# How do you factor 25b^2 - 20by + 4y^2?

Sep 18, 2015

$25 {b}^{2} - 20 b y + 4 {y}^{2} = {\left(5 b - 2 y\right)}^{2}$

#### Explanation:

$25 {b}^{2} - 20 b y + 4 {y}^{2}$

The first term is a perfect square: ${\left(5 b\right)}^{2} = 25 {b}^{2}$

The third term is a perfect square: ${\left(2 y\right)}^{2} = 4 {y}^{2}$

So we MIGHT have ${\left(5 b \pm 2 y\right)}^{2}$

We know that ${\left(u \pm v\right)}^{2} = {u}^{2} \pm 2 u v + {v}^{2}$, so we ask ourselves, "Is the middle term ($20 b y$) equal to twice the product of the first and second terms of the binomial?"

Yes, it is, so we have:

${\left(5 b - 2 y\right)}^{2} = {\left(5 b\right)}^{2} - 2 \left(5 b\right) \left(2 y\right) + {\left(2 y\right)}^{2}$

$= 25 {b}^{2} - 20 b y + 4 {y}^{2}$