# How do you factor completely 8u^2 (u+1) + 2u (u+1) -3 (u+1)?

Jun 11, 2018

$\left(u + 1\right) \left(4 u + 3\right) \left(2 u - 1\right)$

#### Explanation:

The key realization here is that every term has a $u + 1$ in common, so we can factor that our. We're left with:

$\left(u + 1\right) \textcolor{\mathrm{da} r k b l u e}{\left(8 {u}^{2} + 2 u - 3\right)}$

What I have in blue, we can factor by grouping. Here, I'll split the $b$ term up so that we have two expressions. Here's what I mean:

$\left(u + 1\right) \textcolor{\mathrm{da} r k b l u e}{\left(8 {u}^{2} \underline{- 4 u + 6 u} - 3\right)}$

What I have underlined is the same as $2 u$, so I didn't change the value of this expression.

$\left(u + 1\right) \left(\textcolor{b l u e}{8 {u}^{2} - 4 u} + \textcolor{red}{6 u - 3}\right)$

I can factor a $4 u$ out of the blue term, and a $3$ out of the red term. We now have:

$\left(u + 1\right) \cdot \textcolor{b l u e}{4 u \underline{\left(2 u - 1\right)}} \cdot \textcolor{red}{3 \underline{\left(2 u - 1\right)}}$

Both the underlined terms have a $2 u - 1$ in common, so we can factor that out. We get:

$\left(u + 1\right) \left(4 u + 3\right) \left(2 u - 1\right)$

Hope this helps!

Jun 11, 2018

$\left(u + 1\right) \left(2 u - 1\right) \left(4 u + 3\right)$

#### Explanation:

$\text{take out the "color(blue)"common factor } \left(u + 1\right)$

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

$\text{factor the quadratic using the a-c method}$

$\text{the factors of the product } 8 \times - 3 = - 24$

$\text{which sum to + 2 are - 4 and + 6}$

$\text{split the middle term using these factors}$

$8 {u}^{2} - 4 u + 6 u - 3 \leftarrow \textcolor{b l u e}{\text{factor by grouping}}$

$= \textcolor{red}{4 u} \left(2 u - 1\right) \textcolor{red}{+ 3} \left(2 u - 1\right)$

$\text{take out the "color(blue)"common factor } \left(2 u - 1\right)$

$= \left(2 u - 1\right) \left(\textcolor{red}{4 u + 3}\right)$

$\text{putting it together}$

$= \left(u + 1\right) \left(2 u - 1\right) \left(4 u + 3\right)$