Name: Factorisation Using Difference of Two Squares
Uploaded: 2019-07-19
Duration: 5 min 14 s
Description: Sometimes writing a polynomial as the difference of two squares is the only way to factorise it. So let’s learn how to do it in this simple video!

Question

Factorise: $x^{3} - 3 x^{2} - 9 x - 5$

Accepted Answer

According to factor theorem, if $f (x)$ is a polynomial of degree $n \geq 1$ and $' a'$ is any real number, then, $(x - a)$ is a factor of $f (x)$ , if $f (a) = 0$ .

Let $p (x) = x^{3} - 3 x^{2} - 9 x - 5$

Putting $x = - 1$ in $p (x)$ , we obtain,

$p (- 1) = {(- 1)}^{3} - 3 {(- 1)}^{2} - 9 (- 1) - 5 = - 1 - 3 + 9 - 5 = 0$

Hence, $(x + 1)$ is a factor of $p (x)$ .

We shall find the other factor by dividing $p (x) = x^{3} - 3 x^{2} - 9 x - 5$ by $(x + 1)$ using the long division method.

$\begin{matrix} x^{2} & - & 4 x & - & 5 \\ x + 1 & x^{3} & - & 3 x^{2} & - & 9 x & - & 5 \\ x^{3} & + & x^{2} \\ - & - \\ - & 4 x^{2} & - & 9 x & - & 5 \\ - & 4 x^{2} & - & 4 x \\ + & + \\ - & 5 x & - & 5 \\ - & 5 x & - & 5 \\ + & + \\ 0 \end{matrix}$

So, the other factor of $p (x)$ is $x^{2} - 4 x - 5$ .

Thus, we have, $p (x) = (x + 1) (x^{2} - 4 x - 5)$

$= (x + 1) (x^{2} - 5 x + x - 5) = (x + 1) [x (x - 5) + 1 (x - 5)] = (x + 1) (x - 5) (x + 1)$

Therefore, the required answer is $(x + 1) (x + 1) (x - 5)$ .

Factorise: $x^{3} - 3 x^{2} - 9 x - 5$

Important Questions on Polynomials

Learn and Prepare for any exam you want !

Factorise: x3-3x2-9x-5

Important Questions on Polynomials

Learn and Prepare for any exam you want !

Factorise: $x^{3} - 3 x^{2} - 9 x - 5$