How do you factor ${(x + 5)}^{2} - 81$ $(x+5)^2 - 81$ ?

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Answer 1 · 2015-05-11T00:09:33

${(x + 5)}^{2} - 81$ $(x+5)^2-81$
is the same as ${(x + 5)}^{2} - 9^{2}$ $(x+5)^2-9^2$

Any expression of the form $a^{2} - b^{2}$ $a^2-b^2$ can be factored as
$a^{2} - b^{2} = (a + b) (a - b)$ $a^2-b^2 = (a+b)(a-b)$

For the given example $a = (x + 5)$ $a=(x+5)$ and $b = 9$ $b=9$

So
${(x + 5)}^{2} - 81 = ((x + 5) + 9) (x + 5 - 9)$ $(x+5)^2 -81 = ((x+5)+9)(x+5-9)$
which simplifies as
$(x + 14) (x - 4)$ $(x+14)(x-4)$

How do you factor (x+5)^2 - 81(x+5)2−81(x+5)^2 - 81?