Question

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

Answer 1

`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.

Answer 2

`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)`