Find the Roots (Zeros) y=1/4x^3(x^2-9)

$y = \frac{1}{4} x^{3} (x^{2} - 9)$

Step 1

Set $\frac{1}{4} \cdot (x^{3} (x^{2} - 9))$ equal to $0$ .

$\frac{1}{4} \cdot (x^{3} (x^{2} - 9)) = 0$

Step 2

Solve for $x$ .

Multiply both sides of the equation by $4$ .

$4 (\frac{1}{4} \cdot (x^{3} (x^{2} - 9))) = 4 \cdot 0$

Simplify both sides of the equation.

Simplify the left side.

Simplify $4 (\frac{1}{4} \cdot (x^{3} (x^{2} - 9)))$ .

Apply the distributive property.

$4 (\frac{1}{4} \cdot (x^{3} x^{2} + x^{3} \cdot - 9)) = 4 \cdot 0$

Multiply $x^{3}$ by $x^{2}$ by adding the exponents.

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$4 (\frac{1}{4} \cdot (x^{3 + 2} + x^{3} \cdot - 9)) = 4 \cdot 0$

Add $3$ and $2$ .

$4 (\frac{1}{4} \cdot (x^{5} + x^{3} \cdot - 9)) = 4 \cdot 0$

Move $- 9$ to the left of $x^{3}$ .

$4 (\frac{1}{4} \cdot (x^{5} - 9 \cdot x^{3})) = 4 \cdot 0$

Apply the distributive property.

$4 (\frac{1}{4} x^{5} + \frac{1}{4} (- 9 x^{3})) = 4 \cdot 0$

Combine $\frac{1}{4}$ and $x^{5}$ .

$4 (\frac{x^{5}}{4} + \frac{1}{4} (- 9 x^{3})) = 4 \cdot 0$

Multiply $\frac{1}{4} (- 9 x^{3})$ .

Combine $- 9$ and $\frac{1}{4}$ .

$4 (\frac{x^{5}}{4} + \frac{- 9}{4} x^{3}) = 4 \cdot 0$

Combine $\frac{- 9}{4}$ and $x^{3}$ .

$4 (\frac{x^{5}}{4} + \frac{- 9 x^{3}}{4}) = 4 \cdot 0$

Move the negative in front of the fraction.

$4 (\frac{x^{5}}{4} - \frac{9 x^{3}}{4}) = 4 \cdot 0$

Simplify terms.

Apply the distributive property.

$4 \frac{x^{5}}{4} + 4 (- \frac{9 x^{3}}{4}) = 4 \cdot 0$

Cancel the common factor of $4$ .

Cancel the common factor.

$4 \frac{x^{5}}{4} + 4 (- \frac{9 x^{3}}{4}) = 4 \cdot 0$

Rewrite the expression.

$x^{5} + 4 (- \frac{9 x^{3}}{4}) = 4 \cdot 0$

Cancel the common factor of $4$ .

Move the leading negative in $- \frac{9 x^{3}}{4}$ into the numerator.

$x^{5} + 4 \frac{- 9 x^{3}}{4} = 4 \cdot 0$

Cancel the common factor.

$x^{5} + 4 \frac{- 9 x^{3}}{4} = 4 \cdot 0$

Rewrite the expression.

$x^{5} - 9 x^{3} = 4 \cdot 0$

Simplify the right side.

Multiply $4$ by $0$ .

$x^{5} - 9 x^{3} = 0$

Factor the left side of the equation.

Factor $x^{3}$ out of $x^{5} - 9 x^{3}$ .

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

$x^{3} x^{2} - 9 x^{3} = 0$

Factor $x^{3}$ out of $- 9 x^{3}$ .

$x^{3} x^{2} + x^{3} \cdot - 9 = 0$

Factor $x^{3}$ out of $x^{3} x^{2} + x^{3} \cdot - 9$ .

$x^{3} (x^{2} - 9) = 0$

Rewrite $9$ as $3^{2}$ .

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

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 $b = 3$ .

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

Remove unnecessary parentheses.

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

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

$x^{3} = 0$

$x + 3 = 0$

$x - 3 = 0$

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 $x + 3$ 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 $x - 3$ equal to $0$ and solve for $x$ .

Set $x - 3$ equal to $0$ .

$x - 3 = 0$

Add $3$ to both sides of the equation.

$x = 3$

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

$x = 0, - 3, 3$

Step 3

Find the Roots (Zeros) y=1/4x^3(x^2-9)

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