Question

Is it possible to factor `y=4x^3-13x-6`? If so, what are the factors?

Answer 1

`4x^3-13x-6=(x-2)(2x+3)(2x+1)`

Explanation:

Since there is nothing obvious to factor out at the beginning, start by plugging in small numbers for `x` to see if you can get the equation to equal `0`.

Plugging in `x=2` gives `0`, so `x-2` is a factor.

Now divide `4x^3-13x-6` by `x-2` using polynomial long division or synthetic division. If you need help with this step, say something.

`(4x^3-13x-6)/(x-2)=4x^2+8x+3`

So, from here we know that

`4x^3-13x-6=(x-2)(4x^2+8x+3)`

So now all we need to do is factor `4x^2+8x+3`. Looking for a pair of factors of `12` whose sum is `8`, we see that `2,6` works:

`4x^2+8x+3`

`=4x^2+2x+6x+3`

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

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

Thus,

`4x^3-13x-6=(x-2)(2x+3)(2x+1)`