Identify the Zeros and Their Multiplicities f(x)=3x^2-5x-13

$f (x) = 3 x^{2} - 5 x - 13$

Step 1

Set $3 x^{2} - 5 x - 13$ equal to $0$ .

$3 x^{2} - 5 x - 13 = 0$

Step 2

Solve for $x$ .

Use the quadratic formula to find the solutions.

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

Substitute the values $a = 3$ , $b = - 5$ , and $c = - 13$ into the quadratic formula and solve for $x$ .

$\frac{5 \pm \sqrt{{(- 5)}^{2} - 4 \cdot (3 \cdot - 13)}}{2 \cdot 3}$

Simplify.

Simplify the numerator.

Raise $- 5$ to the power of $2$ .

$x = \frac{5 \pm \sqrt{25 - 4 \cdot 3 \cdot - 13}}{2 \cdot 3}$

Multiply $- 4 \cdot 3 \cdot - 13$ .

Multiply $- 4$ by $3$ .

$x = \frac{5 \pm \sqrt{25 - 12 \cdot - 13}}{2 \cdot 3}$

Multiply $- 12$ by $- 13$ .

$x = \frac{5 \pm \sqrt{25 + 156}}{2 \cdot 3}$

Add $25$ and $156$ .

$x = \frac{5 \pm \sqrt{181}}{2 \cdot 3}$

Multiply $2$ by $3$ .

$x = \frac{5 \pm \sqrt{181}}{6}$

Simplify the expression to solve for the $+$ portion of the $\pm$ .

Simplify the numerator.

Raise $- 5$ to the power of $2$ .

$x = \frac{5 \pm \sqrt{25 - 4 \cdot 3 \cdot - 13}}{2 \cdot 3}$

Multiply $- 4 \cdot 3 \cdot - 13$ .

Multiply $- 4$ by $3$ .

$x = \frac{5 \pm \sqrt{25 - 12 \cdot - 13}}{2 \cdot 3}$

Multiply $- 12$ by $- 13$ .

$x = \frac{5 \pm \sqrt{25 + 156}}{2 \cdot 3}$

Add $25$ and $156$ .

$x = \frac{5 \pm \sqrt{181}}{2 \cdot 3}$

Multiply $2$ by $3$ .

$x = \frac{5 \pm \sqrt{181}}{6}$

Change the $\pm$ to $+$ .

$x = \frac{5 + \sqrt{181}}{6}$

Simplify the expression to solve for the $-$ portion of the $\pm$ .

Simplify the numerator.

Raise $- 5$ to the power of $2$ .

$x = \frac{5 \pm \sqrt{25 - 4 \cdot 3 \cdot - 13}}{2 \cdot 3}$

Multiply $- 4 \cdot 3 \cdot - 13$ .

Multiply $- 4$ by $3$ .

$x = \frac{5 \pm \sqrt{25 - 12 \cdot - 13}}{2 \cdot 3}$

Multiply $- 12$ by $- 13$ .

$x = \frac{5 \pm \sqrt{25 + 156}}{2 \cdot 3}$

Add $25$ and $156$ .

$x = \frac{5 \pm \sqrt{181}}{2 \cdot 3}$

Multiply $2$ by $3$ .

$x = \frac{5 \pm \sqrt{181}}{6}$

Change the $\pm$ to $-$ .

$x = \frac{5 - \sqrt{181}}{6}$

The final answer is the combination of both solutions.

$x = \frac{5 + \sqrt{181}}{6}$ (Multiplicity of $1$ )

$x = \frac{5 - \sqrt{181}}{6}$ (Multiplicity of $1$ )

$x = \frac{5 + \sqrt{181}}{6}$ (Multiplicity of $1$ )

$x = \frac{5 - \sqrt{181}}{6}$ (Multiplicity of $1$ )

Step 3

Identify the Zeros and Their Multiplicities f(x)=3x^2-5x-13

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