# How do you factor n^2+16n+64?

Mar 16, 2018

The factored form of the polynomial is ${\left(n + 8\right)}^{2}$.

#### Explanation:

We need to find two numbers that multiply to $64$ (the $c$ value of the quadratic) and add up to $16$ (the $b$ value of the quadratic).

It helps to list out the factor pairs of $64$ and their sums:

color(white){color(black)( (1,+qquad 64 quad=65), (2,+qquad32 quad=34), (4,+qquad16 quad=20), (8,+qquad8 qquad=16color(red)larr), (16,+qquad4 qquad=20), (32,+qquad2 qquad=34), (64,+qquad1 qquad=65):}

The only factors of $64$ that add up to $16$ are $8$ and $8$ (highlighted with a red arrow).

Now, split up the $n$ terms in the quadratic to $8$ and $8$:

$\textcolor{w h i t e}{=} {n}^{2} + 16 n + 64$

$= {n}^{2} + 8 n + 8 n + 64$

Now, factor the first two terms and the last two terms separately, then combine them:

$\textcolor{w h i t e}{=} {n}^{2} \quad + \quad 8 n \quad + \quad 8 n \quad + \quad 64$

$= \overbrace{\textcolor{red}{n} \cdot n} + \overbrace{\textcolor{red}{n} \cdot 8} + \overbrace{\textcolor{b l u e}{8} \cdot n} + \overbrace{\textcolor{b l u e}{8} \cdot 8}$

$= \textcolor{red}{n} \left(n + 8\right) + \textcolor{b l u e}{8} \left(n + 8\right)$

$= \left(\textcolor{red}{n} + \textcolor{b l u e}{8}\right) \left(n + 8\right)$

$= {\left(n + 8\right)}^{2}$

That's the most factored it can get. Hope this helped!