# How do you factor 9x^2 +18x + 8?

Jul 15, 2017

$9 {x}^{2} + 18 x + 8 = \left(3 x + 2\right) \left(3 x + 4\right)$

#### Explanation:

$9 {x}^{2} + 18 x + 8$

is in the form:

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

with $a = 9$, $b = 18$ and $c = 8$.

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

$\Delta = {b}^{2} - 4 a c = {\textcolor{b l u e}{18}}^{2} - 4 \left(\textcolor{b l u e}{9}\right) \left(\textcolor{b l u e}{8}\right) = 324 - 288 = 36 = {6}^{2}$

Since this is positive and a perfect square, this quadratic will factor with integer coefficients.

Let's use an AC method:

Find a pair of factors of $A C = 9 \cdot 8 = 72$ with sum $B = 18$. (We look for a sum rather than a difference since the sign of $c$ is positive).

The pair $12 , 6$ works in that $12 \cdot 6 = 72$ and $12 + 6 = 18$.

Use this pair to split the middle term and factor by grouping:

$9 {x}^{2} + 18 x + 8 = \left(9 {x}^{2} + 12 x\right) + \left(6 x + 8\right)$

$\textcolor{w h i t e}{9 {x}^{2} + 18 x + 8} = 3 x \left(3 x + 4\right) + 2 \left(3 x + 4\right)$

$\textcolor{w h i t e}{9 {x}^{2} + 18 x + 8} = \left(3 x + 2\right) \left(3 x + 4\right)$

$\textcolor{w h i t e}{}$
We can find the same factorisation by completing the square:

Note that $9 {x}^{2} = {\left(3 x\right)}^{2}$ and $18 x = 2 \left(3 x\right) \left(3\right)$.

Hence we find:

$9 {x}^{2} + 18 x + 8 = {\left(3 x\right)}^{2} + 2 \left(3 x\right) \left(3\right) + {3}^{2} - 1$

$\textcolor{w h i t e}{9 {x}^{2} + 18 x + 8} = {\left(3 x + 3\right)}^{2} - {1}^{2}$

$\textcolor{w h i t e}{9 {x}^{2} + 18 x + 8} = \left(\left(3 x + 3\right) - 1\right) \left(\left(3 x + 3\right) + 1\right)$

$\textcolor{w h i t e}{9 {x}^{2} + 18 x + 8} = \left(3 x + 2\right) \left(3 x + 4\right)$