Question

How do you factor `8x^6 + 7x^3 - 1`?

Answer 1

`(1 x^3 +1)(8 x^3-1)`

Explanation:

This is in the form of a quadratic because.

`8(x^3)^2 + 7(x^3)^1 -1`

Find the factors of 8 (and 1) which subtract to give 7

`8-1 = 7`

The signs will be different (because of -1)
There must be more positives (because of +7)

The factors of -1 can only be +1 and -1

`(? x^3 +1)(? x^3-1)`

Place the 8 and the 1 so that we will get +8 and -1. when we multiply.

`(1 x^3 +1)(8 x^3-1)`

Answer 2

Set `t=x^3` then the given equation is transformed to

`8t^2+7t-1`

`8t^2+(8-1)t-1`

`8t^2+8t-t-1`

`8t(t+1)-(t+1)`

`(t+1)*(8*t-1)`

Reverse to x we have that

`(1+x^3)*(8*x^3-1)`

Now `x^3+1` can be factored further as follows

`x^3+1=(x+1)(x^2-x+1)`

and `8*x^3-1` can be factored as follows

`8x^3-1=(2 x-1) (4 x^2+2 x+1)`

Hence finally we get

`8x^6+7x^3-1=(x+1)*(x^2-x+1)*(2*x-1)*(4*x^2+2 x+1)`