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

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

Find the x-intercepts.

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

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

Solve the equation.

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

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

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

$4 x^{2} + 4 = 0$

$x - 5 = 0$

$x + 1 = 0$

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

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

$4 x^{2} + 4 = 0$

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

Subtract $4$ from both sides of the equation.

$4 x^{2} = - 4$

Divide each term in $4 x^{2} = - 4$ by $4$ and simplify.

Divide each term in $4 x^{2} = - 4$ by $4$ .

$\frac{4 x^{2}}{4} = \frac{- 4}{4}$

Simplify the left side.

Cancel the common factor of $4$ .

Cancel the common factor.

$\frac{4 x^{2}}{4} = \frac{- 4}{4}$

Divide $x^{2}$ by $1$ .

$x^{2} = \frac{- 4}{4}$

Simplify the right side.

Divide $- 4$ by $4$ .

$x^{2} = - 1$

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

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

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

$x = \pm i$

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.

$x = i$

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

$x = - i$

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

$x = i, - i$

Set $x - 5$ equal to $0$ and solve for $x$ .

Set $x - 5$ equal to $0$ .

$x - 5 = 0$

Add $5$ to both sides of the equation.

$x = 5$

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.

$x = - 1$

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

$x = i, - i, 5, - 1$

x-intercept(s) in point form.

x-intercept(s): $(5, 0), (- 1, 0)$

Find the y-intercepts.

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

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

Solve the equation.

Remove parentheses.

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

Remove parentheses.

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

Remove parentheses.

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

Remove parentheses.

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

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

Simplify each term.

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

$y = - 2 (4 \cdot 0 + 4) ((0) - 5) ((0) + 1)$

Multiply $4$ by $0$ .

$y = - 2 (0 + 4) ((0) - 5) ((0) + 1)$

Simplify the expression.

Add $0$ and $4$ .

$y = - 2 \cdot 4 ((0) - 5) ((0) + 1)$

Multiply $- 2$ by $4$ .

$y = - 8 ((0) - 5) ((0) + 1)$

Subtract $5$ from $0$ .

$y = - 8 \cdot - 5 ((0) + 1)$

Multiply $- 8$ by $- 5$ .

$y = 40 ((0) + 1)$

Add $0$ and $1$ .

$y = 40 \cdot 1$

Multiply $40$ by $1$ .

$y = 40$

y-intercept(s) in point form.

y-intercept(s): $(0, 40)$

List the intersections.

x-intercept(s): $(5, 0), (- 1, 0)$

y-intercept(s): $(0, 40)$

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

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