Identify the Zeros and Their Multiplicities f(x)=-x^5+9x^4-18x^3

$f (x) = - x^{5} + 9 x^{4} - 18 x^{3}$

To find the roots/zeros, set $- x^{5} + 9 x^{4} - 18 x^{3}$ equal to $0$ and solve.

$- x^{5} + 9 x^{4} - 18 x^{3} = 0$

Factor the left side of the equation.

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Factor $- x^{3}$ out of $- x^{5} + 9 x^{4} - 18 x^{3}$ .

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

$- x^{3} x^{2} + 9 x^{4} - 18 x^{3} = 0$

Factor $- x^{3}$ out of $9 x^{4}$ .

$- x^{3} x^{2} - x^{3} (- 9 x) - 18 x^{3} = 0$

Factor $- x^{3}$ out of $- 18 x^{3}$ .

$- x^{3} x^{2} - x^{3} (- 9 x) - x^{3} \cdot 18 = 0$

Factor $- x^{3}$ out of $- x^{3} (x^{2}) - x^{3} (- 9 x)$ .

$- x^{3} (x^{2} - 9 x) - x^{3} \cdot 18 = 0$

Factor $- x^{3}$ out of $- x^{3} (x^{2} - 9 x) - x^{3} (18)$ .

$- x^{3} (x^{2} - 9 x + 18) = 0$

Factor.

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Factor $x^{2} - 9 x + 18$ using the AC method.

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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 $18$ and whose sum is $- 9$ .

$- 6, - 3$

Write the factored form using these integers.

$- x^{3} ((x - 6) (x - 3)) = 0$

Remove unnecessary parentheses.

$- x^{3} (x - 6) (x - 3) = 0$

Multiply each term in $- x^{3} (x - 6) (x - 3) = 0$ by $- 1$

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Multiply each term in $- x^{3} (x - 6) (x - 3) = 0$ by $- 1$ .

$(- x^{3} (x - 6) (x - 3)) \cdot - 1 = 0 \cdot - 1$

Simplify $(- x^{3} (x - 6) (x - 3)) \cdot - 1$ .

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Apply the distributive property.

$(- x^{3} x - x^{3} \cdot - 6) (x - 3) \cdot - 1 = 0 \cdot - 1$

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

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Move $x$ .

$(- (x \cdot x^{3}) - x^{3} \cdot - 6) (x - 3) \cdot - 1 = 0 \cdot - 1$

Multiply $x$ by $x^{3}$ .

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Raise $x$ to the power of $1$ .

$(- (x^{1} x^{3}) - x^{3} \cdot - 6) (x - 3) \cdot - 1 = 0 \cdot - 1$

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

$(- x^{1 + 3} - x^{3} \cdot - 6) (x - 3) \cdot - 1 = 0 \cdot - 1$

Add $1$ and $3$ .

$(- x^{4} - x^{3} \cdot - 6) (x - 3) \cdot - 1 = 0 \cdot - 1$

Multiply $- 6$ by $- 1$ .

$(- x^{4} + 6 x^{3}) (x - 3) \cdot - 1 = 0 \cdot - 1$

Expand $(- x^{4} + 6 x^{3}) (x - 3)$ using the FOIL Method.

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Apply the distributive property.

$(- x^{4} (x - 3) + 6 x^{3} (x - 3)) \cdot - 1 = 0 \cdot - 1$

Apply the distributive property.

$(- x^{4} x - x^{4} \cdot - 3 + 6 x^{3} (x - 3)) \cdot - 1 = 0 \cdot - 1$

Apply the distributive property.

$(- x^{4} x - x^{4} \cdot - 3 + 6 x^{3} x + 6 x^{3} \cdot - 3) \cdot - 1 = 0 \cdot - 1$

Simplify and combine like terms.

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Simplify each term.

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Multiply $x^{4}$ by $x$ by adding the exponents.

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Move $x$ .

$(- (x \cdot x^{4}) - x^{4} \cdot - 3 + 6 x^{3} x + 6 x^{3} \cdot - 3) \cdot - 1 = 0 \cdot - 1$

Multiply $x$ by $x^{4}$ .

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Raise $x$ to the power of $1$ .

$(- (x^{1} x^{4}) - x^{4} \cdot - 3 + 6 x^{3} x + 6 x^{3} \cdot - 3) \cdot - 1 = 0 \cdot - 1$

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

$(- x^{1 + 4} - x^{4} \cdot - 3 + 6 x^{3} x + 6 x^{3} \cdot - 3) \cdot - 1 = 0 \cdot - 1$

Add $1$ and $4$ .

$(- x^{5} - x^{4} \cdot - 3 + 6 x^{3} x + 6 x^{3} \cdot - 3) \cdot - 1 = 0 \cdot - 1$

Multiply $- 3$ by $- 1$ .

$(- x^{5} + 3 x^{4} + 6 x^{3} x + 6 x^{3} \cdot - 3) \cdot - 1 = 0 \cdot - 1$

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

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Move $x$ .

$(- x^{5} + 3 x^{4} + 6 (x \cdot x^{3}) + 6 x^{3} \cdot - 3) \cdot - 1 = 0 \cdot - 1$

Multiply $x$ by $x^{3}$ .

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Raise $x$ to the power of $1$ .

$(- x^{5} + 3 x^{4} + 6 (x^{1} x^{3}) + 6 x^{3} \cdot - 3) \cdot - 1 = 0 \cdot - 1$

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

$(- x^{5} + 3 x^{4} + 6 x^{1 + 3} + 6 x^{3} \cdot - 3) \cdot - 1 = 0 \cdot - 1$

Add $1$ and $3$ .

$(- x^{5} + 3 x^{4} + 6 x^{4} + 6 x^{3} \cdot - 3) \cdot - 1 = 0 \cdot - 1$

Multiply $- 3$ by $6$ .

$(- x^{5} + 3 x^{4} + 6 x^{4} - 18 x^{3}) \cdot - 1 = 0 \cdot - 1$

Add $3 x^{4}$ and $6 x^{4}$ .

$(- x^{5} + 9 x^{4} - 18 x^{3}) \cdot - 1 = 0 \cdot - 1$

Apply the distributive property.

$- x^{5} \cdot - 1 + 9 x^{4} \cdot - 1 - 18 x^{3} \cdot - 1 = 0 \cdot - 1$

Simplify.

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Multiply $- x^{5} \cdot - 1$ .

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Multiply $- 1$ by $- 1$ .

$1 x^{5} + 9 x^{4} \cdot - 1 - 18 x^{3} \cdot - 1 = 0 \cdot - 1$

Multiply $x^{5}$ by $1$ .

$x^{5} + 9 x^{4} \cdot - 1 - 18 x^{3} \cdot - 1 = 0 \cdot - 1$

Multiply $- 1$ by $9$ .

$x^{5} - 9 x^{4} - 18 x^{3} \cdot - 1 = 0 \cdot - 1$

Multiply $- 1$ by $- 18$ .

$x^{5} - 9 x^{4} + 18 x^{3} = 0 \cdot - 1$

Multiply $0$ by $- 1$ .

$x^{5} - 9 x^{4} + 18 x^{3} = 0$

Factor the left side of the equation.

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Factor $x^{3}$ out of $x^{5} - 9 x^{4} + 18 x^{3}$ .

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

$x^{3} x^{2} - 9 x^{4} + 18 x^{3} = 0$

Factor $x^{3}$ out of $- 9 x^{4}$ .

$x^{3} x^{2} + x^{3} (- 9 x) + 18 x^{3} = 0$

Factor $x^{3}$ out of $18 x^{3}$ .

$x^{3} x^{2} + x^{3} (- 9 x) + x^{3} \cdot 18 = 0$

Factor $x^{3}$ out of $x^{3} x^{2} + x^{3} (- 9 x)$ .

$x^{3} (x^{2} - 9 x) + x^{3} \cdot 18 = 0$

Factor $x^{3}$ out of $x^{3} (x^{2} - 9 x) + x^{3} \cdot 18$ .

$x^{3} (x^{2} - 9 x + 18) = 0$

Factor.

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Factor $x^{2} - 9 x + 18$ using the AC method.

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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 $18$ and whose sum is $- 9$ .

$- 6, - 3$

Write the factored form using these integers.

$x^{3} ((x - 6) (x - 3)) = 0$

Remove unnecessary parentheses.

$x^{3} (x - 6) (x - 3) = 0$

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

$x^{3} = 0$

$x - 6 = 0$

$x - 3 = 0$

Set the first factor equal to $0$ and solve.

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Set the first factor equal to $0$ .

$x^{3} = 0$

Take the cube root of both sides of the equation to eliminate the exponent on the left side.

$x = \sqrt[3]{0}$

Simplify $\sqrt[3]{0}$ .

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Rewrite $0$ as $0^{3}$ .

$x = \sqrt[3]{0^{3}}$

Pull terms out from under the radical, assuming real numbers.

$x = 0$

Set the next factor equal to $0$ and solve.

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Set the next factor equal to $0$ .

$x - 6 = 0$

Add $6$ to both sides of the equation.

$x = 6$

Set the next factor equal to $0$ and solve.

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Set the next factor 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 $x^{3} (x - 6) (x - 3) = 0$ true. The multiplicity of a root is the number of times the root appears.

$x = 0$ (Multiplicity of $3$ )

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

$x = 3$ (Multiplicity of $1$ )

Identify the Zeros and Their Multiplicities f(x)=-x^5+9x^4-18x^3

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