The Factor Theorem

A2 · P3 · 16 min

The factor theorem turns factorising a cubic or quartic from guesswork into a short search: test a few likely numbers, and the moment one gives zero you have a factor. Divide it out, factorise the quadratic that is left, and you have every root. On P3 this is often a complete question in itself, and it also sits inside partial fractions, integration and equation-solving questions whenever a cubic denominator or a cubic equation appears.

The theorem

The remainder theorem says the remainder on dividing $p(x)$ by $(x - a)$ is $p(a)$ . A factor is exactly a divisor that leaves remainder zero. Put those together:

Key result

Factor theorem. $(x - a)$ is a factor of the polynomial $p(x)$ if and only if $p(a) = 0$ .

More generally, $(ax - b)$ is a factor of $p(x)$ if and only if $p\!\left(\dfrac{b}{a}\right) = 0$ .

Proof

Write $p(x) \equiv (x - a)q(x) + R$ . By the remainder theorem, $R = p(a)$ .

If $p(a) = 0$ , then $p(x) \equiv (x - a)q(x)$ , so $(x - a)$ is a factor.

Conversely, if $(x - a)$ is a factor, then $p(x) \equiv (x - a)q(x)$ , and substituting $x = a$ gives $p(a) = 0$ .

Notice the "if and only if": the theorem works in both directions. " $p(3) = 0$ , so $(x - 3)$ is a factor" and " $(x - 3)$ is a factor, so $p(3) = 0$ " are both correct uses.

In graph terms, $(x - a)$ is a factor exactly when the curve $y = p(x)$ crosses or touches the $x$ -axis at $x = a$ .

y = 2x^3 + 3x^2 - 8x + 3 (-3, 0) (0.5, 0) (1, 0)

The cubic $y = 2x^3 + 3x^2 - 8x + 3 = (x - 1)(2x - 1)(x + 3)$ meets the axis at $x = -3$ , $x = \tfrac{1}{2}$ and $x = 1$ , one crossing for each linear factor.

Which values to try

You cannot test every number, but you do not need to. If $p(x)$ has integer coefficients and $(ax - b)$ is a factor (with $a$ , $b$ integers and no common factor), then $b$ divides the constant term and $a$ divides the leading coefficient. So:

Searching for a linear factor

List the factors of the constant term: these are the candidates for $b$ .
List the factors of the leading coefficient: these are the candidates for $a$ .
Try $x = \pm 1$ , then $\pm 2$ , $\pm 3, \ldots$ (values $\pm\dfrac{b}{a}$ ), until $p(x) = 0$ .
When $p\!\left(\tfrac{b}{a}\right) = 0$ , write down the factor $(ax - b)$ .
Divide it out (long division or comparing coefficients) to get a quadratic quotient.
Factorise the quadratic, or use the discriminant to show it does not factorise.

Tip

$p(1)$ is just the sum of the coefficients, so it takes a second to check. $p(-1)$ is the sum with the signs of the odd powers flipped. Always try these two first.

Show, then factorise

Show that $(x - 1)$ is a factor of $p(x) = 2x^3 + 3x^2 - 8x + 3$ , and hence factorise $p(x)$ completely.

Solution

p(1) = 2 + 3 - 8 + 3 = 0,

so by the factor theorem $(x - 1)$ is a factor.

Compare coefficients in $2x^3 + 3x^2 - 8x + 3 \equiv (x - 1)(2x^2 + cx - 3)$ . The leading $2x^2$ and the constant $-3$ are forced by the $2x^3$ and the $+3$ . The $x^2$ coefficient gives $c - 2 = 3$ , so $c = 5$ . Check the $x$ coefficient: $-3 - c = -8$ , which holds.

p(x) = (x - 1)(2x^2 + 5x - 3) = (x - 1)(2x - 1)(x + 3).

A factor with coefficient not 1

Show that $(2x + 1)$ is a factor of $4x^3 + 8x^2 - x - 2$ , and hence solve the equation $4x^3 + 8x^2 - x - 2 = 0$ .

Solution

$2x + 1 = 0$ when $x = -\tfrac{1}{2}$ :

p\!\left(-\tfrac{1}{2}\right) = 4\left(-\tfrac{1}{8}\right) + 8\left(\tfrac{1}{4}\right) + \tfrac{1}{2} - 2 = -\tfrac{1}{2} + 2 + \tfrac{1}{2} - 2 = 0.

So $(2x + 1)$ is a factor. Compare coefficients in $(2x + 1)(2x^2 + cx - 2)$ : the $x^2$ coefficient gives $2c + 2 = 8$ , so $c = 3$ ; check the $x$ coefficient, $c - 4 = -1$ .

4x^3 + 8x^2 - x - 2 = (2x + 1)(2x^2 + 3x - 2) = (2x + 1)(2x - 1)(x + 2).

Solutions: $x = -\tfrac{1}{2}$ , $x = \tfrac{1}{2}$ , $x = -2$ .

Searching for the first factor

Solve $3x^3 - 4x^2 - 17x + 6 = 0$ .

Solution

Let $p(x) = 3x^3 - 4x^2 - 17x + 6$ . Candidates: $\pm 1, \pm 2, \pm 3, \pm 6$ and these divided by $3$ .

$p(1) = 3 - 4 - 17 + 6 = -12$ . $p(-1) = -3 - 4 + 17 + 6 = 16$ . $p(2) = 24 - 16 - 34 + 6 = -20$ .

$p(-2) = -24 - 16 + 34 + 6 = 0$ . So $(x + 2)$ is a factor.

Compare coefficients in $(x + 2)(3x^2 + cx + 3)$ : the $x^2$ coefficient gives $c + 6 = -4$ , so $c = -10$ ; check the $x$ coefficient, $3 + 2c = -17$ .

3x^2 - 10x + 3 = (3x - 1)(x - 3).

So $p(x) = (x + 2)(3x - 1)(x - 3)$ and the solutions are $x = -2$ , $x = \tfrac{1}{3}$ , $x = 3$ .

Watch out

Once you have found one factor, stop searching and divide. Finding the other roots by more trial and error is slow and misses irrational roots; the quadratic quotient gives them directly, by factorising or by the formula.

Unknown coefficients

If you are told that certain expressions are factors, each one gives an equation $p(\ldots) = 0$ .

Two given factors

$(x - 3)$ and $(x + 1)$ are both factors of $p(x) = x^3 + ax^2 + bx + 6$ . Find $a$ and $b$ , and the third factor.

Solution

$p(3) = 0$ : $27 + 9a + 3b + 6 = 0$ , so $9a + 3b = -33$ , i.e. $3a + b = -11$ .

$p(-1) = 0$ : $-1 + a - b + 6 = 0$ , so $a - b = -5$ .

Adding: $4a = -16$ , $a = -4$ , and then $b = a + 5 = 1$ .

$p(x) = x^3 - 4x^2 + x + 6$ . The product of the two known factors is $(x - 3)(x + 1) = x^2 - 2x - 3$ . The third factor must be $(x + c)$ with $-3c = 6$ (constant terms), so $c = -2$ : the third factor is $(x - 2)$ .

Check: $(x - 3)(x + 1)(x - 2)$ has $x^2$ coefficient $-3 + 1 - 2 = -4$ .

Quadratic factors and quartics

A quadratic $x^2 + px + q$ that factorises as $(x - \alpha)(x - \beta)$ is a factor of $p(x)$ exactly when both $(x - \alpha)$ and $(x - \beta)$ are factors, so you need $p(\alpha) = 0$ and $p(\beta) = 0$ .

A quartic with a quadratic factor

The polynomial $p(x) = x^4 + ax^3 + bx^2 - 8x - 4$ has $x^2 - x - 2$ as a factor.

(a) Find $a$ and $b$ .

(b) Hence factorise $p(x)$ completely and solve $p(x) = 0$ .

Solution

(a) $x^2 - x - 2 = (x - 2)(x + 1)$ , so $p(2) = 0$ and $p(-1) = 0$ .

$p(2) = 16 + 8a + 4b - 16 - 4 = 0 \Rightarrow 8a + 4b = 4 \Rightarrow 2a + b = 1$ .

$p(-1) = 1 - a + b + 8 - 4 = 0 \Rightarrow -a + b = -5$ .

Subtract: $3a = 6$ , so $a = 2$ and $b = -3$ .

(b) $p(x) = x^4 + 2x^3 - 3x^2 - 8x - 4$ . Write $p(x) \equiv (x^2 - x - 2)(x^2 + cx + 2)$ , where the $2$ makes the constant $(-2)(2) = -4$ . The $x^3$ coefficient: $c - 1 = 2$ , so $c = 3$ . Check the $x$ coefficient: $-2 - 2c = -8$ .

p(x) = (x^2 - x - 2)(x^2 + 3x + 2) = (x - 2)(x + 1)(x + 1)(x + 2) = (x - 2)(x + 1)^2(x + 2).

Solutions: $x = 2$ , $x = -1$ (a repeated root) and $x = -2$ .

A repeated factor shows on the graph as a point where the curve touches the axis instead of crossing it.

y = x^4 + 2x^3 - 3x^2 - 8x - 4 (-1, 0) (-2, 0) (2, 0)

How many real roots?

Once a cubic is written as (linear factor) $\times$ (quadratic), the discriminant of the quadratic tells you how many more real roots there are.

Key result

If $p(x) = (x - a)(x^2 + px + q)$ :

$p^2 - 4q$	Real roots of $p(x) = 0$
$> 0$	three (two from the quadratic, plus $a$ , unless one coincides with $a$ )
$= 0$	the quadratic gives one repeated root, plus $a$
$< 0$	only $x = a$

A cubic with one real root

Show that $x^3 + 3x^2 + 4x + 2 = 0$ has exactly one real root, and find it.

Solution

Try $x = -1$ : $-1 + 3 - 4 + 2 = 0$ . So $(x + 1)$ is a factor.

$x^3 + 3x^2 + 4x + 2 \equiv (x + 1)(x^2 + cx + 2)$ : the $x^2$ coefficient gives $c + 1 = 3$ , so $c = 2$ ; check $x$ : $2 + c = 4$ .

$x^2 + 2x + 2$ has discriminant $4 - 8 = -4 < 0$ , so it has no real roots.

The equation has exactly one real root, $x = -1$ .

"Hence" with a substitution

Cambridge likes to reuse a factorised polynomial in a disguised equation. If $p(x) = 0$ has roots $\alpha, \beta, \gamma$ , then $p(e^y) = 0$ means $e^y = \alpha$ , $\beta$ or $\gamma$ , and $p(\cos\theta) = 0$ means $\cos\theta = \alpha$ , $\beta$ or $\gamma$ . Some of these will be impossible.

Hence with exponentials

Using the factorisation $2x^3 + 3x^2 - 8x + 3 = (x - 1)(2x - 1)(x + 3)$ , solve the equation

2e^{3y} + 3e^{2y} - 8e^{y} + 3 = 0.

Solution

This is $p(x) = 0$ with $x = e^y$ , since $e^{3y} = (e^y)^3$ and $e^{2y} = (e^y)^2$ . So

e^y = 1, \quad e^y = \tfrac{1}{2}, \quad\text{or}\quad e^y = -3.

$e^y = 1 \Rightarrow y = 0$ . $e^y = \tfrac{1}{2} \Rightarrow y = \ln\tfrac{1}{2} = -\ln 2$ . $e^y = -3$ is impossible since $e^y > 0$ .

Solutions: $y = 0$ and $y = -\ln 2$ .

Common mistakes

Watch out

Sign of the factor. $p(-2) = 0$ means $(x + 2)$ is a factor, not $(x - 2)$ . And $p\!\left(\tfrac{1}{3}\right) = 0$ means $(3x - 1)$ is a factor.

Watch out

"Factorise completely" but stopping at the quadratic. $(x + 2)(3x^2 - 10x + 3)$ is not complete; the quadratic factorises further. If it does not factorise, say so (show its discriminant is negative, or note it has irrational roots).

Watch out

Not writing the conclusion. "Show that $(x - 1)$ is a factor" needs the working $p(1) = 2 + 3 - 8 + 3 = 0$ and the sentence "so $(x - 1)$ is a factor". The answer is given in the question, so the mark is entirely for the working and the conclusion.

Exam tip

For "show that" questions, substitute, show every term, reach $= 0$ , conclude.
"Hence" means use the factor you have just found or been given. Starting again with a different factor can lose the method marks.
Division by comparing coefficients is quick: fix the leading term and the constant first, then find the middle coefficient from one power and check it with another.
In equations like $p(e^y) = 0$ or $p(\sin\theta) = 0$ , reject impossible values ( $e^y \le 0$ , $|\sin\theta| > 1$ ) explicitly, with the reason.

Summary

$(x - a)$ is a factor of $p(x) \iff p(a) = 0$ ; $(ax - b)$ is a factor $\iff p\!\left(\tfrac{b}{a}\right) = 0$ .
Try $x = \pm 1$ first, then other factors of the constant term divided by factors of the leading coefficient.
After finding one factor, divide to get a quadratic, then factorise it or use its discriminant.
Given factors produce equations $p(\ldots) = 0$ for unknown coefficients.
A quadratic factor $(x - \alpha)(x - \beta)$ needs $p(\alpha) = p(\beta) = 0$ .
A repeated factor means the curve touches the axis there.
In "hence" questions, substitute $x = e^y$ , $2^y$ , $\cos\theta$ , etc., and reject impossible values.

Practice

Question

Factorise $x^3 - 6x^2 + 11x - 6$ completely.
Given that $(x + 2)$ is a factor of $x^3 + kx^2 - 4x - 8$ , find $k$ and factorise the cubic completely.
Solve $2x^3 - x^2 - 13x - 6 = 0$ .
Show that $(3x - 2)$ is a factor of $3x^3 - 2x^2 - 6x + 4$ , and hence find the exact roots of $3x^3 - 2x^2 - 6x + 4 = 0$ .
Show that $(x - 1)$ and $(x + 2)$ are factors of $x^4 + x^3 - x^2 + x - 2$ , and hence show that the equation $x^4 + x^3 - x^2 + x - 2 = 0$ has exactly two real roots.
Given that $2x^3 + 3x^2 - 8x + 3 = (x - 1)(2x - 1)(x + 3)$ , solve $2\cos^3\theta + 3\cos^2\theta - 8\cos\theta + 3 = 0$ for $0^\circ \le \theta \le 360^\circ$ .
The polynomial $p(x) = 6x^3 + ax^2 + bx - 2$ has $(2x + 1)$ as a factor, and leaves remainder $-6$ when divided by $(x + 1)$ . (a) Find $a$ and $b$ . (b) Factorise $p(x)$ completely. (c) Hence solve $6 \times 8^y - 7 \times 4^y - 9 \times 2^y - 2 = 0$ .
The cubic $x^3 + 3x^2 - 4$ has a repeated factor. Find it, factorise the cubic fully, and describe what happens to the graph $y = x^3 + 3x^2 - 4$ at each root.

Answers

$p(1) = 1 - 6 + 11 - 6 = 0$ , so $(x - 1)$ is a factor. $(x - 1)(x^2 - 5x + 6) = (x - 1)(x - 2)(x - 3)$ .
$p(-2) = -8 + 4k + 8 - 8 = 4k - 8 = 0$ , so $k = 2$ . Then $x^3 + 2x^2 - 4x - 8 = x^2(x + 2) - 4(x + 2) = (x + 2)(x^2 - 4) = (x + 2)^2(x - 2)$ .
$p(3) = 54 - 9 - 39 - 6 = 0$ , so $(x - 3)$ is a factor. Comparing coefficients: $(x - 3)(2x^2 + 5x + 2) = (x - 3)(2x + 1)(x + 2)$ . Roots $x = 3$ , $-\tfrac{1}{2}$ , $-2$ .
$p\!\left(\tfrac{2}{3}\right) = 3 \cdot \tfrac{8}{27} - 2 \cdot \tfrac{4}{9} - 4 + 4 = \tfrac{8}{9} - \tfrac{8}{9} = 0$ , so $(3x - 2)$ is a factor. Grouping: $x^2(3x - 2) - 2(3x - 2) = (3x - 2)(x^2 - 2)$ . Roots $x = \tfrac{2}{3}$ , $\sqrt{2}$ , $-\sqrt{2}$ .
$p(1) = 1 + 1 - 1 + 1 - 2 = 0$ and $p(-2) = 16 - 8 - 4 - 2 - 2 = 0$ , so both are factors. Their product is $x^2 + x - 2$ , and $x^4 + x^3 - x^2 + x - 2 \equiv (x^2 + x - 2)(x^2 + 1)$ ; expanding the right-hand side gives $x^4 + x^3 - 2x^2 + x^2 + x - 2$ , which confirms it. $x^2 + 1 = 0$ has no real roots (discriminant $-4 < 0$ ), so the only real roots are $x = 1$ and $x = -2$ : exactly two.
$\cos\theta = 1$ , $\tfrac{1}{2}$ or $-3$ . $\cos\theta = -3$ is impossible since $-1 \le \cos\theta \le 1$ . $\cos\theta = 1$ : $\theta = 0^\circ, 360^\circ$ . $\cos\theta = \tfrac{1}{2}$ : $\theta = 60^\circ, 300^\circ$ . Solutions $0^\circ, 60^\circ, 300^\circ, 360^\circ$ .
(a) $p\!\left(-\tfrac{1}{2}\right) = -\tfrac{3}{4} + \tfrac{a}{4} - \tfrac{b}{2} - 2 = 0$ ; multiply by $4$ : $a - 2b = 11$ . $p(-1) = -6 + a - b - 2 = -6 \Rightarrow a - b = 2$ . Subtracting: $-b = 9$ , so $b = -9$ , $a = -7$ . (b) $6x^3 - 7x^2 - 9x - 2 \equiv (2x + 1)(3x^2 + cx - 2)$ ; $x^2$ : $2c + 3 = -7$ , $c = -5$ ; check $x$ : $-4 + c = -9$ . Then $3x^2 - 5x - 2 = (3x + 1)(x - 2)$ , so $p(x) = (2x + 1)(3x + 1)(x - 2)$ . (c) With $x = 2^y$ (since $8^y = (2^y)^3$ and $4^y = (2^y)^2$ ): $2^y = -\tfrac{1}{2}$ or $-\tfrac{1}{3}$ (impossible, $2^y > 0$ ) or $2^y = 2$ . So $y = 1$ .
$p(1) = 1 + 3 - 4 = 0$ , so $(x - 1)$ is a factor: $x^3 + 3x^2 - 4 = (x - 1)(x^2 + 4x + 4) = (x - 1)(x + 2)^2$ . The repeated factor is $(x + 2)$ . The graph crosses the $x$ -axis at $x = 1$ and touches it at $x = -2$ (a local maximum on the axis, since the curve is negative on both sides of $-2$ near there: $p(-3) = -4$ , $p(-1) = -2$ ).

The theorem

Which values to try

Unknown coefficients

Quadratic factors and quartics

How many real roots?

"Hence" with a substitution

Common mistakes

Summary

Practice

Console