Question

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

Answer 1

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

`color(white)(2x^4+2x^3-2x^2-x) = 2x(x-x_0)(x-x_1)(x-x_2)`

where:

`x_n = 1/3(4 cos(1/3 cos^(-1)(5/32) + (2npi)/3) - 1)" "`

Explanation:

All of the terms are divisible by `x`, so we can separate that out as a factor first:

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

Let:

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

This cubic is not easy to factor, so I have posted it as a Precalculus question, with solution https://socratic.org/s/az7KUZgd

`f(x)` has zeros:

`x_n = 1/3(4 cos(1/3 cos^(-1)(5/32) + (2npi)/3) - 1)" "` for `n = 0, 1, 2`

and hence factorises as:

`f(x) = 2 (x-x_0)(x-x_1)(x-x_2)`

So putting it all together, we find:

`2x^4+2x^3-2x^2-x = 2x(x-x_0)(x-x_1)(x-x_2)`

where:

`x_n = 1/3(4 cos(1/3 cos^(-1)(5/32) + (2npi)/3) - 1)" "`