Without dividing, find the remainder when p(x) = x³ + 1 is divided by (x + 1).

0. (x + 1) → x = −1. p(−1) = (−1)³ + 1 = −1 + 1 = 0. The remainder is 0, so (x + 1) is a factor.

Without dividing, find the remainder when $p (x) = x^{3} + 1$ is divided by $(x + 1)$ .

Class 9PolynomialsPolynomials

(x + 1)

→

x = - 1. p (- 1) = (- 1)^{3} + 1 = - 1 + 1 = 0

. The remainder is 0, so

(x + 1)

is a factor.

Answer: D) 0

(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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