# How do you factor 4x^2 -20xy+25y^2?

Oct 15, 2015

(2x-5y)(2x-5y).

#### Explanation:

$4 {x}^{2} - 20 x y + 25 {y}^{2}$
$= 4 {x}^{2} - 10 x y - 10 x y + 25 {y}^{2}$
$= 2 x \left(2 x - 5 y\right) - 5 y \left(2 x - 5 y\right)$
$= \left(2 x - 5 y\right) \left(2 x - 5 y\right)$

Oct 15, 2015

$4 {x}^{2} + 20 x y + 25 {y}^{2} = {\left(2 x + 5 y\right)}^{2}$
Use the formula for the square of a binomial: ${\left(a + b\right)}^{2} = {a}^{2} + 2 a b + {b}^{2}$.
Both $4$ and $25$, the coefficient of ${x}^{2}$ and ${y}^{2}$, are perfect squares. This makes us think that the whole expression could be a perfect square: $4$ is ${2}^{2}$, and $25$ is ${5}^{2}$. So, our claim is that
$4 {x}^{2} - 20 x y + 25 {y}^{2}$ is ${\left(2 x - 5 y\right)}^{2}$. Is it true? The only term to verify is $- 20 x y$, and it is indeed twice the product of $2 x$ and $- 5 y$. So, the conjecture was right.