Find the Roots (Zeros) f(x)=4x^2+4x-24

$f (x) = 4 x^{2} + 4 x - 24$

Set $4 x^{2} + 4 x - 24$ equal to $0$ .

$4 x^{2} + 4 x - 24 = 0$

Solve for $x$ .

Factor the left side of the equation.

Factor $4$ out of $4 x^{2} + 4 x - 24$ .

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

$4 (x^{2}) + 4 x - 24 = 0$

Factor $4$ out of $4 x$ .

$4 (x^{2}) + 4 (x) - 24 = 0$

Factor $4$ out of $- 24$ .

$4 (x^{2}) + 4 x + 4 \cdot - 6 = 0$

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

$4 (x^{2} + x) + 4 \cdot - 6 = 0$

Factor $4$ out of $4 (x^{2} + x) + 4 \cdot - 6$ .

$4 (x^{2} + x - 6) = 0$

Factor.

Factor $x^{2} + x - 6$ 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 $- 6$ and whose sum is $1$ .

$- 2, 3$

Write the factored form using these integers.

$4 ((x - 2) (x + 3)) = 0$

Remove unnecessary parentheses.

$4 (x - 2) (x + 3) = 0$

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

$x - 2 = 0$

$x + 3 = 0$

Set $x - 2$ equal to $0$ and solve for $x$ .

Set $x - 2$ equal to $0$ .

$x - 2 = 0$

Add $2$ to both sides of the equation.

$x = 2$

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

Set $x + 3$ 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 $4 (x - 2) (x + 3) = 0$ true.

$x = 2, - 3$

Find the Roots (Zeros) f(x)=4x^2+4x-24

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