Find the x and y Intercepts f(x)=-5x(2x+1)^2(x^2-6)(x^2+9)

$f (x) = - 5 x {(2 x + 1)}^{2} (x^{2} - 6) (x^{2} + 9)$

Find the x-intercepts.

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

$0 = - 5 x {(2 x + 1)}^{2} (x^{2} - 6) (x^{2} + 9)$

Solve the equation.

Rewrite the equation as $- 5 x {(2 x + 1)}^{2} (x^{2} - 6) (x^{2} + 9) = 0$ .

$- 5 x {(2 x + 1)}^{2} (x^{2} - 6) (x^{2} + 9) = 0$

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

$x = 0$

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

$x^{2} - 6 = 0$

$x^{2} + 9 = 0$

Set $x$ equal to $0$ .

$x = 0$

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 $x^{2} - 6$ equal to $0$ and solve for $x$ .

Set $x^{2} - 6$ equal to $0$ .

$x^{2} - 6 = 0$

Solve $x^{2} - 6 = 0$ for $x$ .

Add $6$ to both sides of the equation.

$x^{2} = 6$

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

$x = \pm \sqrt{6}$

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.

$x = \sqrt{6}$

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

$x = - \sqrt{6}$

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

$x = \sqrt{6}, - \sqrt{6}$

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.

$x = 3 i, - 3 i$

The final solution is all the values that make $- 5 x {(2 x + 1)}^{2} (x^{2} - 6) (x^{2} + 9) = 0$ true.

$x = 0, - \frac{1}{2}, \sqrt{6}, - \sqrt{6}, 3 i, - 3 i$

x-intercept(s) in point form.

x-intercept(s): $(0, 0), (- \frac{1}{2}, 0), (\sqrt{6}, 0), (- \sqrt{6}, 0)$

Find the y-intercepts.

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

$y = - 5 \cdot 0 {(2 (0) + 1)}^{2} ({(0)}^{2} - 6) ({(0)}^{2} + 9)$

Solve the equation.

Remove parentheses.

$y = - 5 \cdot (0 {(2 (0) + 1)}^{2} (0^{2} - 6) ({(0)}^{2} + 9))$

Remove parentheses.

$y = - 5 \cdot (0 {(2 (0) + 1)}^{2} (0^{2} - 6) (0^{2} + 9))$

Remove parentheses.

$y = - 5 \cdot (0 {(2 (0) + 1)}^{2} ({(0)}^{2} - 6) ({(0)}^{2} + 9))$

Simplify $- 5 \cdot (0 {(2 (0) + 1)}^{2} ({(0)}^{2} - 6) ({(0)}^{2} + 9))$ .

Multiply $2$ by $0$ .

$y = - 5 \cdot (0 {(0 + 1)}^{2} ({(0)}^{2} - 6) ({(0)}^{2} + 9))$

Add $0$ and $1$ .

$y = - 5 \cdot (0 \cdot 1^{2} ({(0)}^{2} - 6) ({(0)}^{2} + 9))$

One to any power is one.

$y = - 5 \cdot (0 \cdot 1 ({(0)}^{2} - 6) ({(0)}^{2} + 9))$

Multiply $0$ by $1$ .

$y = - 5 \cdot (0 ({(0)}^{2} - 6) ({(0)}^{2} + 9))$

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

$y = - 5 \cdot (0 (0 - 6) ({(0)}^{2} + 9))$

Subtract $6$ from $0$ .

$y = - 5 \cdot (0 \cdot - 6 ({(0)}^{2} + 9))$

Multiply $0$ by $- 6$ .

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

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

$y = - 5 \cdot (0 (0 + 9))$

Add $0$ and $9$ .

$y = - 5 \cdot (0 \cdot 9)$

Multiply $0$ by $9$ .

$y = - 5 \cdot 0$

Multiply $- 5$ by $0$ .

$y = 0$

y-intercept(s) in point form.

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

List the intersections.

x-intercept(s): $(0, 0), (- \frac{1}{2}, 0), (\sqrt{6}, 0), (- \sqrt{6}, 0)$

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

Find the x and y Intercepts f(x)=-5x(2x+1)^2(x^2-6)(x^2+9)

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