Factorise x³ + 2x² − 5x − 6 completely using the factor theorem.

(x + 1)(x − 2)(x + 3). Test small values: p(−1) = −1 + 2 + 5 − 6 = 0, so (x + 1) is a factor. Dividing gives the quotient x² + x − 6 = (x − 2)(x + 3). Hence p(x) = (x + 1)(x − 2)(x + 3), with zeros −1, 2 and −3.

Factorise $x^{3} + 2 x^{2} - 5 x - 6$ completely using the factor theorem.

Class 9PolynomialsPolynomials

A $(x - 1) (x + 2) (x + 3)$
B $(x + 1) (x - 2) (x + 3)$
C $(x + 1) (x + 2) (x - 3)$
D $(x - 1) (x - 2) (x + 3)$

Step-by-Step Solution

Test small values:

p (- 1) = - 1 + 2 + 5 - 6 = 0

, so

(x + 1)

is a factor. Dividing gives the quotient

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

. Hence

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

, with zeros −1, 2 and −3.

Answer: B)

(x + 1) (x - 2) (x + 3)

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 x3+2x2−5x−6 completely using the factor theorem.

Step-by-Step Solution

Key idea: Factor Theorem

Master this chapter in 30 minutes a day.

More Class 9 Polynomials practice

Factorise $x^{3} + 2 x^{2} - 5 x - 6$ completely using the factor theorem.