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

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Answer 1 · 2015-10-29T20:39:17

$x^{2} - 5 = (x - \sqrt{5}) (x + \sqrt{5})$ $x^2-5 = (x-sqrt(5))(x+sqrt(5))$

The difference of squares identity can be written:

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

In order to treat $x^{2} - 5$ $x^2-5$ as a difference of squares, we need to recognise that $5 = {(\sqrt{5})}^{2}$ $5 = (sqrt(5))^2$ , then we find:

$x^{2} - 5 = x^{2} - ({\sqrt{5}}^{2}) = (x - \sqrt{5}) (x + \sqrt{5})$ $x^2-5 = x^2-(sqrt(5)^2) = (x-sqrt(5))(x+sqrt(5))$

In other words, we let $a = x$ $a=x$ and $b = \sqrt{5}$ $b=sqrt(5)$

How do you factor (x^2-5)(x2−5)(x^2-5)?