The Factor Theorem: Free Response

Each candidate factor names the number you are supposed to try: the one that makes that factor equal to zero. For $x - 2$ that number is $2$, and for $x + 2$ it is $-2$, since $x + 2 = x - (-2)$.

Start with $x - 2$ and evaluate at $2$:

$P(2) = 8 + 4(4) - 7(2) - 10 = 8 + 16 - 14 - 10 = 0$

The evaluation is zero, so $x - 2$ is a factor of $P(x)$, and the factorization $P(x) = (x - 2)\,Q(x)$ exists whether or not you go on to compute $Q$.

Now $x + 2$, evaluated at $-2$. Watch the odd power: $(-2)^3 = -8$, while $(-2)^2 = 4$.

$P(-2) = -8 + 4(4) - 7(-2) - 10 = -8 + 16 + 14 - 10 = 12$

That is not zero, so $x + 2$ is not a factor. Notice how little was needed for the second verdict. You did not divide, and you did not have to inspect the quotient for anything: one number decided it.

The sign is the whole trap here. Evaluating $P$ at $2$ a second time would answer the question you already answered, and evaluating at $-2$ to test $x - 2$ answers a question nobody asked. Write the candidate as $x - c$ before you substitute, and $c$ is the number you want.

This is the theorem read in the direction that costs nothing. A factorization is an identity, true at every value of $x$, so you may evaluate it wherever it is convenient. Choose the values that switch one of the displayed factors off.

At $x = 6$ the first factor vanishes and takes the whole product with it:

$R(6) = (6 - 6)\bigl(2(6) + 3\bigr)S(6) = 0 \cdot 15 \cdot S(6) = 0$

The second factor is not monic, so its root is not the number written inside it. Solve $2x + 3 = 0$ to get $x = -\frac{3}{2}$, and evaluate there:

$R\!\left(-\frac{3}{2}\right) = \left(-\frac{3}{2} - 6\right)(0)\,S\!\left(-\frac{3}{2}\right) = 0$

So $6$ and $-\frac{3}{2}$ are roots of $R$, guaranteed, without knowing a single coefficient of $R$ or anything at all about $S$.

What the form does not do is close the list. $S$ is unknown, and any root of $S$ is also a root of $R$, since the product is zero wherever any factor is. If $S(x) = x^2 - 1$ then $R$ picks up the extra roots $1$ and $-1$; if $S$ is the constant $5$ it picks up none; and $S$ may even vanish at $6$ again. So the displayed factors give a guarantee in one direction only. They certify two roots and are silent about everything else, and reading such a form as a complete root list is a habit worth breaking now, before the polynomials get longer.

The theorem splits into two implications pointing opposite ways. The first, call it (i), reads

$P(c) = 0 \implies x - c \text{ is a factor of } P(x),$

and the second, (ii), reads

$x - c \text{ is a factor of } P(x) \implies P(c) = 0.$

Statement (i) is the manufacturing direction. It starts from the division identity $P(x) = (x - c)\,Q(x) + P(c)$ and observes that a zero value collapses it to $P(x) = (x - c)\,Q(x)$, which is what being a factor means. In part A, $P(2) = 0$ fed straight into (i) and returned the factor $x - 2$.

Statement (ii) is the direction that produced the refusal, but it did not produce it directly. Read as it stands, (ii) begins by assuming something is a factor, and in part A nobody was willing to assume that about $x + 2$. What was used is its contrapositive.

An implication and its contrapositive say the same thing. "If it is a factor then the evaluation is zero" and "if the evaluation is not zero then it is not a factor" are the same claim expressed from opposite ends, so proving either one proves the other, and no extra work is owed:

$P(c) \neq 0 \implies x - c \text{ is not a factor of } P(x)$

That is how an implication produces a refusal: not by being reversed, which would be a different and unproved claim, but by being read backwards through its negations, which is free. Since $P(-2) = 12$ is not zero, $x + 2$ cannot be a factor, and no amount of dividing would change that.

The reason a single evaluation is always enough is now visible. Whatever number $P(c)$ turns out to be, it is either zero or it is not. If it is zero, (i) fires and there is a factor. If it is not, the contrapositive of (ii) fires and there is not. A one-directional theorem would leave one of the two outcomes uncovered, and you would be left dividing to find out. Both directions together are what make the test decisive.

Confirming a given root costs one evaluation and is worth doing, because every later step inherits the mistake if the root is wrong:

$P(3) = 81 - 54 - 225 + 78 + 120 = 0$

So $3$ is a root, and the factor theorem promises $P(x) = (x - 3)\,Q(x)$ for some cubic $Q$. Divide by $x - 3$ to find it:

$\begin{array}{c|ccccc} 3 & 1 & -2 & -25 & 26 & 120 \\ & & 3 & 3 & -66 & -120 \\ \hline & 1 & 1 & -22 & -40 & 0 \end{array}$

The last entry is the remainder, and it is zero, which confirms the root a second time. The entries before it are the coefficients of the quotient, one degree lower than $P$:

$Q(x) = x^3 + x^2 - 22x - 40$

A quartic problem has become a cubic problem, and that is the entire transaction. Multiplying $(x - 3)$ by $Q$ returns $P$ if you want the check.

The root $5$ supplies the factor $x - 5$, and the place to spend it is the cubic, not the quartic. The cubic has smaller numbers, one fewer root to account for, and it is where the remaining factors are hiding. Divide $Q(x) = x^3 + x^2 - 22x - 40$ by $x - 5$:

$\begin{array}{c|cccc} 5 & 1 & 1 & -22 & -40 \\ & & 5 & 30 & 40 \\ \hline & 1 & 6 & 8 & 0 \end{array}$

The remainder is zero again, so $Q(x) = (x - 5)\left(x^2 + 6x + 8\right)$ and therefore

$P(x) = (x - 3)(x - 5)\left(x^2 + 6x + 8\right).$

What is left is a quadratic, and a quadratic is always finishable. Two numbers multiplying to $8$ and adding to $6$ are $4$ and $2$, so $x^2 + 6x + 8 = (x + 4)(x + 2)$, giving

$P(x) = (x - 3)(x - 5)(x + 4)(x + 2).$

The roots of $P$ are therefore $3$, $5$, $-4$ and $-2$. Two of them were given and two were earned, and the earning took one division and a quadratic factorization rather than any search. Note that both earned roots are negative although both given roots were positive: nothing about the first two roots predicts the sign of the rest.

A quick check without expanding all four brackets: the constant term of the product is $(-3)(-5)(4)(2) = 120$, which matches $P$, and the leading coefficient is $1$ in both.

The classmate's first sentence is sound: $P(5) = 0$ does give $x - 5$ as a factor of $P$, straight from the theorem. The second sentence is where the gap opens, and the gap is bigger than it looks.

The step being taken is: $x - 5$ divides a product, therefore it divides one of the two things multiplied. That inference is not licensed in general. A divisor of a product need not divide either factor, and a one-line example settles it: $x^2 - 1$ divides $(x - 1)(x + 1)$ exactly, since it IS that product, yet it divides neither $x - 1$ nor $x + 1$, both of which have too small a degree to be multiples of it. So the classmate is invoking a rule that does not exist. That the conclusion happens to be true here is something still to be earned.

What earns it is the same move the ceiling proof uses: treat the factorization as an identity and evaluate it at the second root. From part A, $P(x) = (x - 3)\,Q(x)$ holds for every $x$, so it holds at $x = 5$:

$0 = P(5) = (5 - 3)\,Q(5) = 2\,Q(5)$

Now the product of the two numbers $2$ and $Q(5)$ is zero, and a product of numbers is zero only when one of them is zero. The first one is not: it is $2$. Therefore $Q(5) = 0$, and the factor theorem applied to $Q$ this time gives $x - 5$ as a factor of $Q$, which is what was wanted.

The place where the two numbers being different matters is exactly one line: the claim that $5 - 3$ is not zero. That is the whole content of $5 \neq 3$ here, and everything downstream leans on it. Suppose instead the second root had been $3$ over again. Then the same evaluation would read $0 = (3 - 3)\,Q(3) = 0$, which is true for every cubic $Q$ whatsoever and therefore tells you nothing about $Q(3)$. A repeated root has to be rechecked inside the quotient rather than inferred from the original, and this is the reason why.

So the classmate's instinct was right and their licence was missing. The corrected version is worth carrying: if $P(a) = 0$ and $P(b) = 0$ with $a \neq b$, then after peeling $x - a$ off, the number $b$ is still a root of what remains.

Each prescribed root $r$ contributes the factor $x - r$, minding the sign: the root $-3$ gives $x + 3$, not $x - 3$. Three roots and degree three leave room for nothing but a constant out front, so the polynomial must have the shape

$P(x) = a\,(x + 3)(x - 2)(x - 6), \qquad a \neq 0.$

Every member of that family already has the three roots asked for, so the point is what chooses $a$. Substitute $x = 1$ and set the value to $-40$:

$P(1) = a\,(4)(-1)(-5) = 20a = -40 \implies a = -2$

That is the factored answer, $P(x) = -2(x + 3)(x - 2)(x - 6)$. Expand in stages rather than all at once. The two brackets with plain numbers first:

$(x - 2)(x - 6) = x^2 - 8x + 12$

Then multiply by $(x + 3)$:

$(x + 3)\left(x^2 - 8x + 12\right) = x^3 - 8x^2 + 12x + 3x^2 - 24x + 36 = x^3 - 5x^2 - 12x + 36$

Finally scale by $a = -2$:

$P(x) = -2x^3 + 10x^2 + 24x - 72$

Check both requirements against the expanded form, since a sign slip in the expansion would otherwise survive. The point: $P(1) = -2 + 10 + 24 - 72 = -40$. A root: $P(2) = -16 + 40 + 48 - 72 = 0$. The negative leading coefficient is not a mistake; it is what the given point asked for.

There are two halves here, and they pull in opposite directions.

One half is a genuine restriction, and it is the ceiling read backwards. A nonzero polynomial of degree $n$ has at most $n$ distinct roots, so a polynomial with three distinct roots cannot have degree $0$, $1$ or $2$:

$3 \leq n$

The zero polynomial is excluded automatically, since every number is a root of it and its root list is not the three named numbers. So the degree is at least three, and that much the roots really do force.

The other half is that nothing pushes back from above. A factor may be repeated, and repeating one changes the degree without adding a single new root:

$(x + 3)^2(x - 2)(x - 6)$

has degree $4$. A product of numbers is zero only when one of its factors is zero, so this polynomial vanishes exactly at $-3$, $2$ and $6$, the same three numbers as before. It is not a cubic, and it meets every condition stated.

The trick is not limited to repetition either. Multiplying by any quadratic that never reaches zero on the real line adds degree and no real roots at all: $x^2 + 2x + 3$ has discriminant $4 - 12 = -8 < 0$, so $(x + 3)(x - 2)(x - 6)\left(x^2 + 2x + 3\right)$ has degree $5$ and still exactly those three real roots.

So the honest statement is one-sided. A root list is a floor on the degree and never a ceiling on it, and the word "cubic" in part A was doing work that the three roots alone could not do.

Two claims are bundled here, and each has to be met separately.

The direction from a constant multiple to shared roots is true, and it is quick. Suppose $G(x) = k\,F(x)$ with $k \neq 0$. Then at any number $r$,

$G(r) = k\,F(r),$

and since $k \neq 0$, the product $k\,F(r)$ is zero exactly when $F(r)$ is. So $F$ and $G$ vanish at precisely the same numbers, and a scale factor never moves a root.

The direction from shared roots to a constant multiple is false. Part B already supplies the raw material, and the sharpest version of it uses two polynomials of the same degree so that a difference in degree cannot be blamed:

$F(x) = (x + 3)^2(x - 2)(x - 6), \qquad G(x) = (x + 3)(x - 2)^2(x - 6)$

Both are quartics, and both vanish exactly at $-3$, $2$ and $6$. If $G = kF$ held for some constant $k$, that equation would have to hold at every value of $x$. Test two of them. At $x = 0$:

$F(0) = 9(-2)(-6) = 108, \qquad G(0) = 3(4)(-6) = -72 \implies k = -\frac{72}{108} = -\frac{2}{3}$

At $x = 1$:

$F(1) = 16(-1)(-5) = 80, \qquad G(1) = 4(1)(-5) = -20 \implies k = -\frac{20}{80} = -\frac{1}{4}$

No single constant can be both, so no such $k$ exists and the direction fails. The claim as a whole is therefore not an equivalence: it is one true implication with a false converse attached.

What repairs it is a condition strong enough to leave no room for the roots to be arranged differently. Suppose $F$ and $G$ are nonzero, both of degree $n$, and both have the same $n$ distinct roots $r_1, \ldots, r_n$. Peel them off $F$ one at a time. Each root removes one linear factor and drops the quotient's degree by exactly one, and after all $n$ peels the leftover has degree $n - n = 0$, so it is a nonzero constant $a$:

$F(x) = a\,(x - r_1)(x - r_2)\cdots(x - r_n)$

The same argument on $G$ gives the same product with some nonzero constant $b$ in front, so $G = \frac{b}{a}F$, a nonzero constant multiple. The counterexample above is not a threat to this, and it is worth seeing why: those quartics have only three distinct roots between them, not four, so the repaired hypothesis was never satisfied. What the extra condition buys is that the number of distinct roots uses up the entire degree, leaving nothing for the polynomials to differ by except a scale factor.

Argue by ruling out the alternative. Suppose, for contradiction, that at least one of $a$, $b$, $c$, $d$ is not zero. Then $P$ is a nonzero polynomial, and its degree is whichever of $3$, $2$, $1$, $0$ is fixed by the highest surviving coefficient. In every case:

$\deg P \leq 3$

The theorem to appeal to is the ceiling proved in this lesson: a nonzero polynomial of degree $n$ has at most $n$ distinct roots. It has two hypotheses and both must be checked rather than assumed.

The first is that the polynomial is not the zero polynomial, and that is exactly what the contradiction hypothesis grants: some coefficient is nonzero. The second lives in the phrase "distinct roots", and it is about the numbers supplied, not about $P$. The four given numbers are $-3$, $-1$, $2$ and $5$, and no two of them are equal, so they are four distinct roots and each one peels off a genuinely new factor.

So $P$ is a nonzero polynomial of degree at most $3$ with at least $4$ distinct roots, and the ceiling says it has at most $\deg P$ of them:

$4 \leq \#\{\text{distinct roots of } P\} \leq \deg P \leq 3$

That chain ends in $4 \leq 3$, which is false. The assumption that produced it was that some coefficient is nonzero, so that assumption fails, and every one of $a$, $b$, $c$, $d$ is zero.

Notice what was not needed. Nobody expanded anything, nobody solved four simultaneous equations, and no property of the particular numbers $-3$, $-1$, $2$, $5$ was used beyond their being four and being different from one another. Any four distinct numbers would have done the same work.

Comparing two polynomials point by point goes nowhere, because four agreements are four equations in eight unknowns. The move that works is to stop looking at two objects and start looking at one. Define their difference:

$D(x) = f(x) - g(x) = (p - u)x^3 + (q - v)x^2 + (r - w)x + (s - m)$

Subtracting coefficient by coefficient shows $D$ has exactly the shape part A was about: a cubic expression whose four coefficients are real numbers, any of which may be zero.

Now translate the hypothesis. Saying $f$ and $g$ agree at a number $t$ says $f(t) - g(t) = 0$, which says $D(t) = 0$. So the four agreement points are four roots of the single polynomial $D$, and they are four different numbers because the agreement points were:

$D(t_1) = D(t_2) = D(t_3) = D(t_4) = 0, \qquad t_1, t_2, t_3, t_4 \text{ all different}$

Part A applies verbatim and forces every coefficient of $D$ to be zero. Reading those four statements back gives $p = u$, $q = v$, $r = w$ and $s = m$, so $f$ and $g$ are the same polynomial, and in particular $f(x) = g(x)$ at every value of $x$, not just at the four where agreement was assumed.

The result is worth keeping in this form: a polynomial of this shape is completely pinned down by its values at four different numbers. Three points would not do it, and the failure is easy to see, since $x^3$ and $x^3 + (x + 1)x(x - 1)$ agree at $-1$, $0$ and $1$ without being equal. Four is exactly the number that closes the gap for degree three, because it is one more than the ceiling.

The theorem is an inequality, and the classmate has read it as an equation. What it says is

$\#\{\text{distinct roots}\} \leq \deg P$

for a nonzero $P$, and an inequality in that direction can only ever forbid roots. It never produces one. That is visible in the proof itself: every step starts from a root you already have and converts it into a factor. Nothing in the argument creates a root out of nothing, so nothing in the conclusion can promise one.

There are two separate ways the actual count can come in under the degree, and one example of each makes the point.

The first way is a repeated factor. Take

$F(x) = (x - 4)^2(x + 1)^2.$

Multiplying out would give a genuine quartic, and a product is zero only when one of its factors is zero, so $F$ vanishes at $4$ and at $-1$ and nowhere else. Degree four, two distinct roots. The degree is spent on factors, and factors are allowed to repeat, while roots are counted as the distinct numbers they are.

The second way is that a factor may never reach zero at all. Take

$G(x) = x^2 + 2x + 5, \qquad \Delta = 2^2 - 4(1)(5) = -16 < 0.$

A negative discriminant means no real number satisfies $G(x) = 0$, so this quadratic has no roots among the real numbers. Degree two, no real roots, and the ceiling is perfectly content, since zero is comfortably at most two.

So the licensed reading is one-directional. If someone hands you four different numbers at which a nonzero quartic vanishes, the ceiling lets you conclude the list is complete and stop looking. It never lets you start by asserting that four such numbers exist. "At most" is a permission to stop searching, not a guarantee that the search would succeed.

Take the two roots one at a time, and take the second one inside what the first one leaves behind.

Since $P(a) = 0$, the factor theorem supplies a polynomial $Q$ with

$P(x) = (x - a)\,Q(x),$

an identity holding at every value of $x$. The temptation now is to apply the factor theorem to $P$ a second time and declare $x - b$ a factor of $P$. That is true, but it is not what is wanted: the goal is the product, so $x - b$ has to be found inside $Q$.

Evaluate the identity at $x = b$, which is legitimate precisely because it is an identity:

$0 = P(b) = (b - a)\,Q(b)$

Here is where the hypothesis is spent, and it is spent at exactly one step. The left side is a product of two numbers, $b - a$ and $Q(b)$, and a product of numbers is zero only when one of them is zero. To conclude anything about $Q(b)$ you must know the other number is not zero, and $b - a \neq 0$ is nothing other than the assumption $a \neq b$. With it, the conclusion is forced:

$Q(b) = 0$

Now apply the factor theorem to $Q$, which is a perfectly good polynomial in its own right: $Q(x) = (x - b)\,S(x)$ for some polynomial $S$. Substituting back,

$P(x) = (x - a)(x - b)\,S(x),$

which says exactly that $(x - a)(x - b)$ is a factor of $P(x)$, with $S$ as the other factor.

The shape of the argument is worth naming, because it is the engine of the whole chapter. Each root is converted into a factor and then evicted from the problem, and the next root is chased inside the smaller polynomial rather than the original. Repeat it and you get the ceiling on the number of roots; stop after two and you get this.

With $a = b$ the claim of part A becomes: if $P(a) = 0$ then $(x - a)^2$ is a factor of $P$. So a counterexample needs a polynomial with a root whose factor appears only once.

Take

$P(x) = x^2 - 3x - 28 = (x - 7)(x + 4), \qquad a = b = 7.$

Check what this choice does and does not satisfy. The two root equations hold, and cheaply: $P(7) = 49 - 21 - 28 = 0$, so $P(a) = 0$ and $P(b) = 0$ are both true. What it deliberately violates is part A's other assumption, the distinctness of $a$ and $b$, which is the entire point of the construction.

Now demonstrate that the conclusion fails. Suppose $(x - 7)^2$ were a factor, so that

$P(x) = (x - 7)^2\,T(x)$

for some polynomial $T$. Degrees add across a product, and $P$ has degree $2$ while $(x - 7)^2$ has degree $2$, so $T$ has degree $0$: it is a constant. Comparing the coefficients of $x^2$ on the two sides makes that constant $1$, which leaves

$P(x) = (x - 7)^2 = x^2 - 14x + 49.$

But $P(x) = x^2 - 3x - 28$, and those are not the same polynomial: their constant terms are $49$ and $-28$. The supposition is impossible, so $(x - 7)^2$ is not a factor of $P$ and the conclusion of part A genuinely fails.

The reason it fails is visible in the proof. With $a = b = 7$, the step that carried the whole argument reads

$0 = P(7) = (7 - 7)\,Q(7) = 0 \cdot Q(7),$

which is true no matter what $Q(7)$ is, and therefore tells you nothing about it. The proof did not merely become harder; it lost its content at exactly the line where the hypothesis used to sit.