# How do you factor 7n ^ { 2} + 45n + 18?

Jun 7, 2018

$\left(n - 6\right) \left(n - \frac{3}{7}\right)$

#### Explanation:

After dividing by $7$ we get
${n}^{2} + \frac{45}{7} n + \frac{18}{7} = 0$

${n}_{1 , 2} = - \frac{45}{14} \pm \setminus \sqrt{\frac{1521}{196}}$
this is
${n}_{1 , 2} = \frac{45}{14} \pm \frac{39}{14}$
so we get
${n}_{1} = 6$
${n}_{2} = \frac{3}{7}$
and we get
the product as
$\left(n - 6\right) \left(n - \frac{3}{7}\right)$

Jun 7, 2018

$\left(7 n + 3\right) \left(n + 6\right)$

#### Explanation:

We can start by factoring by grouping. Here, we will rewrite the $b$ term as the sum of two terms. We get

$7 {n}^{2} + \textcolor{b l u e}{42 n + 3 n} + 18$

Notice, what I have in blue is the same as $45 n$, so I didn't change the value of the expression.

$\textcolor{\mathrm{da} r k b l u e}{7 {n}^{2} + 42 n} + \textcolor{p u r p \le}{3 n + 18}$

I can factor a $7 n$ out of the dark blue terms, and a $3$ out of the purple term. Doing this, we get

$7 n \textcolor{red}{\left(n + 6\right)} + 3 \textcolor{red}{\left(n + 6\right)}$

Both terms have a $n + 6$ in common, so we can factor that out. Doing this, we get

$\left(7 n + 3\right) \left(n + 6\right)$