Question

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

Answer 1

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

Explanation:

The first step is to remove the common factor a.

`rArra(10x^2-23x-5)`

To factorise `10x^2-23x-5` consider the following.

The `color(blue)"standard form of the quadratic function"` is

`color(red)(|bar(ul(color(white)(a/a)color(black)(ax^2+bx+c)color(white)(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 `=10xx-5=-50`

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

Now express `10x^2-23x-5` as

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

and factorising each 'pair' to obtain.

`color(blue)"2x"(5x+1)-color(blue)"5"(5x+1)`

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

`(5x+1)(color(blue)"2x-5")`

`rArr10ax^2-23ax-5a=a(5x+1)(2x-5)`