Factor over the Complex Numbers x^4+x^3+7x^2+9x-18

$x^{4} + x^{3} + 7 x^{2} + 9 x - 18$

Step 1

Regroup terms.

$x^{3} + 9 x + x^{4} + 7 x^{2} - 18$

Step 2

Factor $x$ out of $x^{3} + 9 x$ .

Factor $x$ out of $x^{3}$ .

$x \cdot x^{2} + 9 x + x^{4} + 7 x^{2} - 18$

Factor $x$ out of $9 x$ .

$x \cdot x^{2} + x \cdot 9 + x^{4} + 7 x^{2} - 18$

Factor $x$ out of $x \cdot x^{2} + x \cdot 9$ .

$x (x^{2} + 9) + x^{4} + 7 x^{2} - 18$

Step 3

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

$x (x^{2} + 9) + {(x^{2})}^{2} + 7 x^{2} - 18$

Step 4

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

$x (x^{2} + 9) + u^{2} + 7 u - 18$

Step 5

Factor $u^{2} + 7 u - 18$ 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 $- 18$ and whose sum is $7$ .

$- 2, 9$

Write the factored form using these integers.

$x (x^{2} + 9) + (u - 2) (u + 9)$

Step 6

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

$x (x^{2} + 9) + (x^{2} - 2) (x^{2} + 9)$

Step 7

Factor $x^{2} + 9$ out of $x (x^{2} + 9) + (x^{2} - 2) (x^{2} + 9)$ .

Factor $x^{2} + 9$ out of $x (x^{2} + 9)$ .

$(x^{2} + 9) x + (x^{2} - 2) (x^{2} + 9)$

Factor $x^{2} + 9$ out of $(x^{2} - 2) (x^{2} + 9)$ .

$(x^{2} + 9) x + (x^{2} + 9) (x^{2} - 2)$

Factor $x^{2} + 9$ out of $(x^{2} + 9) x + (x^{2} + 9) (x^{2} - 2)$ .

$(x^{2} + 9) (x + x^{2} - 2)$

Step 8

Let $u = x$ . Substitute $u$ for all occurrences of $x$ .

$(x^{2} + 9) (u + u^{2} - 2)$

Step 9

Factor $u + u^{2} - 2$ 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 $- 2$ and whose sum is $1$ .

$- 1, 2$

Write the factored form using these integers.

$(x^{2} + 9) ((u - 1) (u + 2))$

Step 10

Factor.

Replace all occurrences of $u$ with $x$ .

$(x^{2} + 9) ((x - 1) (x + 2))$

Remove unnecessary parentheses.

$(x^{2} + 9) (x - 1) (x + 2)$

Step 11

Multiply the constant in the polynomial $x^{2} + 9$ by $- i^{2}$ where $i^{2}$ is equal to $- 1$ .

$(x^{2} - 9 i^{2}) (x - 1) (x + 2)$

Step 12

Rewrite $9 i^{2}$ as ${(3 i)}^{2}$ .

$(x^{2} - {(3 i)}^{2}) (x - 1) (x + 2)$

Step 13

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 $i = 3 i$ .

$(x + 3 i) (x - (3 i)) (x - 1) (x + 2)$

Step 14

Multiply $3$ by $- 1$ .

$(x + 3 i) (x - 3 i) (x - 1) (x + 2)$

Factor over the Complex Numbers x^4+x^3+7x^2+9x-18

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