Precalculus Examples

$f (x) = x^{4} - 8 x^{3} + 16 x^{2}$

Step 1

Set $x^{4} - 8 x^{3} + 16 x^{2}$ equal to $0$ .

$x^{4} - 8 x^{3} + 16 x^{2} = 0$

Step 2

Solve for $x$ .

Tap for more steps...

Step 2.1

Factor the left side of the equation.

Tap for more steps...

Step 2.1.1

Factor $x^{2}$ out of $x^{4} - 8 x^{3} + 16 x^{2}$ .

Tap for more steps...

Step 2.1.1.1

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

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

Step 2.1.1.2

Factor $x^{2}$ out of $- 8 x^{3}$ .

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

Step 2.1.1.3

Factor $x^{2}$ out of $16 x^{2}$ .

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

Step 2.1.1.4

Factor $x^{2}$ out of $x^{2} x^{2} + x^{2} (- 8 x)$ .

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

Step 2.1.1.5

Factor $x^{2}$ out of $x^{2} (x^{2} - 8 x) + x^{2} \cdot 16$ .

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

Step 2.1.2

Factor using the perfect square rule.

Tap for more steps...

Step 2.1.2.1

Rewrite $16$ as $4^{2}$ .

$x^{2} (x^{2} - 8 x + 4^{2}) = 0$

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.

$8 x = 2 \cdot x \cdot 4$

Step 2.1.2.3

Rewrite the polynomial.

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

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 $b = 4$ .

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

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 $0$ .

$x^{2} = 0$

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

Step 2.3

Set $x^{2}$ equal to $0$ and solve for $x$ .

Tap for more steps...

Step 2.3.1

Set $x^{2}$ equal to $0$ .

$x^{2} = 0$

Step 2.3.2

Solve $x^{2} = 0$ for $x$ .

Tap for more steps...

Step 2.3.2.1

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

$x = \pm \sqrt{0}$

Step 2.3.2.2

Simplify $\pm \sqrt{0}$ .

Tap for more steps...

Step 2.3.2.2.1

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

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

Step 2.3.2.2.2

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

$x = \pm 0$

Step 2.3.2.2.3

Plus or minus $0$ is $0$ .

$x = 0$

Step 2.4

Set ${(x - 4)}^{2}$ equal to $0$ and solve for $x$ .

Tap for more steps...

Step 2.4.1

Set ${(x - 4)}^{2}$ equal to $0$ .

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

Step 2.4.2

Solve ${(x - 4)}^{2} = 0$ for $x$ .

Tap for more steps...

Step 2.4.2.1

Set the $x - 4$ equal to $0$ .

$x - 4 = 0$

Step 2.4.2.2

Add $4$ to both sides of the equation.

$x = 4$

Step 2.5

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

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

$x = 4$ (Multiplicity of $2$ )

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

$x = 4$ (Multiplicity of $2$ )

Step 3