How do you factor #4x^3 + 32#?

1 Answer
Aug 25, 2016

Answer:

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

Explanation:

First step is to take out a #color(blue)"common factor"# of 4.

#rArr4x^3+32=4(x^3+8)........ (A)#

Now, #x^3+8 color(blue)" is a sum of cubes"# and in general is factorised as follows.

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

now, #(x)^3=x^3" and " (2)^3=8rArra=x" and "b=2#

substitute these values for a and b into (B)

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

substitute back into (A)

#rArr4x^3+32=4(x+2)(x^2-2x+4)#