How do you factor by grouping $b^2 - 8b + 16 - c^2$ ?

Question

2950 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2015-05-17T20:28:05

$b^2-8b+16-c^2$
$= (b^2-8b+16)-c^2$
$= (b-4)^2 - c^2$
$= ((b-4)+c)((b-4)-c)$
$= (b+c-4)(b-c-4)$

The factoring of $(b-4)^2 - c^2$ as $((b-4)+c)((b-4)-c)$ is an instance of the identity:

$m^2-n^2=(m+n)(m-n)$

with $m=b-4$ and $n=c$ .

How do you factor by grouping b^2 - 8b + 16 - c^2 b^2 - 8b + 16 - c^2 ?