Find the Roots (Zeros) f(x)=1/3(x+5)(x-1)

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

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

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

Solve for $x$ .

Multiply both sides of the equation by $3$ .

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

Simplify both sides of the equation.

Simplify the left side.

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

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

Apply the distributive property.

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

Apply the distributive property.

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

Apply the distributive property.

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

Simplify and combine like terms.

Simplify each term.

Multiply $x$ by $x$ .

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

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

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

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

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

Multiply $5$ by $- 1$ .

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

Add $- x$ and $5 x$ .

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

Apply the distributive property.

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

Simplify.

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

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

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

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

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

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

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

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

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

Move the negative in front of the fraction.

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

Apply the distributive property.

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

Simplify.

Cancel the common factor of $3$ .

Cancel the common factor.

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

Rewrite the expression.

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

Cancel the common factor of $3$ .

Cancel the common factor.

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

Rewrite the expression.

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

Cancel the common factor of $3$ .

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

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

Cancel the common factor.

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

Rewrite the expression.

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

Simplify the right side.

Multiply $3$ by $0$ .

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

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

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 $4$ .

$- 1, 5$

Write the factored form using these integers.

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

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

$x - 1 = 0$

$x + 5 = 0$

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.

$x = 1$

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

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

$x + 5 = 0$

Subtract $5$ from both sides of the equation.

$x = - 5$

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

$x = 1, - 5$

Find the Roots (Zeros) f(x)=1/3(x+5)(x-1)

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