# How do you factor m^2-10my+25y^2?

May 11, 2016

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

#### Explanation:

One of our first clues in factoring this expression is that the first and last terms are perfect squares:

$\sqrt{{m}^{2}} = m$

and

$\sqrt{25 {y}^{2}} = 5 y$

In the most general case, we are looking for a solution in the form of

$\left(a m + b y\right) \left(c m + \mathrm{dy}\right) = {m}^{2} - 10 m y + 25 {y}^{2}$
$a c {m}^{2} + \left(b c + a d\right) m y + b {\mathrm{dy}}^{2} = {m}^{2} - 10 m y + 25 {y}^{2}$

From the first term we can see that $a \cdot c = 1$. Assuming $a$ and $c$ are integers, they must be both either $+ 1$ or $- 1$. Let's make them $+ 1$ for now and continue (it actually doesn't matter which we choose at this point - can you see why?):

${m}^{2} + \left(b + d\right) m y + b {\mathrm{dy}}^{2} = {m}^{2} - 10 m y + 25 {y}^{2}$

From the last term we see that $b \cdot d = 25$. From the second term, we must also have $\left(b + d\right) = - 10$. The obvious solution for this is to have $b = d = - 5$. Therefore we have the solution

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

Once we have worked through this type of problem, we could have probably started by guessing this solution from the first two observations.