How do you factor: y= 32x^3 - 4 ?

2 Answers
Apr 14, 2017

y=4(2x-1)(4x^2+2x+1)

Explanation:

Recall:

(a^3-b^3)=(a-b)(a^2+ab+b^2)

Here we go.

y=32x^3-4

By factoring out 4,

Rightarrow y=4(8x^3-1)

By rewriting a bit,

Rightarrow y=4((2x)^3-1^3)

By applying the formula at the top with a=2x and b=1,

Rightarrow y=4(2x-1)((2x)^2+2x cdot 1+1^2)

By cleaning up a bit,

y=4(2x-1)(4x^2+2x+1)

I hope that this was clear.

Apr 14, 2017

y=4(2x-1)(4x^2+2x+1)

Explanation:

The first step is to factor out color(blue)"common factor" of 4

rArry=4(8x^3-1)to(1)

now 8x^3-1 is a color(blue)"difference of cubes" and factorises in general as.

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)|)))

8x^3=(2x)^3" and " 1^3=1

"using " a=2x" and " b=1

rArr8x^3-1=(2x-1)(4x^2+2x+1)

"going back to " (1)

rArry=4(2x-1)(4x^2+2x+1)