Identify the Zeros and Their Multiplicities f(x)=-2x^4-17x^3-36x^2

$f (x) = - 2 x^{4} - 17 x^{3} - 36 x^{2}$

To find the roots/zeros, set $- 2 x^{4} - 17 x^{3} - 36 x^{2}$ equal to $0$ and solve.

$- 2 x^{4} - 17 x^{3} - 36 x^{2} = 0$

Factor the left side of the equation.

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Factor $- x^{2}$ out of $- 2 x^{4} - 17 x^{3} - 36 x^{2}$ .

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Factor $- x^{2}$ out of $- 2 x^{4}$ .

$- x^{2} (2 x^{2}) - 17 x^{3} - 36 x^{2} = 0$

Factor $- x^{2}$ out of $- 17 x^{3}$ .

$- x^{2} (2 x^{2}) - x^{2} (17 x) - 36 x^{2} = 0$

Factor $- x^{2}$ out of $- 36 x^{2}$ .

$- x^{2} (2 x^{2}) - x^{2} (17 x) - x^{2} \cdot 36 = 0$

Factor $- x^{2}$ out of $- x^{2} (2 x^{2}) - x^{2} (17 x)$ .

$- x^{2} (2 x^{2} + 17 x) - x^{2} \cdot 36 = 0$

Factor $- x^{2}$ out of $- x^{2} (2 x^{2} + 17 x) - x^{2} (36)$ .

$- x^{2} (2 x^{2} + 17 x + 36) = 0$

Factor.

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Factor by grouping.

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For a polynomial of the form $a x^{2} + b x + c$ , rewrite the middle term as a sum of two terms whose product is $a \cdot c = 2 \cdot 36 = 72$ and whose sum is $b = 17$ .

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Factor $17$ out of $17 x$ .

$- x^{2} (2 x^{2} + 17 (x) + 36) = 0$

Rewrite $17$ as $8$ plus $9$

$- x^{2} (2 x^{2} + (8 + 9) x + 36) = 0$

Apply the distributive property.

$- x^{2} (2 x^{2} + 8 x + 9 x + 36) = 0$

Factor out the greatest common factor from each group.

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Group the first two terms and the last two terms.

$- x^{2} ((2 x^{2} + 8 x) + 9 x + 36) = 0$

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

$- x^{2} (2 x (x + 4) + 9 (x + 4)) = 0$

Factor the polynomial by factoring out the greatest common factor, $x + 4$ .

$- x^{2} ((x + 4) (2 x + 9)) = 0$

Remove unnecessary parentheses.

$- x^{2} (x + 4) (2 x + 9) = 0$

Multiply each term in $- x^{2} (x + 4) (2 x + 9) = 0$ by $- 1$

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Multiply each term in $- x^{2} (x + 4) (2 x + 9) = 0$ by $- 1$ .

$(- x^{2} (x + 4) (2 x + 9)) \cdot - 1 = 0 \cdot - 1$

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

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Apply the distributive property.

$(- x^{2} x - x^{2} \cdot 4) (2 x + 9) \cdot - 1 = 0 \cdot - 1$

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

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Move $x$ .

$(- (x \cdot x^{2}) - x^{2} \cdot 4) (2 x + 9) \cdot - 1 = 0 \cdot - 1$

Multiply $x$ by $x^{2}$ .

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

$(- (x^{1} x^{2}) - x^{2} \cdot 4) (2 x + 9) \cdot - 1 = 0 \cdot - 1$

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

$(- x^{1 + 2} - x^{2} \cdot 4) (2 x + 9) \cdot - 1 = 0 \cdot - 1$

Add $1$ and $2$ .

$(- x^{3} - x^{2} \cdot 4) (2 x + 9) \cdot - 1 = 0 \cdot - 1$

Multiply $4$ by $- 1$ .

$(- x^{3} - 4 x^{2}) (2 x + 9) \cdot - 1 = 0 \cdot - 1$

Expand $(- x^{3} - 4 x^{2}) (2 x + 9)$ using the FOIL Method.

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Apply the distributive property.

$(- x^{3} (2 x + 9) - 4 x^{2} (2 x + 9)) \cdot - 1 = 0 \cdot - 1$

Apply the distributive property.

$(- x^{3} (2 x) - x^{3} \cdot 9 - 4 x^{2} (2 x + 9)) \cdot - 1 = 0 \cdot - 1$

Apply the distributive property.

$(- x^{3} (2 x) - x^{3} \cdot 9 - 4 x^{2} (2 x) - 4 x^{2} \cdot 9) \cdot - 1 = 0 \cdot - 1$

Simplify and combine like terms.

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Simplify each term.

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Rewrite using the commutative property of multiplication.

$(- 1 \cdot 2 (x^{3} x) - x^{3} \cdot 9 - 4 x^{2} (2 x) - 4 x^{2} \cdot 9) \cdot - 1 = 0 \cdot - 1$

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

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

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

$(- 1 \cdot 2 (x^{3} x^{1}) - x^{3} \cdot 9 - 4 x^{2} (2 x) - 4 x^{2} \cdot 9) \cdot - 1 = 0 \cdot - 1$

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

$(- 1 \cdot 2 x^{3 + 1} - x^{3} \cdot 9 - 4 x^{2} (2 x) - 4 x^{2} \cdot 9) \cdot - 1 = 0 \cdot - 1$

Add $3$ and $1$ .

$(- 1 \cdot 2 x^{4} - x^{3} \cdot 9 - 4 x^{2} (2 x) - 4 x^{2} \cdot 9) \cdot - 1 = 0 \cdot - 1$

Multiply $- 1$ by $2$ .

$(- 2 x^{4} - x^{3} \cdot 9 - 4 x^{2} (2 x) - 4 x^{2} \cdot 9) \cdot - 1 = 0 \cdot - 1$

Multiply $9$ by $- 1$ .

$(- 2 x^{4} - 9 x^{3} - 4 x^{2} (2 x) - 4 x^{2} \cdot 9) \cdot - 1 = 0 \cdot - 1$

Rewrite using the commutative property of multiplication.

$(- 2 x^{4} - 9 x^{3} - 4 \cdot 2 (x^{2} x) - 4 x^{2} \cdot 9) \cdot - 1 = 0 \cdot - 1$

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

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Multiply $x^{2}$ by $x$ .

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

$(- 2 x^{4} - 9 x^{3} - 4 \cdot 2 (x^{2} x^{1}) - 4 x^{2} \cdot 9) \cdot - 1 = 0 \cdot - 1$

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

$(- 2 x^{4} - 9 x^{3} - 4 \cdot 2 x^{2 + 1} - 4 x^{2} \cdot 9) \cdot - 1 = 0 \cdot - 1$

Add $2$ and $1$ .

$(- 2 x^{4} - 9 x^{3} - 4 \cdot 2 x^{3} - 4 x^{2} \cdot 9) \cdot - 1 = 0 \cdot - 1$

Multiply $- 4$ by $2$ .

$(- 2 x^{4} - 9 x^{3} - 8 x^{3} - 4 x^{2} \cdot 9) \cdot - 1 = 0 \cdot - 1$

Multiply $9$ by $- 4$ .

$(- 2 x^{4} - 9 x^{3} - 8 x^{3} - 36 x^{2}) \cdot - 1 = 0 \cdot - 1$

Subtract $8 x^{3}$ from $- 9 x^{3}$ .

$(- 2 x^{4} - 17 x^{3} - 36 x^{2}) \cdot - 1 = 0 \cdot - 1$

Apply the distributive property.

$- 2 x^{4} \cdot - 1 - 17 x^{3} \cdot - 1 - 36 x^{2} \cdot - 1 = 0 \cdot - 1$

Simplify.

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Multiply $- 1$ by $- 2$ .

$2 x^{4} - 17 x^{3} \cdot - 1 - 36 x^{2} \cdot - 1 = 0 \cdot - 1$

Multiply $- 1$ by $- 17$ .

$2 x^{4} + 17 x^{3} - 36 x^{2} \cdot - 1 = 0 \cdot - 1$

Multiply $- 1$ by $- 36$ .

$2 x^{4} + 17 x^{3} + 36 x^{2} = 0 \cdot - 1$

Multiply $0$ by $- 1$ .

$2 x^{4} + 17 x^{3} + 36 x^{2} = 0$

Factor the left side of the equation.

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Factor $x^{2}$ out of $2 x^{4} + 17 x^{3} + 36 x^{2}$ .

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Factor $x^{2}$ out of $2 x^{4}$ .

$x^{2} (2 x^{2}) + 17 x^{3} + 36 x^{2} = 0$

Factor $x^{2}$ out of $17 x^{3}$ .

$x^{2} (2 x^{2}) + x^{2} (17 x) + 36 x^{2} = 0$

Factor $x^{2}$ out of $36 x^{2}$ .

$x^{2} (2 x^{2}) + x^{2} (17 x) + x^{2} \cdot 36 = 0$

Factor $x^{2}$ out of $x^{2} (2 x^{2}) + x^{2} (17 x)$ .

$x^{2} (2 x^{2} + 17 x) + x^{2} \cdot 36 = 0$

Factor $x^{2}$ out of $x^{2} (2 x^{2} + 17 x) + x^{2} \cdot 36$ .

$x^{2} (2 x^{2} + 17 x + 36) = 0$

Factor.

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Factor by grouping.

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For a polynomial of the form $a x^{2} + b x + c$ , rewrite the middle term as a sum of two terms whose product is $a \cdot c = 2 \cdot 36 = 72$ and whose sum is $b = 17$ .

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Factor $17$ out of $17 x$ .

$x^{2} (2 x^{2} + 17 (x) + 36) = 0$

Rewrite $17$ as $8$ plus $9$

$x^{2} (2 x^{2} + (8 + 9) x + 36) = 0$

Apply the distributive property.

$x^{2} (2 x^{2} + 8 x + 9 x + 36) = 0$

Factor out the greatest common factor from each group.

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Group the first two terms and the last two terms.

$x^{2} ((2 x^{2} + 8 x) + 9 x + 36) = 0$

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

$x^{2} (2 x (x + 4) + 9 (x + 4)) = 0$

Factor the polynomial by factoring out the greatest common factor, $x + 4$ .

$x^{2} ((x + 4) (2 x + 9)) = 0$

Remove unnecessary parentheses.

$x^{2} (x + 4) (2 x + 9) = 0$

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

$x^{2} = 0$

$x + 4 = 0$

$2 x + 9 = 0$

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

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Set the first factor equal to $0$ .

$x^{2} = 0$

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

$x = \pm \sqrt{0}$

The complete solution is the result of both the positive and negative portions of the solution.

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Simplify the right side of the equation.

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Rewrite $0$ as $0^{2}$ .

$x = \pm \sqrt{0^{2}}$

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

$x = \pm 0$

$\pm 0$ is equal to $0$ .

$x = 0$

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

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Set the next factor equal to $0$ .

$x + 4 = 0$

Subtract $4$ from both sides of the equation.

$x = - 4$

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

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Set the next factor equal to $0$ .

$2 x + 9 = 0$

Subtract $9$ from both sides of the equation.

$2 x = - 9$

Divide each term by $2$ and simplify.

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Divide each term in $2 x = - 9$ by $2$ .

$\frac{2 x}{2} = \frac{- 9}{2}$

Cancel the common factor of $2$ .

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Cancel the common factor.

$\frac{2 x}{2} = \frac{- 9}{2}$

Divide $x$ by $1$ .

$x = \frac{- 9}{2}$

Move the negative in front of the fraction.

$x = - \frac{9}{2}$

The final solution is all the values that make $x^{2} (x + 4) (2 x + 9) = 0$ true. The multiplicity of a root is the number of times the root appears.

$x = 0$ (Multiplicity of $2$ )

$x = - 4$ (Multiplicity of $1$ )

$x = - \frac{9}{2}$ (Multiplicity of $1$ )

Identify the Zeros and Their Multiplicities f(x)=-2x^4-17x^3-36x^2

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