# How do you factor completely 10ax^2 - 23ax - 5a?

Jul 6, 2016

a(5x +1)(2x -5)

#### Explanation:

The first step is to remove the common factor a.

$\Rightarrow a \left(10 {x}^{2} - 23 x - 5\right)$

To factorise $10 {x}^{2} - 23 x - 5$ consider the following.

The $\textcolor{b l u e}{\text{standard form of the quadratic function}}$ is

$\textcolor{red}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{a {x}^{2} + b x + c} \textcolor{w h i t e}{\frac{a}{a}} |}}}$
and to factorise this

Consider the factors of the 'product' ac which also sum to give b, the coefficient of the x term.

here a = 10 , b = - 23 and c = -5

hence ac $= 10 \times - 5 = - 50$

The factors of -50 which also sum to give -23 are -25 and +2

Now express $10 {x}^{2} - 23 x - 5$ as

10x^2+(color(red)"2x"-color(red)"25x")-5" [ -23x=2x-25x]"

and factorising each 'pair' to obtain.

$\textcolor{b l u e}{\text{2x"(5x+1)-color(blue)"5}} \left(5 x + 1\right)$

Taking out the common factor of (5x + 1)

$\left(5 x + 1\right) \left(\textcolor{b l u e}{\text{2x-5}}\right)$

$\Rightarrow 10 a {x}^{2} - 23 a x - 5 a = a \left(5 x + 1\right) \left(2 x - 5\right)$