Find the x and y Intercepts f(x)=-4(2x+1)^2(x+2)^2(x^2+9)(x^2-5)

$f (x) = - 4 {(2 x + 1)}^{2} {(x + 2)}^{2} (x^{2} + 9) (x^{2} - 5)$

Find the x-intercepts.

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

$0 = - 4 {(2 x + 1)}^{2} {(x + 2)}^{2} (x^{2} + 9) (x^{2} - 5)$

Solve the equation.

Rewrite the equation as $- 4 {(2 x + 1)}^{2} {(x + 2)}^{2} (x^{2} + 9) (x^{2} - 5) = 0$ .

$- 4 {(2 x + 1)}^{2} {(x + 2)}^{2} (x^{2} + 9) (x^{2} - 5) = 0$

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

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

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

$x^{2} + 9 = 0$

$x^{2} - 5 = 0$

Set ${(2 x + 1)}^{2}$ equal to $0$ and solve for $x$ .

Set ${(2 x + 1)}^{2}$ equal to $0$ .

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

Solve ${(2 x + 1)}^{2} = 0$ for $x$ .

Set the $2 x + 1$ equal to $0$ .

$2 x + 1 = 0$

Solve for $x$ .

Subtract $1$ from both sides of the equation.

$2 x = - 1$

Divide each term in $2 x = - 1$ by $2$ and simplify.

Divide each term in $2 x = - 1$ by $2$ .

$\frac{2 x}{2} = \frac{- 1}{2}$

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

$\frac{2 x}{2} = \frac{- 1}{2}$

Divide $x$ by $1$ .

$x = \frac{- 1}{2}$

Simplify the right side.

Move the negative in front of the fraction.

$x = - \frac{1}{2}$

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

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

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

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

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

$x + 2 = 0$

Subtract $2$ from both sides of the equation.

$x = - 2$

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

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

$x^{2} + 9 = 0$

Solve $x^{2} + 9 = 0$ for $x$ .

Subtract $9$ from both sides of the equation.

$x^{2} = - 9$

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

$x = \pm \sqrt{- 9}$

Simplify $\pm \sqrt{- 9}$ .

Rewrite $- 9$ as $- 1 (9)$ .

$x = \pm \sqrt{- 1 (9)}$

Rewrite $\sqrt{- 1 (9)}$ as $\sqrt{- 1} \cdot \sqrt{9}$ .

$x = \pm \sqrt{- 1} \cdot \sqrt{9}$

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

$x = \pm i \cdot \sqrt{9}$

Rewrite $9$ as $3^{2}$ .

$x = \pm i \cdot \sqrt{3^{2}}$

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

$x = \pm i \cdot 3$

Move $3$ to the left of $i$ .

$x = \pm 3 i$

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.

$x = 3 i$

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

$x = - 3 i$

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

$x = 3 i, - 3 i$

Set $x^{2} - 5$ equal to $0$ and solve for $x$ .

Set $x^{2} - 5$ equal to $0$ .

$x^{2} - 5 = 0$

Solve $x^{2} - 5 = 0$ for $x$ .

Add $5$ to both sides of the equation.

$x^{2} = 5$

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

$x = \pm \sqrt{5}$

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.

$x = \sqrt{5}$

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

$x = - \sqrt{5}$

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

$x = \sqrt{5}, - \sqrt{5}$

The final solution is all the values that make $- 4 {(2 x + 1)}^{2} {(x + 2)}^{2} (x^{2} + 9) (x^{2} - 5) = 0$ true.

$x = - \frac{1}{2}, - 2, 3 i, - 3 i, \sqrt{5}, - \sqrt{5}$

x-intercept(s) in point form.

x-intercept(s): $(- \frac{1}{2}, 0), (- 2, 0), (\sqrt{5}, 0), (- \sqrt{5}, 0)$

Find the y-intercepts.

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

$y = - 4 {(2 (0) + 1)}^{2} {((0) + 2)}^{2} ({(0)}^{2} + 9) ({(0)}^{2} - 5)$

Solve the equation.

Remove parentheses.

$y = - 4 {(2 (0) + 1)}^{2} {(0 + 2)}^{2} ({(0)}^{2} + 9) ({(0)}^{2} - 5)$

Remove parentheses.

$y = - 4 {(2 (0) + 1)}^{2} {(0 + 2)}^{2} (0^{2} + 9) ({(0)}^{2} - 5)$

Remove parentheses.

$y = - 4 {(2 (0) + 1)}^{2} {(0 + 2)}^{2} (0^{2} + 9) (0^{2} - 5)$

Remove parentheses.

$y = - 4 {(2 (0) + 1)}^{2} {((0) + 2)}^{2} ({(0)}^{2} + 9) ({(0)}^{2} - 5)$

Simplify $- 4 {(2 (0) + 1)}^{2} {((0) + 2)}^{2} ({(0)}^{2} + 9) ({(0)}^{2} - 5)$ .

Multiply $2$ by $0$ .

$y = - 4 {(0 + 1)}^{2} {((0) + 2)}^{2} ({(0)}^{2} + 9) ({(0)}^{2} - 5)$

Add $0$ and $1$ .

$y = - 4 \cdot 1^{2} {((0) + 2)}^{2} ({(0)}^{2} + 9) ({(0)}^{2} - 5)$

One to any power is one.

$y = - 4 \cdot 1 {((0) + 2)}^{2} ({(0)}^{2} + 9) ({(0)}^{2} - 5)$

Multiply $- 4$ by $1$ .

$y = - 4 {((0) + 2)}^{2} ({(0)}^{2} + 9) ({(0)}^{2} - 5)$

Add $0$ and $2$ .

$y = - 4 \cdot 2^{2} ({(0)}^{2} + 9) ({(0)}^{2} - 5)$

Raise $2$ to the power of $2$ .

$y = - 4 \cdot 4 ({(0)}^{2} + 9) ({(0)}^{2} - 5)$

Multiply $- 4$ by $4$ .

$y = - 16 ({(0)}^{2} + 9) ({(0)}^{2} - 5)$

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

$y = - 16 (0 + 9) ({(0)}^{2} - 5)$

Add $0$ and $9$ .

$y = - 16 \cdot 9 ({(0)}^{2} - 5)$

Multiply $- 16$ by $9$ .

$y = - 144 ({(0)}^{2} - 5)$

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

$y = - 144 (0 - 5)$

Subtract $5$ from $0$ .

$y = - 144 \cdot - 5$

Multiply $- 144$ by $- 5$ .

$y = 720$

y-intercept(s) in point form.

y-intercept(s): $(0, 720)$

List the intersections.

x-intercept(s): $(- \frac{1}{2}, 0), (- 2, 0), (\sqrt{5}, 0), (- \sqrt{5}, 0)$

y-intercept(s): $(0, 720)$

Find the x and y Intercepts f(x)=-4(2x+1)^2(x+2)^2(x^2+9)(x^2-5)

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