# How do you factor 14x ^ { 2} - 7x - 28?

Aug 19, 2017

$14 {x}^{2} - 7 x - 28 = \frac{7}{8} \left(4 x - 1 - \sqrt{33}\right) \left(4 x - 1 + \sqrt{33}\right)$

#### Explanation:

Given:

$f \left(x\right) = 14 {x}^{2} - 7 x - 28$

I will multiply by $\frac{8}{7}$ in order to complete the square and use the difference of squares identity while avoiding fractions as much as possible...

$\frac{8}{7} f \left(x\right) = 16 {x}^{2} - 8 x - 32$

$\textcolor{w h i t e}{\frac{8}{7} f \left(x\right)} = {\left(4 x\right)}^{2} - 2 \left(4 x\right) + 1 - 33$

$\textcolor{w h i t e}{\frac{8}{7} f \left(x\right)} = {\left(4 x - 1\right)}^{2} - {\left(\sqrt{33}\right)}^{2}$

$\textcolor{w h i t e}{\frac{8}{7} f \left(x\right)} = \left(\left(4 x - 1\right) - \sqrt{33}\right) \left(\left(4 x - 1\right) + \sqrt{33}\right)$

$\textcolor{w h i t e}{\frac{8}{7} f \left(x\right)} = \left(4 x - 1 - \sqrt{33}\right) \left(4 x - 1 + \sqrt{33}\right)$

Multiplying both ends by $\frac{7}{8}$ we find:

$14 {x}^{2} - 7 x - 28 = \frac{7}{8} \left(4 x - 1 - \sqrt{33}\right) \left(4 x - 1 + \sqrt{33}\right)$