Factor x^2+16x+64-x^4

$x^{2} + 16 x + 64 - x^{4}$

Step 1

Factor using the perfect square rule.

Rewrite $64$ as $8^{2}$ .

$x^{2} + 16 x + 8^{2} - x^{4}$

Check that the middle term is two times the product of the numbers being squared in the first term and third term.

$16 x = 2 \cdot x \cdot 8$

Rewrite the polynomial.

$x^{2} + 2 \cdot x \cdot 8 + 8^{2} - x^{4}$

Factor using the perfect square trinomial rule $a^{2} + 2 a b + b^{2} = {(a + b)}^{2}$ , where $a = x$ and $b = 8$ .

${(x + 8)}^{2} - x^{4}$

Step 2

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

${(x + 8)}^{2} - {(x^{2})}^{2}$

Step 3

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = x + 8$ and $b = x^{2}$ .

$(x + 8 + x^{2}) (x + 8 - x^{2})$

Step 4

Simplify.

Reorder terms.

$(x^{2} + x + 8) (x + 8 - x^{2})$

Reorder terms.

$(x^{2} + x + 8) (- x^{2} + x + 8)$

Step 5

Factor.

Factor $- 1$ out of $- x^{2} + x + 8$ .

Factor $- 1$ out of $- x^{2}$ .

$(x^{2} + x + 8) (- (x^{2}) + x + 8)$

Factor $- 1$ out of $x$ .

$(x^{2} + x + 8) (- (x^{2}) - 1 (- x) + 8)$

Rewrite $8$ as $- 1 (- 8)$ .

$(x^{2} + x + 8) (- (x^{2}) - 1 (- x) - 1 (- 8))$

Factor $- 1$ out of $- (x^{2}) - 1 (- x)$ .

$(x^{2} + x + 8) (- (x^{2} - x) - 1 (- 8))$

Factor $- 1$ out of $- (x^{2} - x) - 1 (- 8)$ .

$(x^{2} + x + 8) (- (x^{2} - x - 8))$

Remove unnecessary parentheses.

$(x^{2} + x + 8) \cdot - 1 (x^{2} - x - 8)$

Step 6

Factor out negative.

$- ((x^{2} + x + 8) (x^{2} - x - 8))$

Step 7

Remove unnecessary parentheses.

$- (x^{2} + x + 8) (x^{2} - x - 8)$

Factor x^2+16x+64-x^4

Related Questions

Support LCMGCF.com by disabling your adblocker.