# How do you factor 14x^2+8x+72?

Apr 13, 2017

This quadratic has no factors with real coefficients, but we find:

$14 {x}^{2} + 8 x + 72 = \frac{2}{7} \left(7 x + 2 - 2 \sqrt{62} i\right) \left(7 x + 2 + 2 \sqrt{62} i\right)$

#### Explanation:

This quadratic is not factorisable using real coefficients.

We can factor it using complex coefficients by completing the square and using the difference of squares identity:

#a^2-b^2=(a-b)(a+b)

with $a = \left(7 x + 2\right)$ and $b = 2 \sqrt{62}$ as follows:

$\left(\frac{7}{2}\right) \left(14 {x}^{2} + 8 x + 72\right) = 49 {x}^{2} + 28 x + 252$

$\textcolor{w h i t e}{\left(\frac{7}{2}\right) \left(14 {x}^{2} + 8 x + 72\right)} = {\left(7 x\right)}^{2} + 2 \left(7 x\right) \left(2\right) + {2}^{2} + 248$

$\textcolor{w h i t e}{\left(\frac{7}{2}\right) \left(14 {x}^{2} + 8 x + 72\right)} = {\left(7 x + 2\right)}^{2} + {\left(2 \sqrt{62}\right)}^{2}$

$\textcolor{w h i t e}{\left(\frac{7}{2}\right) \left(14 {x}^{2} + 8 x + 72\right)} = {\left(7 x + 2\right)}^{2} - {\left(2 \sqrt{62} i\right)}^{2}$

$\textcolor{w h i t e}{\left(\frac{7}{2}\right) \left(14 {x}^{2} + 8 x + 72\right)} = \left(\left(7 x + 2\right) - 2 \sqrt{62} i\right) \left(\left(7 x + 2\right) + 2 \sqrt{62} i\right)$

$\textcolor{w h i t e}{\left(\frac{7}{2}\right) \left(14 {x}^{2} + 8 x + 72\right)} = \left(7 x + 2 - 2 \sqrt{62} i\right) \left(7 x + 2 + 2 \sqrt{62} i\right)$

So:

$14 {x}^{2} + 8 x + 72 = \frac{2}{7} \left(7 x + 2 - 2 \sqrt{62} i\right) \left(7 x + 2 + 2 \sqrt{62} i\right)$