Find the Roots (Zeros) h(x)=4x^2-8x-60

$h (x) = 4 x^{2} - 8 x - 60$

Set $4 x^{2} - 8 x - 60$ equal to $0$ .

$4 x^{2} - 8 x - 60 = 0$

Solve for $x$ .

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Factor the left side of the equation.

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Factor $4$ out of $4 x^{2} - 8 x - 60$ .

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Factor $4$ out of $4 x^{2}$ .

$4 (x^{2}) - 8 x - 60 = 0$

Factor $4$ out of $- 8 x$ .

$4 (x^{2}) + 4 (- 2 x) - 60 = 0$

Factor $4$ out of $- 60$ .

$4 x^{2} + 4 (- 2 x) + 4 \cdot - 15 = 0$

Factor $4$ out of $4 x^{2} + 4 (- 2 x)$ .

$4 (x^{2} - 2 x) + 4 \cdot - 15 = 0$

Factor $4$ out of $4 (x^{2} - 2 x) + 4 \cdot - 15$ .

$4 (x^{2} - 2 x - 15) = 0$

Factor.

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Factor $x^{2} - 2 x - 15$ using the AC method.

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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 $- 2$ .

$- 5, 3$

Write the factored form using these integers.

$4 ((x - 5) (x + 3)) = 0$

Remove unnecessary parentheses.

$4 (x - 5) (x + 3) = 0$

Divide each term by $4$ and simplify.

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Divide each term in $4 (x - 5) (x + 3) = 0$ by $4$ .

$\frac{4 (x - 5) (x + 3)}{4} = \frac{0}{4}$

Simplify $\frac{4 (x - 5) (x + 3)}{4}$ .

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Cancel the common factor of $4$ .

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Cancel the common factor.

$\frac{4 (x - 5) (x + 3)}{4} = \frac{0}{4}$

Divide $(x - 5) (x + 3)$ by $1$ .

$(x - 5) (x + 3) = \frac{0}{4}$

Expand $(x - 5) (x + 3)$ using the FOIL Method.

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Apply the distributive property.

$x (x + 3) - 5 (x + 3) = \frac{0}{4}$

Apply the distributive property.

$x \cdot x + x \cdot 3 - 5 (x + 3) = \frac{0}{4}$

Apply the distributive property.

$x \cdot x + x \cdot 3 - 5 x - 5 \cdot 3 = \frac{0}{4}$

Simplify and combine like terms.

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Simplify each term.

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Multiply $x$ by $x$ .

$x^{2} + x \cdot 3 - 5 x - 5 \cdot 3 = \frac{0}{4}$

Move $3$ to the left of $x$ .

$x^{2} + 3 \cdot x - 5 x - 5 \cdot 3 = \frac{0}{4}$

Multiply $- 5$ by $3$ .

$x^{2} + 3 x - 5 x - 15 = \frac{0}{4}$

Subtract $5 x$ from $3 x$ .

$x^{2} - 2 x - 15 = \frac{0}{4}$

Divide $0$ by $4$ .

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

Factor $x^{2} - 2 x - 15$ using the AC method.

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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 $- 2$ .

$- 5, 3$

Write the factored form using these integers.

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

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

$x - 5 = 0$

$x + 3 = 0$

Set the first factor equal to $0$ and solve.

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Set the first factor equal to $0$ .

$x - 5 = 0$

Add $5$ to both sides of the equation.

$x = 5$

Set the next factor equal to $0$ and solve.

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Set the next factor equal to $0$ .

$x + 3 = 0$

Subtract $3$ from both sides of the equation.

$x = - 3$

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

$x = 5, - 3$

Find the Roots (Zeros) h(x)=4x^2-8x-60

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