Factorise 2x³ − 3x² − 3x + 2 completely.

(x + 1)(2x − 1)(x − 2). p(−1) = −2 − 3 + 3 + 2 = 0, so (x + 1) is a factor. Dividing by (x + 1) leaves 2x² − 5x + 2, which splits as 2x² − 4x − x + 2 = (2x − 1)(x − 2). So p(x) = (x + 1)(2x − 1)(x − 2).

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

Class 9PolynomialsPolynomials

A $(x - 1) (2 x + 1) (x + 2)$
B $(x + 1) (2 x + 1) (x - 2)$
C $(x + 1) (2 x - 1) (x - 2)$
D $(x + 1) (x - 1) (2 x - 2)$

Step-by-Step Solution

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

, so

(x + 1)

is a factor. Dividing by

(x + 1)

leaves

2 x^{2} - 5 x + 2

, which splits as

2 x^{2} - 4 x - x + 2 = (2 x - 1) (x - 2)

. So

p (x) = (x + 1) (2 x - 1) (x - 2)

Answer: C)

(x + 1) (2 x - 1) (x - 2)

Key idea: Factor Theorem

(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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Factorise 2x3−3x2−3x+2 completely.

Step-by-Step Solution

Key idea: Factor Theorem

Master this chapter in 30 minutes a day.

More Class 9 Polynomials practice

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