# How do you factor 5x^3+6x^2-45x-54?

May 22, 2016

(5x +6)(x-3)(x+3)

#### Explanation:

Group the terms in 'pairs'

thus $\left[5 {x}^{3} + 6 {x}^{2}\right] + \left[- 45 x - 54\right]$

now factorise each pair

$\Rightarrow {x}^{2} \left(5 x + 6\right) - 9 \left(5 x + 6\right)$

We now have a common factor of (5x + 6) and taking it out leaves.

$\left(5 x + 6\right) \left({x}^{2} - 9\right)$

Now ${x}^{2} - 9 \text{ is a difference of squares}$

In general a difference of squares factorises as.

$\textcolor{red}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{{a}^{2} - {b}^{2} = \left(a - b\right) \left(a + b\right)} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

here ${x}^{2} = {\left(x\right)}^{2} \text{ and } 9 = {\left(3\right)}^{2} \Rightarrow a = x , b = 3$

$\Rightarrow {x}^{2} - 9 = \left(x - 3\right) \left(x + 3\right)$

Putting this altogether to obtain.

$5 {x}^{3} + 6 {x}^{2} - 45 x - 54 = \left(5 x + 6\right) \left(x - 3\right) \left(x + 3\right)$