The Factor Theorem

Intuition

Take $P(x) = x^2 - 5x + 6$ . Plug in $x = 2$ and $P(2) = 4 - 10 + 6 = 0$ , so $2$ is a root. That same polynomial factors as $P(x) = (x - 2)(x - 3)$ , so $x - 2$ is a factor. The root showed up exactly where the factor did, and that is not a coincidence about this one polynomial.

The previous lesson showed why: dividing $P(x)$ by $x - c$ always leaves the remainder $P(c)$ , a number you can find without dividing at all. When $P(c) = 0$ , the division leaves nothing behind, so $x - c$ divides $P(x)$ exactly and is a factor. This lesson proves the rule works both ways: a root gives you a factor, and a factor gives you a root. Find one root of a cubic, divide it out, and what remains is a quadratic you already know how to finish.

Learning goals

State the master identity $P(x) = (x - c)Q(x) + P(c)$
Decide whether any linear expression, monic or not, is a factor of $P(x)$ with one evaluation
Turn one known root into a factor by division, then keep searching inside the smaller quotient
Build a polynomial from prescribed real and nonreal roots plus one extra condition
Explain why a nonzero polynomial of degree $n$ has at most $n$ roots

Roots and factors, two languages

The small example above is the pattern; here is the general rule behind it. Two definitions sit at the center of this chapter, and they live in different worlds. A number $c$ is a root (or zero) of the polynomial $P(x)$ when $P(c) = 0$ . That is an arithmetic statement: plug in one number, get zero out. By contrast, a polynomial $D(x)$ is a factor of $P(x)$ when

P(x) = D(x)\,Q(x) \qquad \text{for some polynomial } Q(x),

exactly as $3$ is a factor of $12$ because $12 = 3 \cdot 4$ with nothing left over. That is an algebraic statement about an identity that must hold for every value of $x$ at once. One definition asks about a single evaluation; the other asks about an exact division. There is no obvious reason these two ideas should have anything to do with each other. Yet the theorem of this lesson says that, for linear divisors, they are the very same thing.

The bridge between the two worlds was built in the last two lessons. Dividing $P(x)$ by $x - c$ produces a quotient and a remainder,

P(x) = (x - c)\,Q(x) + R,

where the remainder has degree smaller than the divisor $x - c$ , so $R$ is a constant. The remainder theorem then identifies which constant: substituting $x = c$ kills the first term and leaves $R = P(c)$ . So for every polynomial $P(x)$ and every number $c$ there is a single master identity,

P(x) = (x - c)\,Q(x) + P(c),

and everything in this lesson is a consequence of reading it carefully.

Check your understanding

Dividing $P(x)$ by $x - 4$ gives a quotient $Q(x)$ and leaves remainder $7$ . What is $P(4)$ ?

P(4) = 7

P(4) = 0

P(4) = Q(4)

P(4) = -7

The Factor Theorem

The Factor Theorem. Let $P(x)$ be a polynomial and let $c$ be a number. Then $x - c$ is a factor of $P(x)$ if and only if $P(c) = 0$ .

Take the phrase “if and only if” seriously: it is two separate claims bundled into one sentence. It claims that a root produces a factor, and it also claims that a factor produces a root. A statement of this shape is only proved when both directions are proved, so the proof below does them one at a time.

Proof of the Factor Theorem, both directions#

Divide $P(x)$ by $x - c$ . As recalled above, the division algorithm and the remainder theorem together give the identity

P(x) = (x - c)\,Q(x) + P(c),

valid for every $x$ , where $Q(x)$ is the quotient.

First suppose $P(c) = 0$ . The identity collapses to $P(x) = (x - c)\,Q(x)$ , which is precisely the definition of ” $x - c$ is a factor of $P(x)$ ”: the division came out exact, and $Q(x)$ is the other factor. This is the direction that manufactures factorizations out of roots.

Now suppose instead that $x - c$ is a factor of $P(x)$ , so that $P(x) = (x - c)\,S(x)$ for some polynomial $S(x)$ . A factorization is an identity, true for every value of $x$ , so it is in particular true at $x = c$ :

P(c) = (c - c)\,S(c) = 0 \cdot S(c) = 0.

So $c$ is a root. Notice this direction did not even need the remainder theorem, only the freedom to evaluate an identity at the most convenient point. You can also see it through remainders: if the division by $x - c$ is exact, the remainder is $0$ , and the remainder theorem says the remainder is $P(c)$ , so $P(c) = 0$ . Both directions are established, so the theorem holds as a genuine biconditional.

The two directions do different jobs, and you will use both constantly. Read left to right, a known factorization hands you roots for free: from $P(x) = (x - 2)(x + 7)Q(x)$ you may write down $P(2) = 0$ and $P(-7) = 0$ without computing anything. Read right to left, a known root hands you a factor, together with a genuine factorization $P(x) = (x - c)\,Q(x)$ whose quotient you can compute by synthetic or long division. And because the theorem is a biconditional, one evaluation settles the factor question in either direction: $P(c) = 0$ means $x - c$ is a factor, and $P(c) \neq 0$ means it is not. Nothing is left undecided.

One small generalization is worth recording now, because factors are not always monic. To test whether $2x - 1$ is a factor of $P(x)$ , find the number that makes $2x - 1$ equal to $0$ : that is $x = \tfrac{1}{2}$ . Evaluate there. $P\!\left(\tfrac{1}{2}\right) = 0$ means $2x - 1$ is a factor; anything else means it is not. The rule of thumb covers every case: test the value that kills the candidate factor. For $x + 5$ that value is $-5$ , because $x + 5 = x - (-5)$ .

The general version, for anyone who wants to see why it works: for $a \neq 0$ ,

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

and multiplying a divisor by a nonzero constant never changes whether a division is exact: if $P(x) = \left(x - \tfrac{b}{a}\right)Q(x)$ , then $P(x) = (ax - b) \cdot \tfrac{1}{a}Q(x)$ , and $\tfrac{1}{a}Q(x)$ is still a polynomial. So $ax - b$ is a factor of $P(x)$ exactly when $P\!\left(\tfrac{b}{a}\right) = 0$ , matching the shortcut above.

Check your understanding

Is $x + 3$ a factor of $P(x) = x^3 + 2x^2 - 5x - 6$ ?

A No, because

P(-3) \neq 0

B Yes, because

P(3) = 0

C Yes, because

P(-3) = 0

D No, because

P(3) \neq 0

Check your understanding

Is $2x - 1$ a factor of $P(x) = 2x^3 + x^2 - 5x + 2$ ?

A Yes, because

P\!\left(\tfrac{1}{2}\right) = 0

B No, because

P\!\left(\tfrac{1}{2}\right) \neq 0

C Yes, because

P(2) = 0

D No, because

P\!\left(-\tfrac{1}{2}\right) \neq 0

One root unlocks the whole factorization

Here is the workflow that makes the factor theorem the workhorse of this chapter. Suppose you know, or are told, a single root $c$ of a polynomial of degree $n$ . The theorem promises the factorization $P(x) = (x - c)\,Q(x)$ , and division delivers $Q(x)$ explicitly, a polynomial of degree $n - 1$ . You have traded one problem for a strictly smaller one. Now hunt for a root of $Q$ , peel again, and keep going until what remains is a quadratic, where factoring, the quadratic formula, and complex numbers finish the job completely.

Worked example 1 Factor a cubic from one known root

Factor $P(x) = x^3 - 4x^2 + x + 6$ completely, given that $2$ is a root.

First confirm the given root, since one evaluation is cheap insurance:

P(2) = 8 - 16 + 2 + 6 = 0.

The factor theorem now guarantees $P(x) = (x - 2)\,Q(x)$ . Find $Q$ by synthetic division with $c = 2$ on the coefficients $1, -4, 1, 6$ :

\begin{array}{c|cccc} 2 & 1 & -4 & 1 & 6 \\ & & 2 & -4 & -6 \\ \hline & 1 & -2 & -3 & 0 \end{array}

The remainder $0$ in the corner re-confirms the root, and the other entries spell out the quotient $Q(x) = x^2 - 2x - 3$ . That quadratic factors by the Chapter 4 methods, since $-3$ and $1$ multiply to $-3$ and add to $-2$ :

x^2 - 2x - 3 = (x - 3)(x + 1),

so the complete factorization is

P(x) = (x - 2)(x - 3)(x + 1),

and the roots of $P$ are $2$ , $3$ , and $-1$ . Multiplying the three factors back together, or evaluating $P(3)$ and $P(-1)$ directly, checks the answer.

Worked example 2 The quotient does not have to cooperate

Factor $P(x) = x^3 - 5x^2 + 9x - 5$ completely, given that $1$ is a root.

Confirm: $P(1) = 1 - 5 + 9 - 5 = 0$ . Peel off the factor $x - 1$ by synthetic division on $1, -5, 9, -5$ :

\begin{array}{c|cccc} 1 & 1 & -5 & 9 & -5 \\ & & 1 & -4 & 5 \\ \hline & 1 & -4 & 5 & 0 \end{array}

So $P(x) = (x - 1)\left(x^2 - 4x + 5\right)$ . The quotient refuses to factor over the reals: its discriminant is

\Delta = (-4)^2 - 4(1)(5) = -4 < 0,

so it has no real roots, and over the real numbers the factorization above is already complete. But the previous chapter taught you exactly what a negative discriminant means. The quadratic formula runs to completion with $\sqrt{-4} = 2i$ and gives the conjugate pair

x = \frac{4 \pm 2i}{2} = 2 \pm i,

so over the complex numbers the cubic splits fully into linear factors:

P(x) = (x - 1)\,\bigl(x - (2 + i)\bigr)\,\bigl(x - (2 - i)\bigr).

The real root $1$ appears as a real linear factor; the nonreal pair hides inside the real quadratic factor until you allow complex factors. Both lines are correct answers, to two different questions: “factor over the reals” and “factor over the complex numbers”.

Check your understanding

The cubic $x^3 - 7x + 6$ has $1$ as a root. What is the quotient when it is divided by $x - 1$ ?

x^2 + x - 6

x^2 - x - 6

x^2 + x + 6

x^2 - 7x + 6

Worked example 3 Solve for an unknown coefficient

Find the value of $k$ so that $x - 2$ is a factor of $P(x) = x^3 - 3x^2 + kx - 8$ .

This runs the factor theorem backward from every example so far. Instead of a finished polynomial, $P$ has an unknown $k$ sitting inside it, and the factor condition is what pins it down. The theorem still says exactly the same thing: $x - 2$ is a factor precisely when $P(2) = 0$ . So evaluate at $x = 2$ , unknown $k$ and all:

P(2) = (2)^3 - 3(2)^2 + k(2) - 8 = 8 - 12 + 2k - 8 = 2k - 12.

Setting the factor condition $P(2) = 0$ turns the question into a linear equation for $k$ :

2k - 12 = 0 \quad\Longrightarrow\quad k = 6.

Check by substituting back: with $k = 6$ , $P(x) = x^3 - 3x^2 + 6x - 8$ , and synthetic division by $c = 2$ on $1, -3, 6, -8$ leaves remainder $0$ , confirming the factor. The move to remember: an unknown coefficient inside $P$ does not change the rule at all. Evaluate at the number that kills the factor, set the result equal to $0$ , and solve the resulting equation for the unknown.

Check your understanding

For which value of $k$ is $x + 2$ a factor of $P(x) = x^3 + kx^2 + 3x + 2$ ?

k = 3

k = -3

k = 6

k = -6

Building a polynomial to order

The factor theorem also runs in reverse: instead of dissecting a given polynomial, you can manufacture one with any roots you choose. Each prescribed root $r$ contributes a factor $x - r$ , and the product of those factors has every prescribed number as a root. The reason is the easy direction of the theorem: evaluate the product at $r$ , and the factor $x - r$ makes the whole product zero.

A graph shows what you are really building. The real roots of a polynomial are exactly the $x$ -intercepts of its graph, because “the graph meets the axis at $c$ ” and " $P(c) = 0$ " are the same sentence. So a factored polynomial is a curve whose axis crossings you have pinned down in advance. The cubic below crosses cleanly through the axis at each root because each factor appears only once; a root can also touch the axis without crossing it when its factor repeats, which the next chapter takes up.

The monic cubic with roots -1, 2, and 4. The graph crosses the x-axis exactly where a linear factor vanishes, so each intercept on the graph is a factor on the page: x + 1, x - 2, and x - 4.

One caution before the examples: a list of roots does not pin down one polynomial. The cubics $(x + 1)(x - 2)(x - 4)$ and $2(x + 1)(x - 2)(x - 4)$ have identical roots, and so does every other nonzero multiple. The construction really produces a whole family,

P(x) = a\,(x - r_1)(x - r_2)\cdots(x - r_k), \qquad a \neq 0,

one polynomial for each choice of the leading coefficient $a$ , every one of them of degree $k$ .

Leaving the degree free is worse than just an open leading coefficient: a higher-degree polynomial can repeat one of the factors and still have the same distinct roots and pass through the same point. Compare $P(x) = (x - 1)(x - 3)$ with $R(x) = -(x - 1)^2(x - 3)$ . Both have exactly the roots $1$ and $3$ , and both satisfy $P(0) = R(0) = 3$ , yet $P$ has degree $2$ and $R$ has degree $3$ . So a problem that wants one specific polynomial has to do two things. It has to hold the degree at $k$ , which is what asking for a cubic with three roots, or for the least degree those roots allow, already does. And it has to supply one more condition to choose $a$ , such as the leading coefficient itself or a point the graph passes through. That point has to earn its keep. If its $x$ -value is one of the prescribed roots, $P = 0$ there no matter what $a$ is, so it settles nothing. And its $y$ -value should not be $0$ at a new $x$ -value either: solving for $a$ would then force $a = 0$ , which breaks the rule $a \neq 0$ and means no polynomial of that exact degree with exactly those roots reaches that point. A usable extra point has an $x$ -value outside the root list and a nonzero $y$ -value.

Worked example 4 Prescribed roots plus one point

Find the cubic polynomial with roots $-1$ , $2$ , and $4$ whose graph passes through $(0, 16)$ .

The three roots force the shape $P(x) = a\,(x + 1)(x - 2)(x - 4)$ for some nonzero constant $a$ , and the extra point pins down $a$ . Evaluate at $x = 0$ :

P(0) = a\,(1)(-2)(-4) = 8a = 16 \quad\Longrightarrow\quad a = 2.

Now expand step by step. First the last two factors:

(x - 2)(x - 4) = x^2 - 6x + 8,

then multiply by $(x + 1)$ :

(x + 1)\left(x^2 - 6x + 8\right) = x^3 - 6x^2 + 8x + x^2 - 6x + 8 = x^3 - 5x^2 + 2x + 8,

and finally scale by $a = 2$ :

P(x) = 2x^3 - 10x^2 + 4x + 16.

Check the point: $P(0) = 16$ . Check a root: $P(2) = 16 - 40 + 8 + 16 = 0$ . The monic cubic in the figure above is this polynomial’s $a = 1$ sibling; the point $(0, 16)$ is what selected $a = 2$ from the whole family sharing those three crossings.

Check your understanding

For which value of $a$ does the quadratic $P(x) = a(x - 1)(x + 3)$ pass through $(0, -12)$ ?

a = 4

a = -4

a = 12

a = -12

Worked example 5 Building with a nonreal root

Find the monic cubic with real coefficients that has $3$ and $1 + 2i$ among its roots.

Two roots are given, but a cubic needs three linear factors, and the missing one comes for free. For a polynomial with real coefficients, a nonreal root always arrives with its conjugate: since $1 + 2i$ is a root, $1 - 2i$ must be one too. (The reason: conjugating both sides of the equation $P(z) = 0$ conjugates nothing but $z$ itself, because conjugation passes through sums and products and leaves real coefficients fixed. That argument, proved for quadratics in the complex numbers chapter, uses nothing about degree, so it applies here as well.)

Multiply the conjugate pair of factors first, because that product is guaranteed to come out real:

\begin{aligned} \bigl(x - (1 + 2i)\bigr)\bigl(x - (1 - 2i)\bigr) &= \bigl((x - 1) - 2i\bigr)\bigl((x - 1) + 2i\bigr) \\ &= (x - 1)^2 - (2i)^2, \end{aligned}

and since $(2i)^2 = -4$ , this is $(x - 1)^2 + 4 = x^2 - 2x + 5$ . Now attach the real root’s factor:

\begin{aligned} P(x) = (x - 3)\left(x^2 - 2x + 5\right) &= x^3 - 2x^2 + 5x - 3x^2 + 6x - 15 \\ &= x^3 - 5x^2 + 11x - 15. \end{aligned}

Every coefficient is real, as promised. The pattern is worth remembering: a real root shows up as a real linear factor. A nonreal root and its conjugate instead merge into a real quadratic factor, here $x^2 - 2x + 5$ , whose discriminant is negative. The next chapter builds its whole theory of real polynomials on this pairing.

Check your understanding

A monic cubic with real coefficients has $5$ and $i$ among its roots. What is its factorization over the reals?

(x - 5)\left(x^2 - 1\right)

(x - 5)\left(x^2 + 1\right)

(x + 5)\left(x^2 + 1\right)

(x + 5)\left(x^2 - 1\right)

A ceiling on the number of roots

The factor theorem’s first big structural consequence is a hard limit. A quadratic never has three roots, a cubic never has four, and the pattern is completely general.

Here is why, in one small case. Take a cubic, degree $3$ , and suppose it has three different roots. Each root peels off one linear factor, and each peel drops the degree by exactly $1$ : start at degree $3$ , peel once to reach degree $2$ , peel again to reach degree $1$ , peel a third time to reach degree $0$ , a nonzero constant with no $x$ left in it to make it vanish anywhere. There is no degree left to peel a fourth time, so a fourth distinct root is impossible, unless the cubic was the zero polynomial to begin with. The general argument below just runs this same budget for any degree $n$ .

A nonzero polynomial of degree $n$ has at most $n$ roots#

Let $P(x)$ be a nonzero polynomial of degree $n$ , and let $r_1, r_2, \ldots, r_k$ be distinct roots of it. Because $r_1$ is a root, the factor theorem writes

P(x) = (x - r_1)\,Q_1(x),

and comparing degrees on the two sides, $Q_1$ has degree $n - 1$ . Now evaluate this identity at the second root:

0 = P(r_2) = (r_2 - r_1)\,Q_1(r_2).

The first factor is not zero, because the roots are distinct, and a product of two numbers is zero only when one of them is zero. So $Q_1(r_2) = 0$ , and the factor theorem applies again, this time to $Q_1$ : we get $Q_1(x) = (x - r_2)\,Q_2(x)$ with $Q_2$ of degree $n - 2$ , hence

P(x) = (x - r_1)(x - r_2)\,Q_2(x).

Repeat the same evaluation argument with $r_3, r_4, \ldots$ , each time inside the newest quotient. Every distinct root peels off one more linear factor, and every peel lowers the quotient’s degree by exactly $1$ . After all $k$ roots are used,

P(x) = (x - r_1)(x - r_2)\cdots(x - r_k)\,Q_k(x),

where $Q_k$ is not the zero polynomial (otherwise $P$ would be) and has degree $n - k$ . A degree is never negative, so $n - k \geq 0$ , which is the claim: $k \leq n$ .

Read the theorem for what it is: a ceiling, not a promise. It says a degree- $n$ polynomial cannot have more than $n$ roots; it is silent about how many actually exist. The quadratic $x^2 + 1$ has no real roots at all, and no theorem in this lesson objects, because zero is comfortably at most two. The factor theorem converts roots into factors and factors into roots, but it never manufactures a root out of nothing. Two questions stay open here: whether roots must exist at all, and how many a degree- $n$ polynomial is guaranteed once complex numbers are allowed. Both are the business of the Fundamental Theorem of Algebra, the headline of the next chapter.

The ceiling still has real bite. If someone hands you a cubic and four distinct numbers where it vanishes, you may conclude the “cubic” is actually the zero polynomial, no computation required. And in Worked Example 1, once the roots $2$ , $3$ , and $-1$ were in hand, the search was legitimately over: a cubic has no room for a fourth root.

Check your understanding

A polynomial $P(x)$ is claimed to have degree at most $3$ , and it is also known to vanish at the five distinct numbers $-2, -1, 0, 1, 2$ . What must be true?

P(x)

is the zero polynomial B

P(x)

must actually have degree

5

C The claim is fine; a cubic can have five roots D Nothing can be concluded without more information

There is one honest gap left in the workflow, and you should notice it. Everything above starts from one known root, and in this lesson the root was always handed to you. Where does the first root come from when nobody supplies it? Testing numbers blindly could take forever. The next lesson closes exactly this gap: for polynomials with integer coefficients, the Rational Root Theorem shrinks the infinite search to a short finite list of rational candidates. The factor theorem then processes each of those candidates with a single evaluation.

Common mistakes

Testing the wrong value for the factor’s sign. The factor $x + 5$ vanishes at $x = -5$ , so test $P(-5)$ , never $P(5)$ . Write $x + 5 = x - (-5)$ if the signs blur: the $c$ in the theorem is $-5$ .
Reporting a factor as if it were the factorization. Learning that $2$ is a root gives $P(x) = (x - 2)\,Q(x)$ , not $P(x) = x - 2$ . The quotient carries the rest of the polynomial; divide to find it and keep factoring.
Hunting the next root in the original instead of the quotient. After peeling off $x - c$ , work inside $Q(x)$ : it has smaller degree, smaller numbers, and one fewer root to find. Returning to $P$ re-finds the root you already used.
Forgetting the zero coefficients in synthetic division. The cubic $x^3 - 7x + 6$ has synthetic row $1, 0, -7, 6$ . Dropping the $0$ silently divides a different polynomial.
Believing a root list determines the polynomial. $(x - 1)(x - 3)$ and $7(x - 1)(x - 3)$ have the same roots. A polynomial of degree $n$ with $n$ different roots is fixed by those roots only up to its leading coefficient. When the degree is left free instead, the roots settle even less than that. A construction problem has to pin the degree AND supply one more condition.
Expecting the theorem to find roots. The factor theorem is a converter, not a search engine: it never asserts a root exists. Candidates come from elsewhere, and the next lesson provides the standard supply for integer coefficients.
Pairing conjugates without real coefficients. The rule “nonreal roots arrive with their conjugates” is proved from the coefficients being real. A polynomial like $x^2 - 2ix - 1$ , with the double root $i$ and no root $-i$ , breaks the rule because its coefficients are not real.

Key takeaways

Every division by a linear factor obeys the master identity $P(x) = (x - c)\,Q(x) + P(c)$ .
The Factor Theorem: $x - c$ is a factor of $P(x)$ if and only if $P(c) = 0$ . Both directions hold, so a single evaluation settles the question either way: zero means factor, nonzero means not a factor.
More generally, for $a \neq 0$ , $ax - b$ is a factor of $P(x)$ exactly when $P\!\left(\tfrac{b}{a}\right) = 0$ : always test the value that kills the factor.
Peel to factor: one known root $c$ gives $P(x) = (x - c)\,Q(x)$ by division, and the hunt continues inside the smaller $Q$ . A quadratic quotient can always be finished over the complex numbers, and over the reals as well whenever its discriminant allows.
Build to order: $a\,(x - r_1)(x - r_2)\cdots(x - r_k)$ has exactly the prescribed roots; the scalar $a$ needs one extra condition. Real roots appear as real linear factors, and a nonreal root of a real polynomial brings its conjugate, the pair merging into a real quadratic factor.
A nonzero polynomial of degree $n$ has at most $n$ roots: a ceiling proved by peeling, not a promise that any exist.

The Factor Theorem

Learning goals

Roots and factors, two languages

The Factor Theorem

One root unlocks the whole factorization

Building a polynomial to order

A ceiling on the number of roots

Common mistakes

Practice

Multiple Choice Questions (MCQ)

Core practice