# How do you factor 16-(y-3)^2?

Jul 12, 2016

(7 -y)(1 +y)

#### Explanation:

This expression is a $\textcolor{b l u e}{\text{difference of squares}}$ which, in general factorises as.

$\textcolor{red}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{{a}^{2} - {b}^{2} = \left(a - b\right) \left(a + b\right)} \textcolor{w h i t e}{\frac{a}{a}} |}}} \ldots \ldots . . \left(A\right)$

Now $16 = {4}^{2} \Rightarrow a = 4 \text{ and } {\left(y - 3\right)}^{2} \Rightarrow b = y - 3$

Substitute these values for a and b into (A)

$\Rightarrow 16 - {\left(y - 3\right)}^{2} = \left(4 - \left(y - 3\right)\right) \left(4 + y - 3\right)$

Simplifying the brackets gives.

$\left(4 - y + 3\right) \left(1 + y\right) = \left(7 - y\right) \left(1 + y\right)$

$\Rightarrow 16 - {\left(y - 3\right)}^{2} = \left(7 - y\right) \left(1 + y\right)$