# How do you factor 3x^2 + 4x - 8?

Jun 13, 2016

Factors of $3 {x}^{2} + 4 x - 8$ are $3 \left(x + \frac{2 - \sqrt{28}}{3}\right) \left(x + \frac{2 + \sqrt{28}}{3}\right)$.

#### Explanation:

In $3 {x}^{2} + 4 x - 8$, the discriminant is ${4}^{2} - 4 \cdot 3 \cdot \left(- 8\right) = \left(16 + 96\right) = - 112$, is not the square of a rational number. Hence we cannot factorize it by splitting middle term.

Hence, the way is to find out zeros of quadratic trinomial $3 {x}^{2} + 4 x - 8$. Zeros of $a {x}^{2} + b x + c$ are given by quadratic formula $\frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$.

So its zeros, which are two irratiional conjugate numbers are given by quadratic formula and are

$\frac{- 4 \pm \sqrt{112}}{2 \times 3}$ or

$\frac{- 4 \pm 2 \sqrt{28}}{6}$ or

$\frac{- 4 \pm 2 \sqrt{28}}{6}$ i.e. $\frac{- 2 - \sqrt{28}}{3}$ and $\frac{- 2 + \sqrt{28}}{3}$

Now, if $\alpha$ and $\beta$ are zeros of quadratic polynomial, then its factors are $\left(x - \alpha\right) \left(x - \beta\right)$. However as we have $3$ as coefficient of ${x}^{2}$, we should multiply it by $3$

Hence factors of $3 {x}^{2} + 4 x - 8$ are $3 \left(x + \frac{2 - \sqrt{28}}{3}\right) \left(x + \frac{2 + \sqrt{28}}{3}\right)$.