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

1 Answer
May 22, 2016

Answer:

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

Explanation:

Group the terms in 'pairs'

thus #[5x^3+6x^2]+[-45x-54]#

now factorise each pair

#rArrx^2(5x+6)-9(5x+6)#

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

#(5x+6)(x^2-9)#

Now #x^2-9" is a difference of squares"#

In general a difference of squares factorises as.

#color(red)(|bar(ul(color(white)(a/a)color(black)(a^2-b^2=(a-b)(a+b))color(white)(a/a)|)))#

here #x^2=(x)^2" and " 9=(3)^2rArra=x ,b=3#

#rArrx^2-9=(x-3)(x+3)#

Putting this altogether to obtain.

#5x^3+6x^2-45x-54=(5x+6)(x-3)(x+3)#