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

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

Find the x-intercepts.

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

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

Solve the equation.

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

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

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

$x^{2} - 3 = 0$

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

Set $x^{2} - 3$ equal to $0$ and solve for $x$ .

Set $x^{2} - 3$ equal to $0$ .

$x^{2} - 3 = 0$

Solve $x^{2} - 3 = 0$ for $x$ .

Add $3$ to both sides of the equation.

$x^{2} = 3$

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

$x = \pm \sqrt{3}$

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.

$x = \sqrt{3}$

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

$x = - \sqrt{3}$

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

$x = \sqrt{3}, - \sqrt{3}$

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.

$x = - 1$

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

$x = \sqrt{3}, - \sqrt{3}, - 1$

x-intercept(s) in point form.

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

Find the y-intercepts.

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

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

Solve the equation.

Remove parentheses.

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

Remove parentheses.

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

Remove parentheses.

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

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

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

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

Subtract $3$ from $0$ .

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

Add $0$ and $1$ .

$y = - 3 \cdot 1^{2}$

One to any power is one.

$y = - 3 \cdot 1$

Multiply $- 3$ by $1$ .

$y = - 3$

y-intercept(s) in point form.

y-intercept(s): $(0, - 3)$

List the intersections.

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

y-intercept(s): $(0, - 3)$

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

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