# How do you factor (x+2)^2 -5(x+2)?

Mar 17, 2018

See a solution process below:

#### Explanation:

First, rewrite the term on the left as:

$\left(x + 2\right) \left(x + 2\right) - 5 \left(x - 2\right)$

Next, factor out a common term of $\left(x - 2\right)$ giving:

$\left(\left(x + 2\right) - 5\right) \left(x - 2\right) \implies$

$\left(x + 2 - 5\right) \left(x - 2\right) \implies$

$\left(x - 3\right) \left(x - 2\right)$

Mar 17, 2018

$\left(x - 3\right) \left(x + 2\right)$

#### Explanation:

First we expand the ${\left(x + 2\right)}^{2}$ which looks like $\left(x + 2\right) \left(x + 2\right)$

Next we just multiply it out which will give us
${x}^{2} + 4 x + 4$

Now that we know
${\left(x + 2\right)}^{2} = {x}^{2} + 4 x + 4$

We can do the other part which is $5 \left(x + 2\right)$
Which is $\left(5 x + 10\right)$
Now we know $5 \left(x + 2\right) = \left(5 x + 10\right)$

Now we can write the problem out expanded
${x}^{2} + 4 x + 4 - \left(5 x + 10\right)$

We have to distribute this negative to all terms in the parentheses.
Which gives us
${x}^{2} + 4 x + 4 - 5 x - 10$

Now we just combine like terms
${x}^{2} - 1 x - 6$

Now we need 2 numbers that multiply to $- 6$ and add up to $- 1$
These numbers are $- 3$ and $2$

Notice $- 3 \cdot 2 = - 6$
And $- 3 + 2 = - 1$

So we then get
$\left(x - 3\right) \left(x + 2\right)$