Factorise the cubic x³ − 3x² − x + 3 completely.

(x − 1)(x + 1)(x − 3). p(1) = 1 − 3 − 1 + 3 = 0, so (x − 1) is a factor. Dividing gives x² − 2x − 3 = (x + 1)(x − 3). So (x − 1)(x + 1)(x − 3).

Factorise the cubic $x^{3} - 3 x^{2} - x + 3$ completely.

Class 9PolynomialsPolynomials

p (1) = 1 - 3 - 1 + 3 = 0

, so

(x - 1)

is a factor. Dividing gives

x^{2} - 2 x - 3 = (x + 1) (x - 3)

. So

(x - 1) (x + 1) (x - 3)

Answer: A)

(x - 1) (x + 1) (x - 3)

(x - a)

is a factor of

p (x)

if and only if

p (a) = 0

.**

It's a special case of the remainder theorem (with remainder 0).

Use it to test factors:
Is

(x + 3)

a factor of

p (x) = x^{2} + 5 x + 6

?
Substitute

x = - 3

(note:

x + 3 = 0

→

x = - 3) . p (- 3) = 9 - 15 + 6 = 0

. ✓ Yes,

(x + 3)

is a factor.

This is the standard way to factor cubics — guess and test small zeros.

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