Pure · Algebra & functions

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Polynomials and the factor theorem

The factor theorem — why f(a) = 0 means (x − a) is a factor — and how to use it with algebraic division to factorise and solve cubics.

Pure · Algebra & functions

Polynomials and the factor theorem

The factor theorem — why f(a) = 0 means (x − a) is a factor — and how to use it with algebraic division to factorise and solve cubics.

Why it works

A quadratic you can usually factorise by inspection. A cubic like

x^3 - 3x^2 + 4

is harder — you need a way to find a first factor before the rest falls out. That's what the factor theorem gives you.

The factor theorem. Divide a polynomial

f(x)

(x - a)

. Whatever happens, you can write

f(x) = (x - a)\,q(x) + r,

where

q(x)

is the quotient and

r

is a constant remainder. Now substitute

x = a

: the

(x-a)

term becomes zero and takes

q(x)

with it, leaving

f(a) = r.

So the value $f(a)$ is the remainder. And

(x - a)

divides exactly — i.e. is a factor — precisely when that remainder is zero. Hence:

(x - a) \text{ is a factor of } f(x) \iff f(a) = 0.

Mind the sign: to test the factor

(x - 3)

you substitute

x = 3

; to test

(x + 2)

you substitute

x = -2

(the value that makes the bracket zero).

Algebraic division then peels that factor off. To divide

f(x)

by a known factor

(x - a)

, write

f(x) = (x - a)(\dots)

and find the missing quadratic by matching coefficients — being careful to keep a placeholder for any missing power (a "

+0x

" you might not see written).

The strategy for a cubic: spot one root by trying small values (

x = \pm1, \pm2, \dots

) until

f(a) = 0

, divide out

(x - a)

to leave a quadratic, then factorise that quadratic the usual way. Don't stop at the first root — a cubic has up to three.

Here is

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

. The repeated factor

(x-2)^2

is why the curve touches the axis at

x = 2

rather than crossing it: