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

1 Answer
Jul 30, 2017

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.