not the original. On $3, -4, -17, 6$ it gives $3, 5, -2$ with remainder $0$, so $(x-3)$ divides a second time:

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

A third division by $3$ would settle whether it divides again. On $3, 5, -2$ it gives $3, 14$ with remainder $40$, which is not zero, so the multiplicity of $3$ is exactly $2$ and no more. The quadratic left over factors:

$3x^2 + 5x - 2 = (3x-1)(x+2)$

So $P(x) = (3x-1)(x+2)(x-3)^2$. The zeros are $\frac{1}{3}$ and $-2$, each simple, and $3$ with multiplicity $2$, and $1 + 1 + 2 = 4$ accounts for the whole degree, so there are no non-real roots.

The leading term is $3x^4$: the degree is even and the leading coefficient is positive, so both arms rise. The y-intercept is the constant term, $P(0) = -18$.

The distinct real zeros are $-2$, $\frac{1}{3}$ and $3$, three cuts making four intervals. Evaluate from the factored form, which is far cheaper than the expanded one, at $-3$, $0$, $1$ and $4$:

$P(-3) = (-10)(-1)(36) = 360, \qquad P(0) = (-1)(2)(9) = -18$

$P(1) = (2)(3)(4) = 24, \qquad P(4) = (11)(6)(1) = 66$

So the signs run positive, negative, positive, positive.

Two checks come free. The value at $0$ is the y-intercept, and it agrees with the constant term of the expanded polynomial, so the factorization multiplies back correctly. And both outer signs are positive, which is exactly what two rising arms demand.

A single exact division establishes one thing and no more. A remainder of $0$ when dividing by $(x-3)$ says that $3$ is a zero, which is the Factor Theorem, and it is silent about how many times the factor comes out. Every zero, simple or repeated, produces that first remainder of $0$.

The second division is what measures. Dividing the quotient by $(x-3)$ again and getting $0$ shows that

$P(x) = (x-3)^2 G(x)$

for a polynomial $G$, so the multiplicity is at least $2$. The third division is what closes it: its remainder was not zero, so $G(3) \neq 0$ and the multiplicity is exactly $2$, not $3$ or more.

The sign row confirms part of that from the other side, and it is worth being exact about which part. An even multiplicity means $(x-3)^2$ is positive on both sides of $3$, so $P$ carries the sign of $G(3)$ on both sides and cannot change sign there. Part B found $P$ positive at $1$ and positive at $4$, which straddle $3$ and lie clear of the other two zeros, so no change of sign is exactly what an even multiplicity predicts.

Read backwards, though, that argument delivers only the PARITY. A multiplicity of $4$ would leave the sign unchanged just as convincingly, and so would $6$. The sign row therefore establishes that the multiplicity is even, and so at least $2$; what pins it at exactly $2$ is the third division and its nonzero remainder. The rule that the sign changes at a zero exactly when the multiplicity is odd is sound in both directions, but what it trades in is parity, never size.

The two routes are worth keeping separate in your head, because they check each other. Division is algebra on the coefficients; the sign row is arithmetic on values. When they disagree, one of them carries a slip, and finding it is cheaper than redrawing a wrong sketch.

Start with the count. The Fundamental Theorem of Algebra gives a cubic exactly three roots counted with multiplicity, and the student's list accounts for only two. The missing root cannot be non-real, because the coefficients are real and non-real roots arrive in conjugate pairs, so a single leftover root would have nobody to pair with:

$3 - (1 + 1) = 1, \qquad \text{and } 1 \text{ is odd}$

So there is a third REAL root: either a zero the search missed, or one of the two found is repeated. Either way the list as written is incomplete.

The second reason never mentions the degree. The student calls $5$ a zero of multiplicity $1$, and an odd multiplicity forces a change of sign there, since $(x-5)$ itself changes sign while the rest of the polynomial holds one sign nearby. Yet the same page reports $P$ positive on $-3 < x < 5$ and positive on $x > 5$, which is no change of sign at all. The multiplicity claim and the sign row contradict each other, and one of them has to give way.

Notice what is NOT wrong. The four evaluations are all correct, and so is the end behaviour: a cubic with a positive leading coefficient falls on the left and rises on the right, and the row does start negative and end positive. The error is structural, not arithmetic, which is exactly why recomputing the same four values would never have found it.

The candidate list was built from the constant term alone. Denominators divide the leading coefficient $3$, so $\pm\frac{1}{3}, \pm\frac{2}{3}, \pm\frac{5}{3}, \pm\frac{10}{3}$ belong on it too, and the student never tested any of them.

There is a cheaper repair than testing more candidates: divide out the zeros already found. Synthetic division by $-3$ on $3, -8, -41, 30$ gives $3, -17, 10$ with remainder $0$, and dividing that quotient by $5$ gives $3, -2$ with remainder $0$:

$P(x) = (x+3)(3x^2 - 17x + 10) = (x+3)(x-5)(3x-2)$

A cubic has only three linear factors, and after two divisions what is left is linear, so it hands over its own zero with no list at all: $3x - 2 = 0$ gives $x = \frac{2}{3}$. All three zeros are simple.

Three zeros make four intervals, not three, and the missing zero sits inside the stretch the student treated as one piece. Test $-4$, $0$, $1$ and $6$:

$P(1) = 3 - 8 - 41 + 30 = -16$

with the other three values already on the page: $-126$, $30$ and $144$. So the corrected row runs negative, positive, negative, positive.

Every value the student computed survives. What failed was the assumption behind them: the guarantee that one test value settles an interval covers only an interval holding no zero, and $\frac{2}{3}$, a zero of odd multiplicity, was sitting inside theirs the whole time.

Suppose the missed zero is $c$, of multiplicity $2$, and write $P(x) = (x-c)^2 G(x)$. The squared factor is positive on both sides of $c$, so $P$ carries the sign of $G$ across the whole neighbourhood: passing $c$ changes nothing.

Follow that through the check line by line. The student's test values are correct, since the arithmetic is unaffected. Their intervals are too coarse, but the one holding $c$ carries a single sign on either side of $c$ and vanishes only at $c$ itself, so whatever value they tested reports that sign correctly. There is one accident that would betray the zero, a test value landing exactly on $c$ and returning $0$, but nothing in the method makes that happen, which is why the check cannot be relied on here even though it can get lucky. The sign changes in their row sit at exactly the odd zeros they listed, matching the multiplicities they assigned. And the outer signs still match the arms. Write $k$ for the total multiplicity they did account for. The missed zero contributes two, and any non-real roots come in pairs, so

$\deg P - k \ \text{is even}$

The parity of the degree is therefore untouched, and parity is exactly what the relative direction of the two arms reports. (The arms report one thing more, the sign of the leading coefficient, and that is untouched as well.)

The root count raises no objection either. The roots left unaccounted for are even in number, so they might be a conjugate pair of non-real roots, or the missed double zero, or both, and none of those is a contradiction.

So every line of the check passes, and the sketch is still wrong: it is missing an intercept where the curve meets the axis and turns back without crossing. Which side it turns back on depends on the sign of the rest of the polynomial near $c$, which the hypothetical never fixes. The lesson generalizes. A sign-based check can only ever notice a zero that changes the sign, so a missed zero of even multiplicity is invisible to it, while an odd one at least stands a chance of announcing itself. Passing the check means the pieces of work agree with one another, not that nothing is missing. Completeness has to come from somewhere else: a degree count with every root accounted for.