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

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

Step 1

Find the x-intercepts.

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

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

Solve the equation.

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

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

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

$x^{2} = 0$

$x - 5 = 0$

$x^{2} + 2 = 0$

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

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

$x^{2} = 0$

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

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

$x = \pm \sqrt{0}$

Simplify $\pm \sqrt{0}$ .

Rewrite $0$ as $0^{2}$ .

$x = \pm \sqrt{0^{2}}$

Pull terms out from under the radical, assuming positive real numbers.

$x = \pm 0$

Plus or minus $0$ is $0$ .

$x = 0$

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

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

$x^{2} + 2 = 0$

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

Subtract $2$ from both sides of the equation.

$x^{2} = - 2$

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

$x = \pm \sqrt{- 2}$

Simplify $\pm \sqrt{- 2}$ .

Rewrite $- 2$ as $- 1 (2)$ .

$x = \pm \sqrt{- 1 (2)}$

Rewrite $\sqrt{- 1 (2)}$ as $\sqrt{- 1} \cdot \sqrt{2}$ .

$x = \pm \sqrt{- 1} \cdot \sqrt{2}$

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

$x = \pm i \sqrt{2}$

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.

$x = i \sqrt{2}$

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

$x = - i \sqrt{2}$

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

$x = i \sqrt{2}, - i \sqrt{2}$

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

$x = 0, 5, i \sqrt{2}, - i \sqrt{2}$

x-intercept(s) in point form.

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

Step 2

Find the y-intercepts.

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

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

Solve the equation.

Remove parentheses.

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

Remove parentheses.

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

Remove parentheses.

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

Remove parentheses.

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

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

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

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

Subtract $5$ from $0$ .

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

Multiply $0$ by $- 5$ .

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

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

$y = 0 (0 + 2)$

Add $0$ and $2$ .

$y = 0 \cdot 2$

Multiply $0$ by $2$ .

$y = 0$

y-intercept(s) in point form.

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

Step 3

List the intersections.

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

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

Step 4

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

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