Algebra Examples

$f (x) = \frac{1}{3} (x + 5) (x - 1)$

Step 1

Set $\frac{1}{3} \cdot ((x + 5) (x - 1))$ equal to $0$ .

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

Step 2

Solve for $x$ .

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Step 2.1

Multiply both sides of the equation by $3$ .

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

Step 2.2

Simplify both sides of the equation.

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Step 2.2.1

Simplify the left side.

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Step 2.2.1.1

Simplify $3 (\frac{1}{3} \cdot ((x + 5) (x - 1)))$ .

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Step 2.2.1.1.1

Expand $(x + 5) (x - 1)$ using the FOIL Method.

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Step 2.2.1.1.1.1

Apply the distributive property.

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

Step 2.2.1.1.1.2

Apply the distributive property.

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

Step 2.2.1.1.1.3

Apply the distributive property.

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

Step 2.2.1.1.2

Simplify and combine like terms.

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Step 2.2.1.1.2.1

Simplify each term.

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Step 2.2.1.1.2.1.1

Multiply $x$ by $x$ .

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

Step 2.2.1.1.2.1.2

Move $- 1$ to the left of $x$ .

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

Step 2.2.1.1.2.1.3

Rewrite $- 1 x$ as $- x$ .

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

Step 2.2.1.1.2.1.4

Multiply $5$ by $- 1$ .

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

Step 2.2.1.1.2.2

Add $- x$ and $5 x$ .

$3 (\frac{1}{3} \cdot (x^{2} + 4 x - 5)) = 3 \cdot 0$

Step 2.2.1.1.3

Apply the distributive property.

$3 (\frac{1}{3} x^{2} + \frac{1}{3} (4 x) + \frac{1}{3} \cdot - 5) = 3 \cdot 0$

Step 2.2.1.1.4

Simplify.

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Step 2.2.1.1.4.1

Combine $\frac{1}{3}$ and $x^{2}$ .

$3 (\frac{x^{2}}{3} + \frac{1}{3} (4 x) + \frac{1}{3} \cdot - 5) = 3 \cdot 0$

Step 2.2.1.1.4.2

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

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Step 2.2.1.1.4.2.1

Combine $4$ and $\frac{1}{3}$ .

$3 (\frac{x^{2}}{3} + \frac{4}{3} x + \frac{1}{3} \cdot - 5) = 3 \cdot 0$

Step 2.2.1.1.4.2.2

Combine $\frac{4}{3}$ and $x$ .

$3 (\frac{x^{2}}{3} + \frac{4 x}{3} + \frac{1}{3} \cdot - 5) = 3 \cdot 0$

Step 2.2.1.1.4.3

Combine $\frac{1}{3}$ and $- 5$ .

$3 (\frac{x^{2}}{3} + \frac{4 x}{3} + \frac{- 5}{3}) = 3 \cdot 0$

Step 2.2.1.1.5

Move the negative in front of the fraction.

$3 (\frac{x^{2}}{3} + \frac{4 x}{3} - \frac{5}{3}) = 3 \cdot 0$

Step 2.2.1.1.6

Apply the distributive property.

$3 \frac{x^{2}}{3} + 3 \frac{4 x}{3} + 3 (- \frac{5}{3}) = 3 \cdot 0$

Step 2.2.1.1.7

Simplify.

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Step 2.2.1.1.7.1

Cancel the common factor of $3$ .

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Step 2.2.1.1.7.1.1

Cancel the common factor.

$3 \frac{x^{2}}{3} + 3 \frac{4 x}{3} + 3 (- \frac{5}{3}) = 3 \cdot 0$

Step 2.2.1.1.7.1.2

Rewrite the expression.

$x^{2} + 3 \frac{4 x}{3} + 3 (- \frac{5}{3}) = 3 \cdot 0$

Step 2.2.1.1.7.2

Cancel the common factor of $3$ .

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Step 2.2.1.1.7.2.1

Cancel the common factor.

$x^{2} + 3 \frac{4 x}{3} + 3 (- \frac{5}{3}) = 3 \cdot 0$

Step 2.2.1.1.7.2.2

Rewrite the expression.

$x^{2} + 4 x + 3 (- \frac{5}{3}) = 3 \cdot 0$

Step 2.2.1.1.7.3

Cancel the common factor of $3$ .

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Step 2.2.1.1.7.3.1

Move the leading negative in $- \frac{5}{3}$ into the numerator.

$x^{2} + 4 x + 3 (\frac{- 5}{3}) = 3 \cdot 0$

Step 2.2.1.1.7.3.2

Cancel the common factor.

$x^{2} + 4 x + 3 (\frac{- 5}{3}) = 3 \cdot 0$

Step 2.2.1.1.7.3.3

Rewrite the expression.

$x^{2} + 4 x - 5 = 3 \cdot 0$

Step 2.2.2

Simplify the right side.

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Step 2.2.2.1

Multiply $3$ by $0$ .

$x^{2} + 4 x - 5 = 0$

Step 2.3

Factor $x^{2} + 4 x - 5$ using the AC method.

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Step 2.3.1

Consider the form $x^{2} + b x + c$ . Find a pair of integers whose product is $c$ and whose sum is $b$ . In this case, whose product is $- 5$ and whose sum is $4$ .

$- 1, 5$

Step 2.3.2

Write the factored form using these integers.

$(x - 1) (x + 5) = 0$

Step 2.4

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

$x - 1 = 0$

$x + 5 = 0$

Step 2.5

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

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Step 2.5.1

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

$x - 1 = 0$

Step 2.5.2

Add $1$ to both sides of the equation.

$x = 1$

Step 2.6

Set $x + 5$ equal to $0$ and solve for $x$ .

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Step 2.6.1

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

$x + 5 = 0$

Step 2.6.2

Subtract $5$ from both sides of the equation.

$x = - 5$

Step 2.7

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

$x = 1, - 5$

Step 3