Identify the Zeros and Their Multiplicities f(x)=5x^3-x^2-180x+36

$f (x) = 5 x^{3} - x^{2} - 180 x + 36$

Step 1

Set $5 x^{3} - x^{2} - 180 x + 36$ equal to $0$ .

$5 x^{3} - x^{2} - 180 x + 36 = 0$

Step 2

Solve for $x$ .

Factor the left side of the equation.

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

$(5 x^{3} - x^{2}) - 180 x + 36 = 0$

Factor out the greatest common factor (GCF) from each group.

$x^{2} (5 x - 1) - 36 (5 x - 1) = 0$

Factor the polynomial by factoring out the greatest common factor, $5 x - 1$ .

$(5 x - 1) (x^{2} - 36) = 0$

Rewrite $36$ as $6^{2}$ .

$(5 x - 1) (x^{2} - 6^{2}) = 0$

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 $b = 6$ .

$(5 x - 1) ((x + 6) (x - 6)) = 0$

Remove unnecessary parentheses.

$(5 x - 1) (x + 6) (x - 6) = 0$

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

$5 x - 1 = 0$

$x + 6 = 0$

$x - 6 = 0$

Set $5 x - 1$ equal to $0$ and solve for $x$ .

Set $5 x - 1$ equal to $0$ .

$5 x - 1 = 0$

Solve $5 x - 1 = 0$ for $x$ .

Add $1$ to both sides of the equation.

$5 x = 1$

Divide each term in $5 x = 1$ by $5$ and simplify.

Divide each term in $5 x = 1$ by $5$ .

$\frac{5 x}{5} = \frac{1}{5}$

Simplify the left side.

Cancel the common factor of $5$ .

Cancel the common factor.

$\frac{5 x}{5} = \frac{1}{5}$

Divide $x$ by $1$ .

$x = \frac{1}{5}$

Set $x + 6$ equal to $0$ and solve for $x$ .

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

$x + 6 = 0$

Subtract $6$ from both sides of the equation.

$x = - 6$

Set $x - 6$ equal to $0$ and solve for $x$ .

Set $x - 6$ equal to $0$ .

$x - 6 = 0$

Add $6$ to both sides of the equation.

$x = 6$

The final solution is all the values that make $(5 x - 1) (x + 6) (x - 6) = 0$ true. The multiplicity of a root is the number of times the root appears.

$x = \frac{1}{5}$ (Multiplicity of $1$ )

$x = - 6$ (Multiplicity of $1$ )

$x = 6$ (Multiplicity of $1$ )

$x = \frac{1}{5}$ (Multiplicity of $1$ )

$x = - 6$ (Multiplicity of $1$ )

$x = 6$ (Multiplicity of $1$ )

Step 3

Identify the Zeros and Their Multiplicities f(x)=5x^3-x^2-180x+36

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