# Is 100 - 20y + y^2 a perfect square trinomial and how do you factor it?

Jul 26, 2015

Yes, your expression is a perfect square trinomial.

#### Explanation:

A trinomial is simply a polynomial that has three terms. Your expression is a trinomial because it has three terms: ${y}^{2}$, $- 20 y$, and $100$.

In order for it to be a perfect square trinomial, it has to be equal to the square of a binomial, which is a polynomial that has two terms.

You can factor your expression in order to check if it meets this criterion. A trinomial can be factored as a product of two binomials, so all you have to do is see if these two binomials are equal

${y}^{2} - 20 y + 100 = \left(y + a\right) \left(y + b\right)$

The right side of this equation is equal to

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

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

Now match the terms of your original expression with these terms

${y}^{2} \textcolor{b l u e}{- 20} y + \textcolor{g r e e n}{100} = {y}^{2} + \textcolor{b l u e}{\left(a + b\right)} y + \textcolor{g r e e n}{{b}^{2}}$

You need to have

${b}^{2} = 100 \implies b = \pm 10$

$\left(a + b\right) = - 20$

• If $b = 10$, then

$a = - 20 - 10 = - 30$

This solution is not valid because the expression would be equal to

$\left(y - 30\right) \cdot \left(y + 10\right) = \textcolor{red}{\cancel{\textcolor{b l a c k}{{y}^{2} - 20 y - 300}}}$

• If $b = - 10$, then

$a = - 20 + 10 = - 10$

This is a valid solution, since

$\left(y - 10\right) \cdot \left(y - 10\right) = {y}^{2} - 20 y + 100$

Therefore, your original expression is a perfect square trinomial because you can write it as

${y}^{2} - 20 y + 100 = \left(y - 10\right) \cdot \left(y - 10\right) = \textcolor{g r e e n}{{\left(y - 10\right)}^{2}}$