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

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

Step 1

Find the x-intercepts.

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

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

Solve the equation.

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

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

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

$x = 0$

$x + 3 = 0$

$x + 2 = 0$

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

Set $x$ equal to $0$ .

$x = 0$

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.

$x = - 3$

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

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

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

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

Factor the left side of the equation.

Factor $2$ out of $4 x + 2$ .

Factor $2$ out of $4 x$ .

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

Factor $2$ out of $2$ .

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

Factor $2$ out of $2 (2 x) + 2 (1)$ .

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

Apply the product rule to $2 (2 x + 1)$ .

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

Divide each term in $2^{2} {(2 x + 1)}^{2} = 0$ by $2^{2}$ and simplify.

Divide each term in $2^{2} {(2 x + 1)}^{2} = 0$ by $2^{2}$ .

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

Simplify the left side.

Cancel the common factor of $2^{2}$ .

Cancel the common factor.

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

Divide ${(2 x + 1)}^{2}$ by $1$ .

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

Simplify the right side.

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

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

Divide $0$ by $4$ .

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

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.

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

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

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

x-intercept(s) in point form.

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

Step 2

Find the y-intercepts.

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

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

Solve the equation.

Remove parentheses.

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

Remove parentheses.

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

Remove parentheses.

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

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

Add $0$ and $3$ .

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

Multiply $0$ by $3$ .

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

Add $0$ and $2$ .

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

Multiply $0$ by $2$ .

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

Multiply $4$ by $0$ .

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

Add $0$ and $2$ .

$y = 0 \cdot 2^{2}$

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

$y = 0 \cdot 4$

Multiply $0$ by $4$ .

$y = 0$

y-intercept(s) in point form.

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

Step 3

List the intersections.

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

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

Step 4

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

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