How do you factor 64x^2 + 112x + 49?

Jun 8, 2016

${\left(7 x + 8\right)}^{2}$

Explanation:

The key to factoring this is noticing that the first and last terms are both squared terms:

• $64 {x}^{2} = {\left(8 x\right)}^{2}$
• $49 = {\left(7\right)}^{2}$

This is a good indicator that the expression is a perfect square binomial, which comes in the form:

${\left(a + b\right)}^{2} = \left(a + b\right) \left(a + b\right) = {a}^{2} + 2 a b + {b}^{2}$

Here, $a = 8 x$ and $b = 7$, as indicated to us by the $64 {x}^{2}$ and $49$ terms. However, we still have to determine whether or not the middle term $112 x$ is equal to $2 a b$.

$2 a b = 2 \left(8 x\right) \left(7\right) = 112 x$

Since $2 a b$ does equal $112 x$, we know that this is a perfect square binomial.

$64 {x}^{2} + 112 x + 49 = {\left(7 x + 8\right)}^{2}$