Find the Roots (Zeros) p(x)=(x+2)(2x^2+3x-9)

$p (x) = (x + 2) (2 x^{2} + 3 x - 9)$

Set $(x + 2) (2 x^{2} + 3 x - 9)$ equal to $0$ .

$(x + 2) (2 x^{2} + 3 x - 9) = 0$

Solve for $x$ .

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

$x + 2 = 0$

$2 x^{2} + 3 x - 9 = 0$

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

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

$x + 2 = 0$

Subtract $2$ from both sides of the equation.

$x = - 2$

Set $2 x^{2} + 3 x - 9$ equal to $0$ and solve for $x$ .

Set $2 x^{2} + 3 x - 9$ equal to $0$ .

$2 x^{2} + 3 x - 9 = 0$

Solve $2 x^{2} + 3 x - 9 = 0$ 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 $a \cdot c = 2 \cdot - 9 = - 18$ and whose sum is $b = 3$ .

Factor $3$ out of $3 x$ .

$2 x^{2} + 3 (x) - 9 = 0$

Rewrite $3$ as $- 3$ plus $6$

$2 x^{2} + (- 3 + 6) x - 9 = 0$

Apply the distributive property.

$2 x^{2} - 3 x + 6 x - 9 = 0$

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

$(2 x^{2} - 3 x) + 6 x - 9 = 0$

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

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

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

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

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

$2 x - 3 = 0$

$x + 3 = 0$

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

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

The final solution is all the values that make $(x + 2) (2 x^{2} + 3 x - 9) = 0$ true.

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

Find the Roots (Zeros) p(x)=(x+2)(2x^2+3x-9)

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