Find the Roots (Zeros) f(x)=(x^2+2x-15)(x^2+8x+17)

$f (x) = (x^{2} + 2 x - 15) (x^{2} + 8 x + 17)$

Step 1

Set $(x^{2} + 2 x - 15) (x^{2} + 8 x + 17)$ equal to $0$ .

$(x^{2} + 2 x - 15) (x^{2} + 8 x + 17) = 0$

Step 2

Solve for $x$ .

If any individual factor on the left side of the equation is equal to $0$ , the entire expression will be equal to $0$ .

$x^{2} + 2 x - 15 = 0$

$x^{2} + 8 x + 17 = 0$

Set $x^{2} + 2 x - 15$ equal to $0$ and solve for $x$ .

Set $x^{2} + 2 x - 15$ equal to $0$ .

$x^{2} + 2 x - 15 = 0$

Solve $x^{2} + 2 x - 15 = 0$ for $x$ .

Factor $x^{2} + 2 x - 15$ 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 $- 15$ and whose sum is $2$ .

$- 3, 5$

Write the factored form using these integers.

$(x - 3) (x + 5) = 0$

If any individual factor on the left side of the equation is equal to $0$ , the entire expression will be equal to $0$ .

$x - 3 = 0$

$x + 5 = 0$

Set $x - 3$ equal to $0$ and solve for $x$ .

Set $x - 3$ equal to $0$ .

$x - 3 = 0$

Add $3$ to both sides of the equation.

$x = 3$

Set $x + 5$ equal to $0$ and solve for $x$ .

Set $x + 5$ equal to $0$ .

$x + 5 = 0$

Subtract $5$ from both sides of the equation.

$x = - 5$

The final solution is all the values that make $(x - 3) (x + 5) = 0$ true.

$x = 3, - 5$

Set $x^{2} + 8 x + 17$ equal to $0$ and solve for $x$ .

Set $x^{2} + 8 x + 17$ equal to $0$ .

$x^{2} + 8 x + 17 = 0$

Solve $x^{2} + 8 x + 17 = 0$ for $x$ .

Use the quadratic formula to find the solutions.

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

Substitute the values $a = 1$ , $b = 8$ , and $c = 17$ into the quadratic formula and solve for $x$ .

$\frac{- 8 \pm \sqrt{8^{2} - 4 \cdot (1 \cdot 17)}}{2 \cdot 1}$

Simplify.

Simplify the numerator.

Raise $8$ to the power of $2$ .

$x = \frac{- 8 \pm \sqrt{64 - 4 \cdot 1 \cdot 17}}{2 \cdot 1}$

Multiply $- 4 \cdot 1 \cdot 17$ .

Multiply $- 4$ by $1$ .

$x = \frac{- 8 \pm \sqrt{64 - 4 \cdot 17}}{2 \cdot 1}$

Multiply $- 4$ by $17$ .

$x = \frac{- 8 \pm \sqrt{64 - 68}}{2 \cdot 1}$

Subtract $68$ from $64$ .

$x = \frac{- 8 \pm \sqrt{- 4}}{2 \cdot 1}$

Rewrite $- 4$ as $- 1 (4)$ .

$x = \frac{- 8 \pm \sqrt{- 1 \cdot 4}}{2 \cdot 1}$

Rewrite $\sqrt{- 1 (4)}$ as $\sqrt{- 1} \cdot \sqrt{4}$ .

$x = \frac{- 8 \pm \sqrt{- 1} \cdot \sqrt{4}}{2 \cdot 1}$

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

$x = \frac{- 8 \pm i \cdot \sqrt{4}}{2 \cdot 1}$

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

$x = \frac{- 8 \pm i \cdot \sqrt{2^{2}}}{2 \cdot 1}$

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

$x = \frac{- 8 \pm i \cdot 2}{2 \cdot 1}$

Move $2$ to the left of $i$ .

$x = \frac{- 8 \pm 2 i}{2 \cdot 1}$

Multiply $2$ by $1$ .

$x = \frac{- 8 \pm 2 i}{2}$

Simplify $\frac{- 8 \pm 2 i}{2}$ .

$x = - 4 \pm i$

The final answer is the combination of both solutions.

$x = - 4 + i, - 4 - i$

The final solution is all the values that make $(x^{2} + 2 x - 15) (x^{2} + 8 x + 17) = 0$ true.

$x = 3, - 5, - 4 + i, - 4 - i$

Step 3

Find the Roots (Zeros) f(x)=(x^2+2x-15)(x^2+8x+17)

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