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

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

Step 1

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

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

Step 2

Solve for $x$ .

Factor the left side of the equation.

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

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

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

Factor $2$ out of $2 x$ .

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

Factor $2$ out of $- 24$ .

$2 (x^{2}) + 2 x + 2 \cdot - 12 = 0$

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

$2 (x^{2} + x) + 2 \cdot - 12 = 0$

Factor $2$ out of $2 (x^{2} + x) + 2 \cdot - 12$ .

$2 (x^{2} + x - 12) = 0$

Factor.

Factor $x^{2} + x - 12$ 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 $- 12$ and whose sum is $1$ .

$- 3, 4$

Write the factored form using these integers.

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

Remove unnecessary parentheses.

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

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

$x - 3 = 0$

$x + 4 = 0$

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 $x + 4$ equal to $0$ and solve for $x$ .

Set $x + 4$ equal to $0$ .

$x + 4 = 0$

Subtract $4$ from both sides of the equation.

$x = - 4$

The final solution is all the values that make $2 (x - 3) (x + 4) = 0$ true.

$x = 3, - 4$

Step 3

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

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