Name: Application of the Remainder Theorem: Example 2
Uploaded: 2019-07-19
Duration: 4 min 59 s
Description: What are the conditions to be a factor of a polynomial? Can you guess what the condition for x+1 to be a Factor of a Polynomial is? Watch the video to learn...

Question

Use factor theorem to factorise the following polynomials completely:

$4 x^{3} + 4 x^{2} - 9 x - 9$ and $x^{3} - 19 x - 30$

Accepted Answer

Let $4 x^{3} + 4 x^{2} - 9 x - 9$ .

$∴ p (- 1) = 4 {(- 1)}^{3} + 4 {(- 1)}^{2} - 9 (- 1) - 9$

$= - 4 + 4 + 9 - 9$

$= 0$

If $f (x)$ is a polynomial and $a$ is a real number, then $x + a$ is a factor of $f (x)$ , if and only if $f (- a) = 0$ .

Therefore, $x + 1$ is a factor of $4 x^{3} + 4 x^{2} - 9 x - 9$ .

$∴ 4 x^{3} + 4 x^{2} - 9 x - 9$

$= 4 x^{2} (x + 1) - 9 (x + 1)$

$= (x + 1) (4 x^{2} - 9)$

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

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

Let $g (x) = x^{3} - 19 x - 30$ .

$∴ g (- 2) = {(- 2)}^{3} - 19 (- 2) - 30$

$= - 8 + 38 - 30$

$= 0$

Therefore, $x + 2$ is a factor of $x^{3} - 19 x - 30$ .

Thus, $x^{3} - 19 x - 30$

$= x^{3} - 2 x^{2} - 15 x + 2 x^{2} - 4 x - 30$

$= x (x^{2} - 2 x - 15) + 2 (x^{2} - 2 x - 15)$

$= (x + 2) (x^{2} - 2 x - 15)$

Hence, $4 x^{3} + 4 x^{2} - 9 x - 9 = (x + 1) (2 x - 3) (2 x + 3)$ and $x^{3} - 19 x - 30 = (x + 2) (x^{2} - 2 x - 15)$ .

Use factor theorem to factorise the following polynomials completely:

$4 x^{3} + 4 x^{2} - 9 x - 9$ and $x^{3} - 19 x - 30$

Important Questions on Factorisation

Learn and Prepare for any exam you want !

Use factor theorem to factorise the following polynomials completely: 4x3+4x2-9x-9 and x3-19x-30

Important Questions on Factorisation

Learn and Prepare for any exam you want !

Use factor theorem to factorise the following polynomials completely:

$4 x^{3} + 4 x^{2} - 9 x - 9$ and $x^{3} - 19 x - 30$