# How do you factor  16y^2-25?

Mar 21, 2018

$\left(4 y + 5\right) \left(4 y - 5\right)$
You need to consider what multiplies to make 16(either $1 \cdot 16 , 2 \cdot 8 ,$or $4 \cdot 4$), and what multiplies to make 25( $5 \cdot 5$). Also notice that this is a binomial, not a trinomial.

#### Explanation:

The only factor of $25$ is $5 \cdot 5 = {5}^{2}$, so the factorization must be of the form $\left(a + 5\right) \left(b - 5\right)$, as a negative times a positive is negative. Now consider that there is no middle term, so it must have been canceled out. This implies that the coefficients of $y$ are the same. This leaves only $\left(4 y + 5\right) \left(4 y - 5\right)$.

Mar 21, 2018

$16 {y}^{2} - 25 = \left(4 y - 5\right) \times \left(4 y + 5\right)$