Find the x and y Intercepts f(x)=x^2(x^2-3)(6x+5)

$f (x) = x^{2} (x^{2} - 3) (6 x + 5)$

Step 1

Find the x-intercepts.

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

$0 = x^{2} (x^{2} - 3) (6 x + 5)$

Solve the equation.

Rewrite the equation as $x^{2} (x^{2} - 3) (6 x + 5) = 0$ .

$x^{2} (x^{2} - 3) (6 x + 5) = 0$

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

$x^{2} = 0$

$x^{2} - 3 = 0$

$6 x + 5 = 0$

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 $x^{2} - 3$ 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 $6 x + 5$ equal to $0$ and solve for $x$ .

Set $6 x + 5$ equal to $0$ .

$6 x + 5 = 0$

Solve $6 x + 5 = 0$ for $x$ .

Subtract $5$ from both sides of the equation.

$6 x = - 5$

Divide each term in $6 x = - 5$ by $6$ and simplify.

Divide each term in $6 x = - 5$ by $6$ .

$\frac{6 x}{6} = \frac{- 5}{6}$

Simplify the left side.

Cancel the common factor of $6$ .

Cancel the common factor.

$\frac{6 x}{6} = \frac{- 5}{6}$

Divide $x$ by $1$ .

$x = \frac{- 5}{6}$

Simplify the right side.

Move the negative in front of the fraction.

$x = - \frac{5}{6}$

The final solution is all the values that make $x^{2} (x^{2} - 3) (6 x + 5) = 0$ true.

$x = 0, \sqrt{3}, - \sqrt{3}, - \frac{5}{6}$

x-intercept(s) in point form.

x-intercept(s): $(0, 0), (\sqrt{3}, 0), (- \sqrt{3}, 0), (- \frac{5}{6}, 0)$

Step 2

Find the y-intercepts.

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

$y = {(0)}^{2} ({(0)}^{2} - 3) (6 (0) + 5)$

Solve the equation.

Remove parentheses.

$y = 0^{2} ({(0)}^{2} - 3) (6 (0) + 5)$

Remove parentheses.

$y = 0^{2} (0^{2} - 3) (6 (0) + 5)$

Remove parentheses.

$y = {(0)}^{2} ({(0)}^{2} - 3) (6 (0) + 5)$

Simplify ${(0)}^{2} ({(0)}^{2} - 3) (6 (0) + 5)$ .

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

$y = 0 ({(0)}^{2} - 3) (6 (0) + 5)$

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

$y = 0 (0 - 3) (6 (0) + 5)$

Subtract $3$ from $0$ .

$y = 0 \cdot - 3 (6 (0) + 5)$

Multiply $0$ by $- 3$ .

$y = 0 (6 (0) + 5)$

Multiply $6$ by $0$ .

$y = 0 (0 + 5)$

Add $0$ and $5$ .

$y = 0 \cdot 5$

Multiply $0$ by $5$ .

$y = 0$

y-intercept(s) in point form.

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

Step 3

List the intersections.

x-intercept(s): $(0, 0), (\sqrt{3}, 0), (- \sqrt{3}, 0), (- \frac{5}{6}, 0)$

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

Step 4

Find the x and y Intercepts f(x)=x^2(x^2-3)(6x+5)

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