Precalculus Examples

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

Step 1

Solve the equation for $y$ .

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Step 1.1

Remove parentheses.

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

Step 1.2

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

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Step 1.2.1

Apply the distributive property.

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

Step 1.2.2

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

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Step 1.2.2.1

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

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

Step 1.2.2.2

Add $3$ and $2$ .

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

Step 1.2.3

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

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

Step 1.2.4

Apply the distributive property.

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

Step 1.2.5

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

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

Step 1.2.6

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

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Step 1.2.6.1

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

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

Step 1.2.6.2

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

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

Step 1.2.7

Move the negative in front of the fraction.

$y = \frac{x^{5}}{4} - \frac{9 x^{3}}{4}$

Step 2

A linear equation is an equation of a straight line, which means that the degree of a linear equation must be $0$ or $1$ for each of its variables. In this case, the degree of variable $y$ is $1$ and the degree of variable $x$ is $5$ .

Not Linear