Identify the Zeros and Their Multiplicities d(x)=x^5+6x^4+9x^3

$d (x) = x^{5} + 6 x^{4} + 9 x^{3}$

Step 1

Set $x^{5} + 6 x^{4} + 9 x^{3}$ equal to $0$ .

$x^{5} + 6 x^{4} + 9 x^{3} = 0$

Step 2

Solve for $x$ .

Factor the left side of the equation.

Factor $x^{3}$ out of $x^{5} + 6 x^{4} + 9 x^{3}$ .

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

$x^{3} x^{2} + 6 x^{4} + 9 x^{3} = 0$

Factor $x^{3}$ out of $6 x^{4}$ .

$x^{3} x^{2} + x^{3} (6 x) + 9 x^{3} = 0$

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

$x^{3} x^{2} + x^{3} (6 x) + x^{3} \cdot 9 = 0$

Factor $x^{3}$ out of $x^{3} x^{2} + x^{3} (6 x)$ .

$x^{3} (x^{2} + 6 x) + x^{3} \cdot 9 = 0$

Factor $x^{3}$ out of $x^{3} (x^{2} + 6 x) + x^{3} \cdot 9$ .

$x^{3} (x^{2} + 6 x + 9) = 0$

Factor using the perfect square rule.

Rewrite $9$ as $3^{2}$ .

$x^{3} (x^{2} + 6 x + 3^{2}) = 0$

Check that the middle term is two times the product of the numbers being squared in the first term and third term.

$6 x = 2 \cdot x \cdot 3$

Rewrite the polynomial.

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

Factor using the perfect square trinomial rule $a^{2} + 2 a b + b^{2} = {(a + b)}^{2}$ , where $a = x$ and $b = 3$ .

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

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

$x^{3} = 0$

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

Set $x^{3}$ equal to $0$ and solve for $x$ .

Set $x^{3}$ equal to $0$ .

$x^{3} = 0$

Solve $x^{3} = 0$ for $x$ .

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

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

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

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

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

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

$x = 0$

Set ${(x + 3)}^{2}$ equal to $0$ and solve for $x$ .

Set ${(x + 3)}^{2}$ equal to $0$ .

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

Solve ${(x + 3)}^{2} = 0$ for $x$ .

Set the $x + 3$ equal to $0$ .

$x + 3 = 0$

Subtract $3$ from both sides of the equation.

$x = - 3$

The final solution is all the values that make $x^{3} {(x + 3)}^{2} = 0$ true. The multiplicity of a root is the number of times the root appears.

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

$x = - 3$ (Multiplicity of $2$ )

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

$x = - 3$ (Multiplicity of $2$ )

Step 3

Identify the Zeros and Their Multiplicities d(x)=x^5+6x^4+9x^3

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