Find the Roots (Zeros) f(x)=(2x+5)(x^2-2x-5)

$f (x) = (2 x + 5) (x^{2} - 2 x - 5)$

Step 1

Set $(2 x + 5) (x^{2} - 2 x - 5)$ equal to $0$ .

$(2 x + 5) (x^{2} - 2 x - 5) = 0$

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 $0$ .

$2 x + 5 = 0$

$x^{2} - 2 x - 5 = 0$

Set $2 x + 5$ equal to $0$ and solve for $x$ .

Set $2 x + 5$ equal to $0$ .

$2 x + 5 = 0$

Solve $2 x + 5 = 0$ for $x$ .

Subtract $5$ from both sides of the equation.

$2 x = - 5$

Divide each term in $2 x = - 5$ by $2$ and simplify.

Divide each term in $2 x = - 5$ by $2$ .

$\frac{2 x}{2} = \frac{- 5}{2}$

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

$\frac{2 x}{2} = \frac{- 5}{2}$

Divide $x$ by $1$ .

$x = \frac{- 5}{2}$

Simplify the right side.

Move the negative in front of the fraction.

$x = - \frac{5}{2}$

Set $x^{2} - 2 x - 5$ equal to $0$ and solve for $x$ .

Set $x^{2} - 2 x - 5$ equal to $0$ .

$x^{2} - 2 x - 5 = 0$

Solve $x^{2} - 2 x - 5 = 0$ for $x$ .

Use the quadratic formula to find the solutions.

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

Substitute the values $a = 1$ , $b = - 2$ , and $c = - 5$ into the quadratic formula and solve for $x$ .

$\frac{2 \pm \sqrt{{(- 2)}^{2} - 4 \cdot (1 \cdot - 5)}}{2 \cdot 1}$

Simplify.

Simplify the numerator.

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

$x = \frac{2 \pm \sqrt{4 - 4 \cdot 1 \cdot - 5}}{2 \cdot 1}$

Multiply $- 4 \cdot 1 \cdot - 5$ .

Multiply $- 4$ by $1$ .

$x = \frac{2 \pm \sqrt{4 - 4 \cdot - 5}}{2 \cdot 1}$

Multiply $- 4$ by $- 5$ .

$x = \frac{2 \pm \sqrt{4 + 20}}{2 \cdot 1}$

Add $4$ and $20$ .

$x = \frac{2 \pm \sqrt{24}}{2 \cdot 1}$

Rewrite $24$ as $2^{2} \cdot 6$ .

Factor $4$ out of $24$ .

$x = \frac{2 \pm \sqrt{4 (6)}}{2 \cdot 1}$

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

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

Pull terms out from under the radical.

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

Multiply $2$ by $1$ .

$x = \frac{2 \pm 2 \sqrt{6}}{2}$

Simplify $\frac{2 \pm 2 \sqrt{6}}{2}$ .

$x = 1 \pm \sqrt{6}$

The final answer is the combination of both solutions.

$x = 1 + \sqrt{6}, 1 - \sqrt{6}$

The final solution is all the values that make $(2 x + 5) (x^{2} - 2 x - 5) = 0$ true.

$x = - \frac{5}{2}, 1 + \sqrt{6}, 1 - \sqrt{6}$

Step 3

The result can be shown in multiple forms.

Exact Form:

$x = - \frac{5}{2}, 1 + \sqrt{6}, 1 - \sqrt{6}$

Decimal Form:

$x = - 2.5, 3.44948974 \dots, - 1.44948974 \dots$

Mixed Number Form:

$x = - 2 \frac{1}{2}, 1 + \sqrt{6}, 1 - \sqrt{6}$

Step 4

Find the Roots (Zeros) f(x)=(2x+5)(x^2-2x-5)

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