Find the x and y Intercepts f(X)=-5x^7+5x^3

$f (X) = - 5 x^{7} + 5 x^{3}$

Find the x-intercepts.

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

$0 = - 5 x^{7} + 5 x^{3}$

Solve the equation.

Rewrite the equation as $- 5 x^{7} + 5 x^{3} = 0$ .

$- 5 x^{7} + 5 x^{3} = 0$

Factor the left side of the equation.

Factor $- 5 x^{3}$ out of $- 5 x^{7} + 5 x^{3}$ .

Factor $- 5 x^{3}$ out of $- 5 x^{7}$ .

$- 5 x^{3} x^{4} + 5 x^{3} = 0$

Factor $- 5 x^{3}$ out of $5 x^{3}$ .

$- 5 x^{3} x^{4} - 5 x^{3} \cdot - 1 = 0$

Factor $- 5 x^{3}$ out of $- 5 x^{3} (x^{4}) - 5 x^{3} (- 1)$ .

$- 5 x^{3} (x^{4} - 1) = 0$

Rewrite $x^{4}$ as ${(x^{2})}^{2}$ .

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

Rewrite $1$ as $1^{2}$ .

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

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = x^{2}$ and $b = 1$ .

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

Factor.

Simplify.

Rewrite $1$ as $1^{2}$ .

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

Factor.

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = x$ and $b = 1$ .

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

Remove unnecessary parentheses.

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

Remove unnecessary parentheses.

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

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

$x^{3} = 0$

$x^{2} + 1 = 0$

$x + 1 = 0$

$x - 1 = 0$

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

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

$x^{3} = 0$

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

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

$x = \sqrt[3]{0}$

Simplify $\sqrt[3]{0}$ .

Rewrite $0$ as $0^{3}$ .

$x = \sqrt[3]{0^{3}}$

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

$x = 0$

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

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

$x^{2} + 1 = 0$

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

Subtract $1$ from both sides of the equation.

$x^{2} = - 1$

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 $x + 1$ equal to $0$ and solve for $x$ .

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

$x + 1 = 0$

Subtract $1$ from both sides of the equation.

$x = - 1$

Set $x - 1$ equal to $0$ and solve for $x$ .

Set $x - 1$ equal to $0$ .

$x - 1 = 0$

Add $1$ to both sides of the equation.

$x = 1$

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

$x = 0, i, - i, - 1, 1$

x-intercept(s) in point form.

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

Find the y-intercepts.

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

$y = - 5 {(0)}^{7} + 5 {(0)}^{3}$

Solve the equation.

Remove parentheses.

$y = - 5 \cdot 0^{7} + 5 {(0)}^{3}$

Remove parentheses.

$y = - 5 \cdot 0^{7} + 5 \cdot 0^{3}$

Remove parentheses.

$y = - 5 {(0)}^{7} + 5 {(0)}^{3}$

Simplify $- 5 {(0)}^{7} + 5 {(0)}^{3}$ .

Simplify each term.

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

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

Multiply $- 5$ by $0$ .

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

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

$y = 0 + 5 \cdot 0$

Multiply $5$ by $0$ .

$y = 0 + 0$

Add $0$ and $0$ .

$y = 0$

y-intercept(s) in point form.

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

List the intersections.

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

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

Find the x and y Intercepts f(X)=-5x^7+5x^3

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