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

1 Answer
Jul 6, 2016

Answer:

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)#