Find the Roots (Zeros) f(x)=(2x^2+5x-12)/(x+4)

$f (x) = \frac{2 x^{2} + 5 x - 12}{x + 4}$

Step 1

Set $\frac{2 x^{2} + 5 x - 12}{x + 4}$ equal to $0$ .

$\frac{2 x^{2} + 5 x - 12}{x + 4} = 0$

Step 2

Solve for $x$ .

Set the numerator equal to zero.

$2 x^{2} + 5 x - 12 = 0$

Solve the equation for $x$ .

Factor by grouping.

For a polynomial of the form $a x^{2} + b x + c$ , rewrite the middle term as a sum of two terms whose product is $a \cdot c = 2 \cdot - 12 = - 24$ and whose sum is $b = 5$ .

Factor $5$ out of $5 x$ .

$2 x^{2} + 5 (x) - 12 = 0$

Rewrite $5$ as $- 3$ plus $8$

$2 x^{2} + (- 3 + 8) x - 12 = 0$

Apply the distributive property.

$2 x^{2} - 3 x + 8 x - 12 = 0$

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

$(2 x^{2} - 3 x) + 8 x - 12 = 0$

Factor out the greatest common factor (GCF) from each group.

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

Factor the polynomial by factoring out the greatest common factor, $2 x - 3$ .

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

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

$2 x - 3 = 0$

$x + 4 = 0$

Set $2 x - 3$ equal to $0$ and solve for $x$ .

Set $2 x - 3$ equal to $0$ .

$2 x - 3 = 0$

Solve $2 x - 3 = 0$ for $x$ .

Add $3$ to both sides of the equation.

$2 x = 3$

Divide each term in $2 x = 3$ by $2$ and simplify.

Divide each term in $2 x = 3$ by $2$ .

$\frac{2 x}{2} = \frac{3}{2}$

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

$\frac{2 x}{2} = \frac{3}{2}$

Divide $x$ by $1$ .

$x = \frac{3}{2}$

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.

$x = \frac{3}{2}, - 4$

Exclude the solutions that do not make $\frac{2 x^{2} + 5 x - 12}{x + 4} = 0$ true.

$x = \frac{3}{2}$

Step 3

Find the Roots (Zeros) f(x)=(2x^2+5x-12)/(x+4)

Related Questions

Support LCMGCF.com by disabling your adblocker.