# How do you factor y^4-14y^3+49y^2?

Sep 18, 2016

${y}^{2} \left(y - 7\right)$

#### Explanation:

Take out a common factor of ${y}^{2}$ first.

${y}^{2} \left({y}^{2} - 14 y \textcolor{red}{+ 49}\right) \text{ } \rightarrow \textcolor{red}{+ 49}$ a clue!

You should recognize $49$ as being a perfect square!

In ${\left(x + y\right)}^{2} = {x}^{2} + 2 x y + {y}^{2} = a {x}^{2} + b x + c$ there is always a special relationship between the coefficient of the second term (b) and the third term (c)

$\left(b \div 2\right) \text{ and then squared gives } c$.

If this relationship exists in the trinomial you will know you have a square of a binomial.

Let's look closer..... (14div2)^2 = 7^2=49!!

$\therefore {y}^{2} \textcolor{red}{\left({y}^{2} - 14 y + 49\right)} = {y}^{2} \textcolor{red}{{\left(y - 7\right)}^{2}}$

If you missed that, you can try to find factors of 49 which add to 14. This will give you factors of 7 and 7 anyway:

${y}^{2} \left(y - 7\right) \left(y - 7\right)$

=${y}^{2} {\left(y - 7\right)}^{2}$