Question

How do you factor the trinomial `6x^2 +8x+ 2`?

Answer 1

You can factor `6x^2+8x+2` as `6(x+1)(3x+1)`

Explanation:

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

1. The trinomial has no solutions. Then, it is not possible to factor it.
2. 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`.
3. 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=8^2-4*6*2=64-48=16>0`

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

`x_{1,2}=\frac{-b\pm\sqrt{\Delta}}{2a}=(-8\pm\sqrt(16))/12=(-8\pm4)/12`

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