# How do you factor by grouping 5n^2 + 19n + 12?

May 9, 2018

$\implies \left(5 n + 4\right) \left(n + 3\right)$

#### Explanation:

Factor by grouping? This is just normal factoring, no grouping is needed:

$5 {n}^{2} + 19 n + 12 = \left(5 n + 4\right) \left(n + 3\right)$

If you want to use factoring by grouping anyway, then you have to break up the middle term with $n$ into pieces that sum to $19 n$ in total.

$= \left(5 {n}^{2} + 15 n\right) + \left(4 n + 12\right)$

$= 5 n \left(n + 3\right) + 4 \left(n + 3\right)$

$= \left(5 n + 4\right) \left(n + 3\right)$

May 9, 2018

$\left(5 n + 4\right) \left(n + 3\right)$

#### Explanation:

When we do factoring by grouping, we want to split the $b$ term (coefficient on $n$) into two terms so we can factor two separate expression.

$5 {n}^{2} + \textcolor{b l u e}{19 n} + 12$ can be rewritten as

$5 {n}^{2} + \textcolor{b l u e}{15 n + 4 n} + 12$

Notice, the blue terms are equivalent, so I didn't change the meaning of this expression.

Now, we can factor!

$\textcolor{red}{5 {n}^{2} + 15 n} + \textcolor{\lim e}{4 n + 12}$

We can factor a $5 n$ out of the red expression, and a $4$ out of the green expression. We get

$\textcolor{red}{5 n \left(n + 3\right)} + \textcolor{\lim e}{4 \left(n + 3\right)}$

which can be rewritten as

$\left(5 n + 4\right) \left(n + 3\right)$

Hope this helps!

May 9, 2018

Use an AC method to split the middle term, then factor by grouping to find:

$5 {n}^{2} + 19 n + 12 = \left(5 n + 4\right) \left(n + 3\right)$

#### Explanation:

Given:

$5 {n}^{2} + 19 n + 12$

Use an AC Method to find the split we need.

Look for a pair of factors of $A C = 5 \cdot 12 = 60$ with sum $B = 19$.

The pair $15 , 4$ works in that $15 \cdot 4 = 60$ and $15 + 4 = 19$.

Using this pair to split the middle term, we can then factor by grouping:

$5 {n}^{2} + 19 n + 12 = \left(5 {n}^{2} + 15 n\right) + \left(4 n + 12\right)$

$\textcolor{w h i t e}{5 {n}^{2} + 19 n + 12} = 5 n \left(n + 3\right) + 4 \left(n + 3\right)$

$\textcolor{w h i t e}{5 {n}^{2} + 19 n + 12} = \left(5 n + 4\right) \left(n + 3\right)$