# Is it possible to factor y=x^2 +8x +16 ? If so, what are the factors?

Jan 24, 2016

${x}^{2} + 8 x + 16 = {\left(x + 4\right)}^{2}$

#### Explanation:

This is a perfect square trinomial.

First notice that both ${x}^{2}$ and $16 = {4}^{2}$ are perfect squares.

Then multiply out ${\left(x + 4\right)}^{2}$ (using FOIL if you find it helpful) to see if the middle term matches:

${\left(x + 4\right)}^{2}$

$= \left(x + 4\right) \left(x + 4\right)$

$= \stackrel{\text{First" overbrace((x*x)) + stackrel "Outside" overbrace((x * 4)) + stackrel "Inside" overbrace((4 * x)) + stackrel "Last}}{\overbrace{\left(4 \cdot 4\right)}}$

$= {x}^{2} + 4 x + 4 x + 16$

$= {x}^{2} + 8 x + 16$

In general note that ${\left(a + b\right)}^{2} = {a}^{2} + 2 a b + {b}^{2}$, so if you recognise $a$ and $b$ then just check that the middle term is $2 a b$.