# How do you factor 6x^2 + 13x + 8?

##### 1 Answer
Aug 26, 2016

$6 {x}^{2} + 13 x + 8 = 6 \left(x + \frac{13}{12} - \frac{\sqrt{23}}{12} i\right) \left(x + \frac{13}{12} + \frac{\sqrt{23}}{12} i\right)$

#### Explanation:

$6 {x}^{2} + 13 x + 8$ is in the form $a {x}^{2} + b x + c$ with $a = 6$, $b = 13$ and $c = 8$

This has discriminant $\Delta$ given by the formula:

$\Delta = {b}^{2} - 4 a c = {13}^{2} - \left(4 \cdot 6 \cdot 8\right) = 169 - 192 = - 23$

Since $\Delta < 0$ this quadratic has no simpler factors with Real coefficients, but it does have Complex zeros given by the quadratic formula:

$x = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$

$= \frac{- b \pm \sqrt{\Delta}}{2 a}$

$= \left(- 13 \pm \frac{\sqrt{- 23}}{12}\right)$

$= - \frac{13}{12} \pm \frac{\sqrt{23}}{12} i$

Hence factorisation:

$6 {x}^{2} + 13 x + 8 = 6 \left(x + \frac{13}{12} - \frac{\sqrt{23}}{12} i\right) \left(x + \frac{13}{12} + \frac{\sqrt{23}}{12} i\right)$

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Footnote

I suspect a sign error in the question since the following is much simpler:

$6 {x}^{2} + 13 x - 8 = \left(3 x + 8\right) \left(2 x - 1\right)$