Factor over the Complex Numbers x^4+290x^2+289

$x^{4} + 290 x^{2} + 289$

Step 1

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

${(x^{2})}^{2} + 290 x^{2} + 289$

Step 2

Let $u = x^{2}$ . Substitute $u$ for all occurrences of $x^{2}$ .

$u^{2} + 290 u + 289$

Step 3

Factor $u^{2} + 290 u + 289$ using the AC method.

Consider the form $x^{2} + i x + c$ . Find a pair of integers whose product is $c$ and whose sum is $i$ . In this case, whose product is $289$ and whose sum is $290$ .

$1, 289$

Write the factored form using these integers.

$(u + 1) (u + 289)$

Step 4

Replace all occurrences of $u$ with $x^{2}$ .

$(x^{2} + 1) (x^{2} + 289)$

Step 5

Multiply the constant in the polynomial $x^{2} + 1$ by $- i^{2}$ where $i^{2}$ is equal to $- 1$ .

$(x^{2} - 1 i^{2}) (x^{2} + 289)$

Step 6

Multiply the constant in the polynomial $x^{2} + 289$ by $- i^{2}$ where $i^{2}$ is equal to $- 1$ .

$(x^{2} - 1 i^{2}) (x^{2} - 289 i^{2})$

Step 7

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

$(x + i) (x - i) (x^{2} - 289 i^{2})$

Step 8

Rewrite $289 i^{2}$ as ${(17 i)}^{2}$ .

$(x + i) (x - i) (x^{2} - {(17 i)}^{2})$

Step 9

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - i^{2} = (a + i) (a - i)$ where $a = x$ and $i = 17 i$ .

$(x + i) (x - i) ((x + 17 i) (x - (17 i)))$

Step 10

Factor.

Multiply $17$ by $- 1$ .

$(x + i) (x - i) ((x + 17 i) (x - 17 i))$

Remove unnecessary parentheses.

$(x + i) (x - i) (x + 17 i) (x - 17 i)$

Factor over the Complex Numbers x^4+290x^2+289

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