Question

How do you factor `6x ^ { 3} + 2x ^ { 2} - 3x - 1`?

Answer 1

In terms of a factorization involving integers, `6x^3+2x^2-3x-1=(3x+1)(2x^{2}-1)`. It can be factored in terms of linear factors if you allow irrational numbers: `6x^3+2x^2-3x-1=2(3x+1)(x-1/sqrt(2))(x+1/sqrt(2))`.

Explanation:

Through graphing the function `f(x)=6x^3+2x^2-3x-1` and the Rational Root Theorem , you'll see that it looks like `x=-1/3` is a rational root of `f(x)`, which you can confirm like this:

`f(-1/3)=6*(-1/3)^3+2(-1/3)^2-3(-1/3)-1`

`=-6/27+2/9+1-1=-2/9+2/9=0`.

Then, after using either long division or synthetic division , you will find that `6x^3+2x^2-3x-1=(3x+1)(2x^{2}-1)`.

If you want to factor in terms of irrational numbers as done above, note that `\pm 1/sqrt(2)` are the roots of `2x^{2}-1`.