Factor x^4-16=0

$x^{4} - 16 = 0$

Step 1

Rewrite $x^{4}$ as ${(x^{2})}^{2}$ .

${(x^{2})}^{2} - 16 = 0$

Step 2

Rewrite $16$ as $4^{2}$ .

${(x^{2})}^{2} - 4^{2} = 0$

Step 3

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = x^{2}$ and $b = 4$ .

$(x^{2} + 4) (x^{2} - 4) = 0$

Step 4

Simplify.

Rewrite $4$ as $2^{2}$ .

$(x^{2} + 4) (x^{2} - 2^{2}) = 0$

Factor.

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = x$ and $b = 2$ .

$(x^{2} + 4) ((x + 2) (x - 2)) = 0$

Remove unnecessary parentheses.

$(x^{2} + 4) (x + 2) (x - 2) = 0$

Factor x^4-16=0

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