The remainder theorem from the previous lesson has a corollary on which all polynomial factoring rests: if f(a) = 0, then x − a is a factor of f(x). That gives a simple recipe for factoring a cubic. First find one root, usually among the small whole numbers. Then divide by the corresponding binomial, which drops the degree to two. What is left is a quadratic, and the quadratic formula finishes the job.

The factor theorem

If the polynomial f(x) is divided by x − a, the remainder is f(a). If f(a) = 0, then x − a is a factor of f(x).

First example

f(x) = 2x^{3} + x^{2} - 13x + 6

(2x^{3} + x^{2} - 13x + 6) : (x - 2)

f(x) = [(2x + 1)x - 13]x + 6

f(2) = [(2\cdot 2 + 1)\cdot 2 - 13]\cdot 2 + 6 = (-3)\cdot 2 + 6 = -6 + 6 = 0

(2x^{3} + x^{2} - 13x + 6) : (x - 2) = 2x^{2} + 5x - 3

$2x^{3}$	$- 4x^{2}$
0	$5x^{2}$	$- 13x$
	$5x^{2}$	$- 10x$
	0	$- 3x$	$+ 6$
		$- 3x$	$+ 6$
			$0$

(x - 2)(2x^{2} + 5x - 3) = 2x^{3} + x^{2} - 13x + 6

f(x) = 2x^{3} + x^{2} - 13x + 6 = (x - 2)(2x^{2} + 5x - 3)

The quadratic equation

f(x) = ax^{2} + bx + c

ax^{2} + bx + c = 0,\quad a \neq 0

x = \frac{-b \pm \sqrt{b^{2} - 4ac}}{2a}

a quadratic equation has two solutions

x = \frac{-5 \pm \sqrt{5^{2} - 4\cdot 2\cdot(-3)}}{2\cdot 2} = \frac{-5 \pm \sqrt{25 + 24}}{4} = \frac{-5 \pm \sqrt{49}}{4} = \frac{-5 \pm 7}{4}

x_{1} = -3

x_{2} = \tfrac{1}{2}

ax^{2} + bx + c = a(x - x_{1})(x - x_{2})

2x^{2} + 5x - 3 = 2(x + 3)\left(x - \tfrac{1}{2}\right) = (x + 3)(2x - 1)

f(x) = 2x^{3} + x^{2} - 13x + 6 = (x - 2)(2x^{2} + 5x - 3) = (x - 2)(x + 3)(2x - 1)

Second example

f(x) = x^{3} - 5x^{2} - 2x + 24

(x^{3} - 5x^{2} - 2x + 24) : (x - 3)

f(x) = [(x - 5)x - 2]x + 24 \quad\Rightarrow\quad f(3) = [(3 - 5)\cdot 3 - 2]\cdot 3 + 24 = -24 + 24 = 0

(x^{3} - 5x^{2} - 2x + 24) : (x - 3) = x^{2} - 2x - 8

$x^{3}$	$- 3x^{2}$
0	$- 2x^{2}$	$- 2x$
	$- 2x^{2}$	$+ 6x$
	0	$- 8x$	$+ 24$
		$- 8x$	$+ 24$
			$0$

(x - 3)(x^{2} - 2x - 8) = x^{3} - 5x^{2} - 2x + 24

f(x) = x^{3} - 5x^{2} - 2x + 24 = (x - 3)(x^{2} - 2x - 8)

x = \frac{2 \pm \sqrt{(-2)^{2} - 4\cdot 1\cdot(-8)}}{2\cdot 1} = \frac{2 \pm \sqrt{4 + 32}}{2} = \frac{2 \pm 6}{2}

x_{1} = -2

x_{2} = 4

x^{2} - 2x - 8 = (x + 2)(x - 4)

f(x) = x^{3} - 5x^{2} - 2x + 24 = (x - 3)(x^{2} - 2x - 8) = (x - 3)(x + 2)(x - 4)

Third example

f(x) = x^{3} - 6x^{2} - 7x + 60

(x^{3} - 6x^{2} - 7x + 60) : (x - 4)

f(x) = [(x - 6)x - 7]x + 60 \quad\Rightarrow\quad f(4) = [(4 - 6)\cdot 4 - 7]\cdot 4 + 60 = -60 + 60 = 0

(x^{3} - 6x^{2} - 7x + 60) : (x - 4) = x^{2} - 2x - 15

$x^{3}$	$- 4x^{2}$
0	$- 2x^{2}$	$- 7x$
	$- 2x^{2}$	$+ 8x$
	0	$- 15x$	$+ 60$
		$- 15x$	$+ 60$
			$0$

(x - 4)(x^{2} - 2x - 15) = x^{3} - 6x^{2} - 7x + 60

x^{2} - 2x - 15 = (x + 3)(x - 5)

(x + a)(x − b) = x² + (a − b)x − ab

f(x) = x^{3} - 6x^{2} - 7x + 60 = (x - 4)(x^{2} - 2x - 15) = (x - 4)(x + 3)(x - 5)

When the first factor is not given

We look for a value k with f(k) = 0, trying the small whole numbers in turn and evaluating with the nested form - that is only a few multiplications per attempt.

f(x) = x^{3} + 5x^{2} - 2x - 24 = [(x + 5)x - 2]x - 24

k = 1 \;\Rightarrow\; [(1 + 5)\cdot 1 - 2]\cdot 1 - 24 = -20 \neq 0

k = -1 \;\Rightarrow\; [(-1 + 5)\cdot(-1) - 2]\cdot(-1) - 24 = -18 \neq 0

k = 2 \;\Rightarrow\; [(2 + 5)\cdot 2 - 2]\cdot 2 - 24 = 0

(x^{3} + 5x^{2} - 2x - 24) : (x - 2) = x^{2} + 7x + 12

$x^{3}$	$- 2x^{2}$
0	$7x^{2}$	$- 2x$
	$7x^{2}$	$- 14x$
	0	$12x$	$- 24$
		$12x$	$- 24$
			$0$

(x - 2)(x^{2} + 7x + 12) = x^{3} + 5x^{2} - 2x - 24

x^{2} + 7x + 12 = (x + 3)(x + 4)

x² + (a + b)x + ab = (x + a)(x + b)

f(x) = x^{3} + 5x^{2} - 2x - 24 = (x - 2)(x + 3)(x + 4)

The same method with a leading coefficient other than one

f(x) = 2x^{3} - 9x^{2} + 7x + 6 = [(2x - 9)x + 7]x + 6

k = 1 \;\Rightarrow\; [(2\cdot 1 - 9)\cdot 1 + 7]\cdot 1 + 6 = 6 \neq 0

k = -1 \;\Rightarrow\; [(2\cdot(-1) - 9)\cdot(-1) + 7]\cdot(-1) + 6 = -12 \neq 0

k = 2 \;\Rightarrow\; [(2\cdot 2 - 9)\cdot 2 + 7]\cdot 2 + 6 = 0

(2x^{3} - 9x^{2} + 7x + 6) : (x - 2) = 2x^{2} - 5x - 3

$2x^{3}$	$- 4x^{2}$
0	$- 5x^{2}$	$+ 7x$
	$- 5x^{2}$	$+ 10x$
	0	$- 3x$	$+ 6$
		$- 3x$	$+ 6$
			$0$

(x - 2)(2x^{2} - 5x - 3) = 2x^{3} - 9x^{2} + 7x + 6

x = \frac{5 \pm \sqrt{(-5)^{2} - 4\cdot 2\cdot(-3)}}{2\cdot 2} = \frac{5 \pm \sqrt{25 + 24}}{4} = \frac{5 \pm 7}{4}

x_{1} = 3

x_{2} = -\tfrac{1}{2}

2x^{2} - 5x - 3 = 2(x - 3)\left(x + \tfrac{1}{2}\right) = (x - 3)(2x + 1)

2x^{3} - 9x^{2} + 7x + 6 = (x - 2)(2x^{2} - 5x - 3) = (x - 2)(x - 3)(2x + 1)

PhiBoard is a free online whiteboard for maths

This lesson is part of it. On the board itself you can draw a number line, a graph or a figure, write a formula and solve a problem together with a teacher or a friend - live, in the browser. Our app is completely free.

Create a free account

See what the board can do

Check whether you really know it

Theory is half the way. Work through a few exercises on the same material - each with a step-by-step solution, no account and no fees.

Go to the exercises

Frequently asked questions

What does the factor theorem say?

That if f(a) = 0, then x − a is a factor of f(x). It follows from the remainder theorem: the remainder on division by x − a equals f(a), so a zero value means a zero remainder.

How do you find the first root when nobody gives it to you?

Try the small whole numbers in turn: 1, −1, 2, −2, 3 and so on. It pays to start with the divisors of the constant term, because only those can be whole-number roots.

What is the quadratic formula?

For ax² + bx + c = 0 with a ≠ 0 we have x = (−b ± √(b² − 4ac)) / (2a). A quadratic equation in general form has two solutions.

How do you write a quadratic as a product once you know its roots?

With ax² + bx + c = a(x − x₁)(x − x₂). For 2x² + 5x − 3 the roots are −3 and 1/2, so 2x² + 5x − 3 = 2(x + 3)(x − 1/2) = (x + 3)(2x − 1).

A lesson from the PhiBoard knowledge base: phiboard.com/en/knowledge-base/from-zero-to-engineer/the-factor-theorem. On the lesson page you will also find games and practice exercises - free and without an account.

← Previous

Horner's method and the remainder theorem

Factoring fourth-degree polynomials