# How do you factor completely 9x^2 + 20x + 100?

Feb 2, 2017

$9 {x}^{2} + 20 x + 100 = \left(3 x + \frac{10}{3} - \frac{20}{3} \sqrt{2} i\right) \left(3 x + \frac{10}{3} + \frac{20}{3} \sqrt{2} i\right)$

#### Explanation:

Given:

$9 {x}^{2} + 20 x + 100$

This is in the form:

$a {x}^{2} + b x + c$

with $a = 9$, $b = 20$ and $c = 100$.

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

$\Delta = {b}^{2} - 4 a c = {20}^{2} - 4 \left(9\right) \left(100\right) = 400 - 3600 = - 3200$

Since $\Delta < 0$ this quadratic has no Real zeros and no linear factors with Real coefficients.

We can factor it using Complex coefficients by completing the square as follows:

$9 {x}^{2} + 20 x + 100 = {\left(3 x\right)}^{2} + 2 \left(3 x\right) \left(\frac{10}{3}\right) + {\left(\frac{10}{3}\right)}^{2} - {\left(\frac{10}{3}\right)}^{2} + 100$

$\textcolor{w h i t e}{9 {x}^{2} + 20 x + 100} = {\left(3 x + \frac{10}{3}\right)}^{2} + \frac{800}{9}$

$\textcolor{w h i t e}{9 {x}^{2} + 20 x + 100} = {\left(3 x + \frac{10}{3}\right)}^{2} - {\left(\frac{20}{3} \sqrt{2} i\right)}^{2}$

$\textcolor{w h i t e}{9 {x}^{2} + 20 x + 100} = \left(\left(3 x + \frac{10}{3}\right) - \frac{20}{3} \sqrt{2} i\right) \left(\left(3 x + \frac{10}{3}\right) + \frac{20}{3} \sqrt{2} i\right)$

$\textcolor{w h i t e}{9 {x}^{2} + 20 x + 100} = \left(3 x + \frac{10}{3} - \frac{20}{3} \sqrt{2} i\right) \left(3 x + \frac{10}{3} + \frac{20}{3} \sqrt{2} i\right)$