Factor x^8-25x^4+144=0

$x^{8} - 25 x^{4} + 144 = 0$

Step 1

Rewrite $x^{8}$ as ${(x^{4})}^{2}$ .

${(x^{4})}^{2} - 25 x^{4} + 144 = 0$

Step 2

Let $u = x^{4}$ . Substitute $u$ for all occurrences of $x^{4}$ .

$u^{2} - 25 u + 144 = 0$

Step 3

Factor $u^{2} - 25 u + 144$ using the AC method.

Consider the form $x^{2} + b x + c$ . Find a pair of integers whose product is $c$ and whose sum is $b$ . In this case, whose product is $144$ and whose sum is $- 25$ .

$- 16, - 9$

Write the factored form using these integers.

$(u - 16) (u - 9) = 0$

Step 4

Replace all occurrences of $u$ with $x^{4}$ .

$(x^{4} - 16) (x^{4} - 9) = 0$

Step 5

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

$({(x^{2})}^{2} - 16) (x^{4} - 9) = 0$

Step 6

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

$({(x^{2})}^{2} - 4^{2}) (x^{4} - 9) = 0$

Step 7

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 $b = 4$ .

$(x^{2} + 4) (x^{2} - 4) (x^{4} - 9) = 0$

Step 8

Simplify.

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

$(x^{2} + 4) (x^{2} - 2^{2}) (x^{4} - 9) = 0$

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 $b = 2$ .

$(x^{2} + 4) ((x + 2) (x - 2)) (x^{4} - 9) = 0$

Remove unnecessary parentheses.

$(x^{2} + 4) (x + 2) (x - 2) (x^{4} - 9) = 0$

Step 9

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

$(x^{2} + 4) (x + 2) (x - 2) ({(x^{2})}^{2} - 9) = 0$

Step 10

Rewrite $9$ as $3^{2}$ .

$(x^{2} + 4) (x + 2) (x - 2) ({(x^{2})}^{2} - 3^{2}) = 0$

Step 11

Factor.

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = x^{2}$ and $b = 3$ .

$(x^{2} + 4) (x + 2) (x - 2) ((x^{2} + 3) (x^{2} - 3)) = 0$

Remove unnecessary parentheses.

$(x^{2} + 4) (x + 2) (x - 2) (x^{2} + 3) (x^{2} - 3) = 0$

Factor x^8-25x^4+144=0

Related Questions

Support LCMGCF.com by disabling your adblocker.