# How do you factor 100+4x^2-16y-40x?

Apr 21, 2017

If the $- 16 y$ should have been $- 16 {y}^{2}$, then we find:

$100 + 4 {x}^{2} - 16 {y}^{2} - 40 x = 4 \left(x - 2 y - 5\right) \left(x + 2 y - 5\right)$

#### Explanation:

For the record: I think the question should have specified $- 16 {y}^{2}$ rather than $- 16 y$. Here's what happens if that is the case:

The difference of squares identity can be written:

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

Use this with $a = \left(x - 5\right)$ and $b = 2 y$ as follows:

$100 + 4 {x}^{2} - 16 {y}^{2} - 40 x = 4 \left({x}^{2} - 10 x + 25 - 4 {y}^{2}\right)$

$\textcolor{w h i t e}{100 + 4 {x}^{2} - 16 {y}^{2} - 40 x} = 4 \left({x}^{2} - 2 x \left(5\right) + {5}^{2} - 4 {y}^{2}\right)$

$\textcolor{w h i t e}{100 + 4 {x}^{2} - 16 {y}^{2} - 40 x} = 4 \left({\left(x - 5\right)}^{2} - {\left(2 y\right)}^{2}\right)$

$\textcolor{w h i t e}{100 + 4 {x}^{2} - 16 {y}^{2} - 40 x} = 4 \left(\left(x - 5\right) - 2 y\right) \left(\left(x - 5\right) + 2 y\right)$

$\textcolor{w h i t e}{100 + 4 {x}^{2} - 16 {y}^{2} - 40 x} = 4 \left(x - 2 y - 5\right) \left(x + 2 y - 5\right)$