Factor over the Complex Numbers 2x^2+18

$2 x^{2} + 18$

Step 1

Factor $2$ out of $2 x^{2} + 18$ .

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

$2 (x^{2}) + 18$

Factor $2$ out of $18$ .

$2 x^{2} + 2 \cdot 9$

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

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

Step 2

Use the quadratic formula to find the roots for $x^{2} + 9 = 0$

Use the quadratic formula to find the solutions.

$\frac{- b \pm \sqrt{b^{2} - 4 (a c)}}{2 a}$

Substitute the values $a = 1$ , $b = 0$ , and $c = 9$ into the quadratic formula and solve for $x$ .

$\frac{0 \pm \sqrt{0^{2} - 4 \cdot (1 \cdot 9)}}{2 \cdot 1}$

Simplify.

Simplify the numerator.

Raising $0$ to any positive power yields $0$ .

$x = \frac{0 \pm \sqrt{0 - 4 \cdot 1 \cdot 9}}{2 \cdot 1}$

Multiply $- 4 \cdot 1 \cdot 9$ .

Multiply $- 4$ by $1$ .

$x = \frac{0 \pm \sqrt{0 - 4 \cdot 9}}{2 \cdot 1}$

Multiply $- 4$ by $9$ .

$x = \frac{0 \pm \sqrt{0 - 36}}{2 \cdot 1}$

Subtract $36$ from $0$ .

$x = \frac{0 \pm \sqrt{- 36}}{2 \cdot 1}$

Rewrite $- 36$ as $- 1 (36)$ .

$x = \frac{0 \pm \sqrt{- 1 \cdot 36}}{2 \cdot 1}$

Rewrite $\sqrt{- 1 (36)}$ as $\sqrt{- 1} \cdot \sqrt{36}$ .

$x = \frac{0 \pm \sqrt{- 1} \cdot \sqrt{36}}{2 \cdot 1}$

Rewrite $\sqrt{- 1}$ as $i$ .

$x = \frac{0 \pm i \cdot \sqrt{36}}{2 \cdot 1}$

Rewrite $36$ as $6^{2}$ .

$x = \frac{0 \pm i \cdot \sqrt{6^{2}}}{2 \cdot 1}$

Pull terms out from under the radical, assuming positive real numbers.

$x = \frac{0 \pm i \cdot 6}{2 \cdot 1}$

Move $6$ to the left of $i$ .

$x = \frac{0 \pm 6 i}{2 \cdot 1}$

Multiply $2$ by $1$ .

$x = \frac{0 \pm 6 i}{2}$

Simplify $\frac{0 \pm 6 i}{2}$ .

$x = \pm 3 i$

Step 3

Find the factors from the roots, then multiply the factors together.

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

Step 4

Simplify the factored form.

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

Factor over the Complex Numbers 2x^2+18

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