Question

How do you factor `x^3-2x^2+4x-8`?

Answer 1

The factored form is `(x^2+4)(x-2)`.

Explanation:

You have to factor by grouping:

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

`=color(blue)(x^2)*x-color(blue)(x^2)*2+4x-8`

`=color(blue)(x^2)(x-2)+4x-8`

`=color(blue)(x^2)(x-2)+color(red)4*x+color(red)4*-2`

`=color(blue)(x^2)(x-2)+color(red)4(x-2)`

`=(color(blue)(x^2)+color(red)4)(x-2)`

Answer 2

This factors to `(x^2+4)(x-2)`

Explanation:

Starting with the left-hand side, we see `x^3`; lets factor that first.

Based on the middle terms, I would believe that the factors are uneven (NOT `x^(3/2)`) This is because that if it WAS even, there wouldn't be two terms in the middle of the equation. I begin to write the factors as such:

`(x^2+a)(x+b)`

Now, expand the above and back-substitute the values of a & b from the starting equation:

`(x^2+a)(x+b)=x^3+bx^2+ax+ab`

Comparing the middle coefficients, it would seem that b=-2 and a=4. This is confirmed with the final term `ab=4*(-2)=-8`

Therefore, the factored solution is:
`(x^2+4)(x-2)`