How do you factor #128x^3 - 1024#?

1 Answer
Feb 6, 2017

Answer:

#128(x-2)(x^2+2x+4)#

Explanation:

The 2 terms have a #color(blue)"common factor"# of 128, which can be taken out.

#rArr128(x^3-8)to(A)#

#x^3-8# is a #color(blue)"difference of cubes"# which, in general, is factorised as shown.

#color(red)(bar(ul(|color(white)(2/2)color(black)(a^3-b^3=(a-b)(a^2+ab+b^2))color(white)(2/2)|)))#

using a = x and b = 2, then

#rArrx^3-8=(x-2)(x^2+2x+2^2)=(x-2)(x^2+2x+4)#

Returning to (A) the complete factorisation is.

#128x^3-1024=128(x-2)(x^2+2x+4)#