How do you factor #6x²+4x-2#?

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Answer 1 · 2016-06-19T10:42:16

You have the factorization #6(x+1)(3x-1)#

To factor a trinomial, you simply have to look for its roots. There are three possible cases:

The trinomial has no solutions. Then, it is not possible to factor it.
The trinomial has only one solution #x_0#. Then, it is a square of a binomial, more precisely, it is #a(x-x_0)^2#.
The trinomial has two different solutions #x_1# and #x_2#. Then, you can factor is as the product of two binomials, i.e. #a(x-x_1)(x-x_2)#.

The quantity that tells us how many solutions a trinomial has is its discriminant: if the trinomial is #ax^2+bx+c#, its discriminant is

#Delta=b^2-4ac#

If #Delta<0# then we are in case one, if it equals zero we are in case two, if #Delta>0# we are in case three.

In your case,

#Delta=b^2-4ac=4^2-4*6*(-2)=16+48=64>0#

Your trinomial has two solutions, which are given by the formula

#x_{1,2}=\frac{-b\pm\sqrt{\Delta}}{2a}=(-4\pm\sqrt(64))/12=(-4\pm8)/12#

The two possible choices given by the #\pm# sign give us #x_1=(-4-8)/12=-12/12=-1# and #x_2=(-4+8)/12=4/12=1/3#