Algebra Examples

$f (x) = x^{5} - 16 x^{4} + 64 x^{3}$

Step 1

Set $x^{5} - 16 x^{4} + 64 x^{3}$ equal to $0$ .

$x^{5} - 16 x^{4} + 64 x^{3} = 0$

Step 2

Solve for $x$ .

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Step 2.1

Factor the left side of the equation.

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Step 2.1.1

Factor $x^{3}$ out of $x^{5} - 16 x^{4} + 64 x^{3}$ .

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Step 2.1.1.1

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

$x^{3} x^{2} - 16 x^{4} + 64 x^{3} = 0$

Step 2.1.1.2

Factor $x^{3}$ out of $- 16 x^{4}$ .

$x^{3} x^{2} + x^{3} (- 16 x) + 64 x^{3} = 0$

Step 2.1.1.3

Factor $x^{3}$ out of $64 x^{3}$ .

$x^{3} x^{2} + x^{3} (- 16 x) + x^{3} \cdot 64 = 0$

Step 2.1.1.4

Factor $x^{3}$ out of $x^{3} x^{2} + x^{3} (- 16 x)$ .

$x^{3} (x^{2} - 16 x) + x^{3} \cdot 64 = 0$

Step 2.1.1.5

Factor $x^{3}$ out of $x^{3} (x^{2} - 16 x) + x^{3} \cdot 64$ .

$x^{3} (x^{2} - 16 x + 64) = 0$

Step 2.1.2

Factor using the perfect square rule.

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Step 2.1.2.1

Rewrite $64$ as $8^{2}$ .

$x^{3} (x^{2} - 16 x + 8^{2}) = 0$

Step 2.1.2.2

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

$16 x = 2 \cdot x \cdot 8$

Step 2.1.2.3

Rewrite the polynomial.

$x^{3} (x^{2} - 2 \cdot x \cdot 8 + 8^{2}) = 0$

Step 2.1.2.4

Factor using the perfect square trinomial rule $a^{2} - 2 a b + b^{2} = {(a - b)}^{2}$ , where $a = x$ and $b = 8$ .

$x^{3} {(x - 8)}^{2} = 0$

Step 2.2

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

$x^{3} = 0$

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

Step 2.3

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

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Step 2.3.1

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

$x^{3} = 0$

Step 2.3.2

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

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Step 2.3.2.1

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

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

Step 2.3.2.2

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

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Step 2.3.2.2.1

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

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

Step 2.3.2.2.2

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

$x = 0$

Step 2.4

Set ${(x - 8)}^{2}$ equal to $0$ and solve for $x$ .

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Step 2.4.1

Set ${(x - 8)}^{2}$ equal to $0$ .

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

Step 2.4.2

Solve ${(x - 8)}^{2} = 0$ for $x$ .

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Step 2.4.2.1

Set the $x - 8$ equal to $0$ .

$x - 8 = 0$

Step 2.4.2.2

Add $8$ to both sides of the equation.

$x = 8$

Step 2.5

The final solution is all the values that make $x^{3} {(x - 8)}^{2} = 0$ true. The multiplicity of a root is the number of times the root appears.

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

$x = 8$ (Multiplicity of $2$ )

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

$x = 8$ (Multiplicity of $2$ )

Step 3