Polynomials and Their Graphs: Free Response

Nothing can be read off the polynomial as written, because two of its six terms are like terms of degree $6$ and two more are like terms of degree $3$. Collect them first.

The two sixth-degree terms are $2x^6$ and $-2x^6$, and they wipe each other out. The two cubic terms are $-8x^3$ and $4x^3$, which combine to $-4x^3$. The remaining terms, $-x$ and $5$, have nothing to pair with, so the tidied polynomial, with exponents descending, is

$q(x) = -4x^3 - x + 5.$

Now every name is safe to read. The highest surviving power is $x^3$, so the degree is $3$ and the leading term is $-4x^3$, making the leading coefficient $-4$, sign included. The constant term is $5$, again sign included, and three terms survive, so this is a cubic trinomial.

The trap that the housekeeping avoids is worth naming. Read before tidying, the expression looks like a polynomial of degree $6$ with leading coefficient $2$, and both of those readings are wrong: once the sum is collected, the sixth-degree terms are not there at all. A degree is a property of the polynomial, not of the way somebody happened to write it down.

Three separate rules do the three jobs, and none of them needs the expansion.

Degrees add across a product, and a squared factor is that factor used twice. The first factor has degree $2$, and the second contributes its degree $3$ twice over, so

$\deg P = 2 + 3 + 3 = 8.$

Leading coefficients multiply, so build the leading term out of the leading pieces alone. The first factor leads with $3x^2$. The factor $2 - x^3$ leads with $-x^3$, and squaring it gives $(-x^3)^2 = x^6$, whose coefficient is $+1$, because the square of a negative is positive. Multiplying the leading pieces,

$(3x^2)(x^6) = 3x^8,$

so the leading coefficient is $3$.

The constant term is the value at $x = 0$, since every other term carries an $x$ and dies there. Each factor contributes its own value at $0$:

$P(0) = (5)(2)^2 = 20.$

So the trinomial factor supplied the $3$ and the $5$, and the repeated factor supplied the $x^6$ and the $4$. The expansion, which would open out into nine products before anything was collected, was never needed.

Rewrite each expression so that what is being done to $x$ is visible, then apply one test.

Dividing by the number $3$ is the same as multiplying every term by $\tfrac{1}{3}$:

$\frac{x^4 - 6x}{3} = \tfrac{1}{3}x^4 - 2x.$

The exponents on $x$ are $4$ and $1$, both whole numbers, so this is a polynomial. The fraction sits exactly where a fraction is allowed, on a coefficient.

The second expression divides by the variable itself. Written as a term it is $3x^{-4}$, and $-4$ is not a whole number, so the expression is not a polynomial. Notice what is at fault: not the $3$ on top, but the variable underneath.

The third looks worse than it is. Its coefficients are $\sqrt{7}$, $-1$ and $\tfrac{1}{2}$, and its exponents on $x$ are $2$, $1$ and $0$. The square root is applied to the number $7$, never to $x$, so nothing is wrong at all: this is a quadratic polynomial that happens to carry an irrational leading coefficient.

The fourth reverses the two roles the definition cares about. In $4^x$ the constant is the base and the variable is the exponent, which no term of the form $c\,x^k$ can imitate. Its second term $-x^4$ is perfectly legal, but one illegal term is enough to disqualify the whole sum.

So one rule decided all four: a polynomial is a finite sum of terms $c\,x^k$ in which every $k$ is a whole number. The rule polices the exponents on the variable and nothing else, which is why a coefficient may be a fraction, a negative number or an irrational number without any difficulty whatever.

One term of each polynomial decides everything asked for, so the work is finding that term.

For $f$ the terms are merely out of order. The highest power present is $x^8$, and the coefficient sitting on it is $-1$:

$f(x) = -x^8 + 12x^5 + 400.$

The degree $8$ is even, so the two ends agree with each other, and the leading coefficient $-1$ is negative, so what they agree on is falling. Both ends fall.

For $g$, degrees add and leading coefficients multiply, so read the factors one at a time. The factor $(3 - x)^2$ has degree $2$ and leads with $(-x)^2 = x^2$; the factor $2x + 1$ has degree $1$ and leads with $2x$; the factor $x^2 + 5$ has degree $2$ and leads with $x^2$. Adding the degrees and multiplying the leading pieces,

$\deg g = 2 + 1 + 2 = 5, \qquad (x^2)(2x)(x^2) = 2x^5.$

The degree $5$ is odd, so the ends disagree, and the leading coefficient $2$ is positive, so the far right rises and the far left falls.

Notice what played no part: the constant $400$ in $f$, and the $+5$ inside a factor of $g$. Both matter near the origin, where they can lift or drop the curve, and both are invisible at the ends.

The classmate has compared the wrong quantities. A coefficient scales a term, but the exponent governs how fast the term grows, and past some input a faster growth overtakes any fixed multiple of a slower one. Comparing $12$ with $1$ answers a question about small inputs, not about the far end.

Evaluate at the two inputs and watch the handover, taking each power before multiplying:

$f(2) = 12(32) - 256 + 400 = 528,$

$f(3) = 12(243) - 6561 + 400 = -3245.$

The outputs are positive at $2$ and negative at $3$, which already contradicts a graph rising away to the right.

One rearrangement shows that nothing can turn it back around. Factor $x^5$ out of the two terms that carry it:

$f(x) = x^5(12 - x^3) + 400.$

Once $x \ge 3$ the bracket is at most $12 - 27 = -15$, so the first piece is at most $-15x^5 \le -3645$, and adding $400$ still leaves $f(x) \le -3245$. The bound only worsens as $x$ grows, since $x^5$ and $x^3 - 12$ both increase, so $f$ is negative from $x = 3$ onward and its size passes every bound. The far right falls.

A bonus falls out of the two evaluations. A polynomial graph is unbroken, and $f$ is positive at $2$ and negative at $3$, so $f$ has a real zero somewhere between those two inputs.

The first claim is the sign-change principle applied to the two ends.

Falling on the far left means the outputs drop below any bound out there, so there is an input $a$ with $f(a) < 0$. Rising on the far right means the outputs pass any bound, so there is an input $b > a$ with $f(b) > 0$. That gives

$f(a) < 0 < f(b),$

and the graph of a polynomial is unbroken, with no gap or jump available, so it cannot get from below the axis to above it without touching the axis somewhere between $a$ and $b$. At that input the value is $0$. This is also why every odd-degree polynomial has a real zero: an odd degree is exactly what makes the two ends disagree.

Beyond that one zero the ends guarantee nothing, and two cubics settle it. Take

$p(x) = x^3 + 9x = x(x^2 + 9).$

A product is zero only when one of its factors is, and $x^2 + 9$ is at least $9$ for every real $x$, so the only real zero is $x = 0$: exactly one. Now take

$s(x) = (x - 1)(x - 4)(x + 2),$

whose real zeros are $1$, $4$ and $-2$: exactly three, and no more, since a cubic can have at most three.

Both are cubics with a positive leading coefficient, $p$ leading with $x^3$ and $s$ leading with $(x)(x)(x) = x^3$, so both graphs fall on the far left and rise on the far right while their zero counts differ. The ends fix the parity of the degree and a floor of one zero. How many zeros there actually are is decided in the middle of the polynomial, which the ends never see.

Build the three dimensions of the box one at a time.

The depth is the height of a folded flap, which is the side of the square that was cut away, so it is $x$. The length runs along the $24$ cm side of the sheet, and a square of side $x$ has been removed from each of its two ends, so it is $24 - 2x$. The width runs along the $15$ cm side and loses $x$ at each end in the same way, so it is $15 - 2x$. Volume is the product of the three dimensions:

$V(x) = x(24 - 2x)(15 - 2x).$

Subtracting $2x$ rather than $x$ is where this construction is most often lost: each side of the sheet meets two corners.

Now read the polynomial off the factors. Each factor has degree $1$, so the degrees add to $3$. The leading coefficients multiply, and they are $1$, $-2$ and $-2$:

$(x)(-2x)(-2x) = 4x^3.$

The constant term is the value at $x = 0$, and the first factor is $0$ there, so the constant term is $0$. In the language of the card, cutting nothing leaves a flat sheet with no volume.

Finally the window. Every dimension of a real box is positive, so all three factors must be positive: $x > 0$ from the first, $x < 12$ from the second and $x < 7.5$ from the third. The strictest of them wins, so, with $x$ measured in centimetres, the box exists exactly when

$0 < x < 7.5.$

Each value is one multiplication of three numbers, taken straight from the factored form:

$V(1) = (1)(22)(13) = 286,$

$V(2) = (2)(20)(11) = 440,$

$V(4) = (4)(16)(7) = 448,$

$V(5) = (5)(14)(5) = 350.$

A volume of exactly $400$ is a zero of $W(x) = V(x) - 400$, which is again a polynomial and so has an unbroken graph of its own. Subtracting $400$ from each value,

$W(1) = -114, \quad W(2) = 40,$

$W(4) = 48, \quad W(5) = -50.$

Between $x = 1$ and $x = 2$ the sign changes from negative to positive, so $W$ takes the value $0$ somewhere in between: some cut between $1$ cm and $2$ cm gives a box of exactly $400$ cubic centimetres. Between $x = 4$ and $x = 5$ the sign changes back from positive to negative, trapping a second such cut, so those two intervals are the answer.

Be exact about what has been shown. The sign change proves that a cut exists inside each of those two intervals; it does not name either one, and no arithmetic here comes close to naming them. Nor does it prove that an interval holds only one such cut, only that it holds at least one, and it says nothing at all about the gap from $2$ to $4$, which these four values never probe. What makes the two conclusions airtight is unbrokenness: a curve that is below the target at one input and above it at another, with no jump available, has to pass through the target on the way.

Subtracting a constant changes no exponent, so

$W(x) = V(x) - 400 = 4x^3 - 78x^2 + 360x - 400$

is a polynomial of degree $3$, and a polynomial of degree $3$ has at most $3$ real zeros. At most three different values of $x$, admissible or not, can therefore give a volume of $400$.

Now the two evaluations, again from the factored form, where both bracketed factors have gone negative:

$V(13) = (13)(-2)(-11) = 286,$

$V(14) = (14)(-4)(-13) = 728.$

That $V(13)$ lands on the same $286$ as $V(1)$ is a coincidence of these two inputs and nothing more; the two have no arithmetic in common.

So $W(13) = -114$ and $W(14) = 328$: the sign changes once more, and a third solution of $V(x) = 400$ lies between $13$ and $14$. Together with the two trapped between $1$ and $5$, that is three solutions, which is exactly the ceiling, so the equation has no others anywhere on the real line.

That third solution is sound mathematics and impossible card. For $x$ between $13$ and $14$ the expression $24 - 2x$ is negative and so is $15 - 2x$, so the two horizontal dimensions are both negative and their product is positive, which is the only reason the formula returns a plausible-looking volume at all. The physical objection is not that a $13$ cm square is too big for the sheet, since one such square would fit; it is that TWO of them have to come out of the $15$ cm side, and $2(13)$ already exceeds $15$.

The general lesson is about the gap between a polynomial and a model. The polynomial is defined for every real number, as every polynomial is, while the model means something only on $0 < x < 7.5$. Outside that window the arithmetic keeps working perfectly and the meaning stops, which is why an answer from a model always has to be read back against the situation that produced it.

Take the clues one at a time and keep the strongest bound.

Three different crossings mean three different real zeros, and a polynomial of degree $n$ has at most $n$ real zeros, so

$n \ge 3.$

The turning points ask for more. A degree-$n$ graph has at most $n - 1$ turning points, so four of them need $n - 1 \ge 4$, that is

$n \ge 5,$

and that is the binding clue. The zeros supply a weaker version of the same demand: between each consecutive pair of the three zeros the graph leaves the axis and comes back, so it must turn at least once there, which gives only $2$ turning points and only $n \ge 3$ again.

Now bring the ends in, where they confirm rather than extend. Falling on the left and rising on the right means the ends disagree, and the ends disagree exactly when the degree is odd; $5$ is odd, so nothing has to move. The turns carry their own parity check: ends that disagree mean the graph finishes travelling the way it started, which takes an even number of turns, and $4$ is even. The crossing count agrees as well, since three crossings flip the sign three times and so leave the two ends pointing opposite ways, exactly as described.

A bound is worth having only if it is attained, so here is a quintic fitting every word of the description:

$w(x) = x(x + 2)(x - 1)\left[(x - 5)^2 + 1\right].$

Its degree is $1 + 1 + 1 + 2 = 5$ and its leading term is $x^5$, so it falls on the left and rises on the right. The bracket is at least $1$ for every real $x$, so the only zeros are $-2$, $0$ and $1$: three of them. A short table shows the rest:

$w(-3) = -780, \quad w(-1) = 74, \quad w(2) = 80,$

$w(3) = 150, \quad w(4) = 144, \quad w(6) = 480.$

For $0 < x < 1$ the factors $x$ and $x + 2$ are positive while $x - 1$ is negative and the bracket is positive, so $w$ is negative across that whole stretch. Reading left to right, then, $w$ climbs from $-780$ to $74$, drops below the axis, climbs to $150$, drops to $144$, and climbs again: five runs, so at least four turning points, and a quintic can have at most four, so it has exactly four. The three sign changes on the way, at $-2$, at $0$ and at $1$, are the three crossings. The description is realised at degree $5$, so the least possible degree is $5$.

Two independent facts each rule the description out, and finding both is the point: a description can fail in more than one way.

First the turning count. Trace the curve from far left to far right. Both ends rise, so out on the left the curve comes down towards the middle, which is to say it is falling as you travel rightward, and out on the right it is climbing away, which is to say it is rising. Every turning point flips between falling and rising, and nothing else does, so the number of turning points must be odd. Two is even, and the description is already impossible.

Second the crossings. Between consecutive crossings the graph keeps one sign, since a change of sign on an unbroken curve would produce another meeting with the axis. So the sign flips at each crossing, and writing $s$ for the sign far to the left, three crossings run

$s \ \to \ -s \ \to \ s \ \to \ -s,$

leaving the far-right sign opposite to the far-left one. But both ends rising means the outputs are large and positive at both ends, so the two signs are the same. The description contradicts itself a second time.

Either contradiction is fatal on its own, and they are not the same objection: the first counts turns against the parity that the ends impose, the second counts crossings against the ends themselves. A described graph can be tested from more than one direction, and one that survives the first test may still fail the second.

Write $f$ for the polynomial, which has degree at least $1$, and take the main case $k \ge 1$ first, listing the inputs where the graph meets the axis in increasing order,

$c_1 < c_2 < \cdots < c_k.$

These are the only inputs at which $f$ is $0$. They cut the number line into $k + 1$ stretches: everything to the left of $c_1$, the pieces between consecutive meeting points, and everything to the right of $c_k$. On any one stretch $f$ is never $0$, and it cannot change sign there either, because an unbroken graph passing from negative to positive would have to touch the axis in between, producing a meeting point that is not on the list. So $f$ keeps a single sign on each stretch.

Crossing at $c_i$ says exactly that the sign on the stretch after $c_i$ is the opposite of the sign on the stretch before it. Writing $s$ for the sign on the leftmost stretch, the signs therefore run

$s, \ -s, \ s, \ -s, \ \ldots$

and after $k$ flips the rightmost stretch carries $s$ when $k$ is even and $-s$ when $k$ is odd.

Now connect signs to ends. An end that rises has outputs growing past every bound, so $f$ is positive far along it; an end that falls has $f$ negative far along it. The far-left sign is the sign on the leftmost stretch and the far-right sign is the sign on the rightmost one, so the two ends point the same way exactly when those two signs agree, and the count of flips just made shows that this happens exactly when $k$ is even. Both directions come from the same count, so the claim holds as stated.

The case $k = 0$ needs a separate sentence and behaves the same way. There is nothing to flip, so the single stretch is the whole line and $f$ keeps one sign everywhere; the two ends therefore agree, matching the claim, since $0$ is even. The polynomial $x^2 + 1$ is the standard instance, positive throughout with both ends rising.

For the consequence, the ends agree exactly when the degree is even. So an even $k$ forces an even degree and an odd $k$ forces an odd degree: the parity of the crossing count and the parity of the degree always match. What is not forced is the degree itself. Beyond that parity, the crossings give only the bound $n \ge k$, because a graph may meet the axis far fewer times than its degree permits: the graph of $x^9$ crosses once, and its crossing count and its degree are both odd, exactly as proved.

At $x = 0$ every term except the constant one carries a factor of $x$ and therefore vanishes:

$f(0) = 8.$

That is the constant term itself, so the graph crosses the $y$-axis at $(0, 8)$.

At $x = 1$ every power of $1$ is $1$, so each term contributes nothing but its own coefficient and the value is the sum of the coefficients. Minding that the missing $x^3$ term contributes a coefficient of $0$,

$f(1) = 3 - 7 + 0 + 1 - 6 + 8 = -1.$

At $x = -1$ the powers alternate in sign: $(-1)^5 = -1$, $(-1)^4 = 1$, $(-1)^2 = 1$ and $(-1)^1 = -1$. Substitute in parentheses and take each power before multiplying:

$f(-1) = 3(-1) - 7(1) + 1 - 6(-1) + 8,$

$f(-1) = -3 - 7 + 1 + 6 + 8 = 5.$

The term $-6x$ is the one that catches people out: at $x = -1$ it contributes $+6$, not $-6$.

Write the polynomial in full, with $a_k$ the coefficient sitting on $x^k$:

$f(x) = a_nx^n + a_{n-1}x^{n-1} + \cdots + a_1x + a_0.$

Evaluating at $1$ gives the sum of the coefficients, since every power of $1$ is $1$. Evaluating at $-1$ gives the same list of coefficients with a sign attached by the parity of the exponent, since $(-1)^k$ is $1$ for even $k$ and $-1$ for odd $k$:

$f(1) = a_n + a_{n-1} + \cdots + a_1 + a_0,$

$f(-1) = (-1)^na_n + (-1)^{n-1}a_{n-1} + \cdots - a_1 + a_0.$

Compare the two lists coefficient by coefficient. A coefficient $a_k$ with $k$ even appears as $+a_k$ in both, so it contributes $2a_k$ to the sum and $0$ to the difference. A coefficient with $k$ odd appears as $+a_k$ in the first and $-a_k$ in the second, so it contributes $0$ to the sum and $2a_k$ to the difference. Gathering the contributions,

$f(1) + f(-1) = 2(a_0 + a_2 + a_4 + \cdots),$

$f(1) - f(-1) = 2(a_1 + a_3 + a_5 + \cdots),$

which are the two claims. Nothing about the degree or the size of the coefficients was used, so both hold for every polynomial.

Now the check on $g$, whose coefficients are $a_4 = 2$, $a_3 = -5$, $a_2 = 0$, $a_1 = 1$ and $a_0 = -6$. The two evaluations are

$g(1) = 2 - 5 + 0 + 1 - 6 = -8,$

$g(-1) = 2 + 5 + 0 - 1 - 6 = 0.$

The even-power coefficients are $2$, $0$ and $-6$, summing to $-4$, and the odd-power ones are $-5$ and $1$, also summing to $-4$. The identities predict a sum of $2(-4) = -8$ and a difference of $2(-4) = -8$, and indeed $g(1) + g(-1) = -8$ and $g(1) - g(-1) = -8$. Both check out, and that the two happen to agree here is an accident of this particular $g$.