Find the x and y Intercepts f(x)=(3x+2)(x+1)(x^2+4)

$f (x) = (3 x + 2) (x + 1) (x^{2} + 4)$

Find the x-intercepts.

To find the x-intercept(s), substitute in $0$ for $y$ and solve for $x$ .

$0 = (3 x + 2) (x + 1) (x^{2} + 4)$

Solve the equation.

Rewrite the equation as $(3 x + 2) (x + 1) (x^{2} + 4) = 0$ .

$(3 x + 2) (x + 1) (x^{2} + 4) = 0$

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

$3 x + 2 = 0$

$x + 1 = 0$

$x^{2} + 4 = 0$

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

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

$3 x + 2 = 0$

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

Subtract $2$ from both sides of the equation.

$3 x = - 2$

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

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

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

Simplify the left side.

Cancel the common factor of $3$ .

Cancel the common factor.

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

Divide $x$ by $1$ .

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

Simplify the right side.

Move the negative in front of the fraction.

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

Set $x + 1$ equal to $0$ and solve for $x$ .

Set $x + 1$ equal to $0$ .

$x + 1 = 0$

Subtract $1$ from both sides of the equation.

$x = - 1$

Set $x^{2} + 4$ equal to $0$ and solve for $x$ .

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

$x^{2} + 4 = 0$

Solve $x^{2} + 4 = 0$ for $x$ .

Subtract $4$ from both sides of the equation.

$x^{2} = - 4$

Take the square root of both sides of the equation to eliminate the exponent on the left side.

$x = \pm \sqrt{- 4}$

Simplify $\pm \sqrt{- 4}$ .

Rewrite $- 4$ as $- 1 (4)$ .

$x = \pm \sqrt{- 1 (4)}$

Rewrite $\sqrt{- 1 (4)}$ as $\sqrt{- 1} \cdot \sqrt{4}$ .

$x = \pm \sqrt{- 1} \cdot \sqrt{4}$

Rewrite $\sqrt{- 1}$ as $i$ .

$x = \pm i \cdot \sqrt{4}$

Rewrite $4$ as $2^{2}$ .

$x = \pm i \cdot \sqrt{2^{2}}$

Pull terms out from under the radical, assuming positive real numbers.

$x = \pm i \cdot 2$

Move $2$ to the left of $i$ .

$x = \pm 2 i$

The complete solution is the result of both the positive and negative portions of the solution.

First, use the positive value of the $\pm$ to find the first solution.

$x = 2 i$

Next, use the negative value of the $\pm$ to find the second solution.

$x = - 2 i$

The complete solution is the result of both the positive and negative portions of the solution.

$x = 2 i, - 2 i$

The final solution is all the values that make $(3 x + 2) (x + 1) (x^{2} + 4) = 0$ true.

$x = - \frac{2}{3}, - 1, 2 i, - 2 i$

x-intercept(s) in point form.

x-intercept(s): $(- \frac{2}{3}, 0), (- 1, 0)$

Find the y-intercepts.

To find the y-intercept(s), substitute in $0$ for $x$ and solve for $y$ .

$y = (3 (0) + 2) ((0) + 1) ({(0)}^{2} + 4)$

Solve the equation.

Remove parentheses.

$y = (3 (0) + 2) (0 + 1) ({(0)}^{2} + 4)$

Remove parentheses.

$y = (3 (0) + 2) (0 + 1) (0^{2} + 4)$

Remove parentheses.

$y = (3 (0) + 2) ((0) + 1) ({(0)}^{2} + 4)$

Simplify $(3 (0) + 2) ((0) + 1) ({(0)}^{2} + 4)$ .

Multiply $3$ by $0$ .

$y = (0 + 2) ((0) + 1) ({(0)}^{2} + 4)$

Add $0$ and $2$ .

$y = 2 ((0) + 1) ({(0)}^{2} + 4)$

Add $0$ and $1$ .

$y = 2 \cdot 1 ({(0)}^{2} + 4)$

Multiply $2$ by $1$ .

$y = 2 ({(0)}^{2} + 4)$

Raising $0$ to any positive power yields $0$ .

$y = 2 (0 + 4)$

Add $0$ and $4$ .

$y = 2 \cdot 4$

Multiply $2$ by $4$ .

$y = 8$

y-intercept(s) in point form.

y-intercept(s): $(0, 8)$

List the intersections.

x-intercept(s): $(- \frac{2}{3}, 0), (- 1, 0)$

y-intercept(s): $(0, 8)$

Find the x and y Intercepts f(x)=(3x+2)(x+1)(x^2+4)

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