# How do you factor completely x^2 - 7x + 30?

Dec 22, 2015

${x}^{2} - 7 x + 30 = \left(x - \frac{7}{2} - \frac{\sqrt{71}}{2} i\right) \left(x - \frac{7}{2} + \frac{\sqrt{71}}{2} i\right)$

#### Explanation:

${x}^{2} - 7 x + 30$ is of the form $a {x}^{2} + b x + c$ with $a = 1$, $b = - 7$ and $c = 30$.

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

$\Delta = {b}^{2} - 4 a c = {\left(- 7\right)}^{2} - \left(4 \times 1 \times 30\right) = 49 - 120 = - 71$

Since this is negative, the quadratic has no Real zeros and no simpler factors with Real coefficients.

If you still want to factor it, you can use the quadratic formula to find the Complex conjugate pair of zeros and hence derive the factors:

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

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

$= \frac{7 \pm \sqrt{- 71}}{2}$

$= \frac{7 \pm i \sqrt{71}}{2}$

$= \frac{7}{2} \pm \frac{\sqrt{71}}{2} i$

Hence:

${x}^{2} - 7 x + 30 = \left(x - \frac{7}{2} - \frac{\sqrt{71}}{2} i\right) \left(x - \frac{7}{2} + \frac{\sqrt{71}}{2} i\right)$