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

Jun 19, 2016

You can factor $6 {x}^{2} + 8 x + 2$ as $6 \left(x + 1\right) \left(3 x + 1\right)$

#### 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 $a {\left(x - {x}_{0}\right)}^{2}$.
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. $a \left(x - {x}_{1}\right) \left(x - {x}_{2}\right)$.

The quantity that tells us how many solutions a trinomial has is its discriminant: if the trinomial is $a {x}^{2} + b x + c$, its discriminant is

$\Delta = {b}^{2} - 4 a c$

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.

$\Delta = {b}^{2} - 4 a c = {8}^{2} - 4 \cdot 6 \cdot 2 = 64 - 48 = 16 > 0$
${x}_{1 , 2} = \setminus \frac{- b \setminus \pm \setminus \sqrt{\setminus \Delta}}{2 a} = \frac{- 8 \setminus \pm \setminus \sqrt{16}}{12} = \frac{- 8 \setminus \pm 4}{12}$
The two possible choices given by the $\setminus \pm$ sign give us ${x}_{1} = \frac{- 8 - 4}{12} = - \frac{12}{12} = - 1$ and ${x}_{2} = \frac{- 8 + 4}{12} = - \frac{4}{12} = - \frac{1}{3}$