Find the x and y Intercepts y=-(x+3)(x+1)^2(2x-5)

$y = - (x + 3) {(x + 1)}^{2} (2 x - 5)$

Find the x-intercepts.

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

$0 = - (x + 3) {(x + 1)}^{2} (2 x - 5)$

Solve the equation.

Rewrite the equation as $- (x + 3) {(x + 1)}^{2} (2 x - 5) = 0$ .

$- (x + 3) {(x + 1)}^{2} (2 x - 5) = 0$

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

$x + 3 = 0$

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

$2 x - 5 = 0$

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

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

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

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

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

$x + 1 = 0$

Subtract $1$ from both sides of the equation.

$x = - 1$

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

Set $2 x - 5$ equal to $0$ .

$2 x - 5 = 0$

Solve $2 x - 5 = 0$ for $x$ .

Add $5$ to both sides of the equation.

$2 x = 5$

Divide each term in $2 x = 5$ by $2$ and simplify.

Divide each term in $2 x = 5$ by $2$ .

$\frac{2 x}{2} = \frac{5}{2}$

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

$\frac{2 x}{2} = \frac{5}{2}$

Divide $x$ by $1$ .

$x = \frac{5}{2}$

The final solution is all the values that make $- (x + 3) {(x + 1)}^{2} (2 x - 5) = 0$ true.

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

x-intercept(s) in point form.

x-intercept(s): $(- 3, 0), (- 1, 0), (\frac{5}{2}, 0)$

Find the y-intercepts.

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

$y = - ((0) + 3) {((0) + 1)}^{2} (2 (0) - 5)$

Solve the equation.

Remove parentheses.

$y = - (0 + 3) {((0) + 1)}^{2} (2 (0) - 5)$

Remove parentheses.

$y = - (0 + 3) {(0 + 1)}^{2} (2 (0) - 5)$

Remove parentheses.

$y = - ((0) + 3) {((0) + 1)}^{2} (2 (0) - 5)$

Simplify $- ((0) + 3) {((0) + 1)}^{2} (2 (0) - 5)$ .

Add $0$ and $3$ .

$y = - 1 \cdot 3 {((0) + 1)}^{2} (2 (0) - 5)$

Multiply $- 1$ by $3$ .

$y = - 3 {((0) + 1)}^{2} (2 (0) - 5)$

Add $0$ and $1$ .

$y = - 3 \cdot 1^{2} (2 (0) - 5)$

One to any power is one.

$y = - 3 \cdot 1 (2 (0) - 5)$

Multiply $- 3$ by $1$ .

$y = - 3 (2 (0) - 5)$

Multiply $2$ by $0$ .

$y = - 3 (0 - 5)$

Subtract $5$ from $0$ .

$y = - 3 \cdot - 5$

Multiply $- 3$ by $- 5$ .

$y = 15$

y-intercept(s) in point form.

y-intercept(s): $(0, 15)$

List the intersections.

x-intercept(s): $(- 3, 0), (- 1, 0), (\frac{5}{2}, 0)$

y-intercept(s): $(0, 15)$

Find the x and y Intercepts y=-(x+3)(x+1)^2(2x-5)

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