# How do you factor 25x^2n - 49 ?

Dec 27, 2015

If $n$ is a positive constant, then this factorises as:

$25 {x}^{2} n - 49 = \left(5 x \sqrt{n} - 7\right) \left(5 x \sqrt{n} + 7\right)$

#### Explanation:

This question seems a little curious. Should the $n$ be there? Should it be ${n}^{2}$?

In any case, we can use the difference of squares identity:

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

with $a = 5 x \sqrt{n}$ and $b = 7$...

$25 {x}^{2} n - 49$

$= {\left(5 x \sqrt{n}\right)}^{2} - {7}^{2}$

$= \left(5 x \sqrt{n} - 7\right) \left(5 x \sqrt{n} + 7\right)$