Identify the Zeros and Their Multiplicities f(x)=x^6+8x^5+16x^4

$f (x) = x^{6} + 8 x^{5} + 16 x^{4}$

Step 1

Set $x^{6} + 8 x^{5} + 16 x^{4}$ equal to $0$ .

$x^{6} + 8 x^{5} + 16 x^{4} = 0$

Step 2

Solve for $x$ .

Factor the left side of the equation.

Factor $x^{4}$ out of $x^{6} + 8 x^{5} + 16 x^{4}$ .

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

$x^{4} x^{2} + 8 x^{5} + 16 x^{4} = 0$

Factor $x^{4}$ out of $8 x^{5}$ .

$x^{4} x^{2} + x^{4} (8 x) + 16 x^{4} = 0$

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

$x^{4} x^{2} + x^{4} (8 x) + x^{4} \cdot 16 = 0$

Factor $x^{4}$ out of $x^{4} x^{2} + x^{4} (8 x)$ .

$x^{4} (x^{2} + 8 x) + x^{4} \cdot 16 = 0$

Factor $x^{4}$ out of $x^{4} (x^{2} + 8 x) + x^{4} \cdot 16$ .

$x^{4} (x^{2} + 8 x + 16) = 0$

Factor using the perfect square rule.

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

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

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

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

Rewrite the polynomial.

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

Factor using the perfect square trinomial rule $a^{2} + 2 a b + b^{2} = {(a + b)}^{2}$ , where $a = x$ and $b = 4$ .

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

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

$x^{4} = 0$

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

Set $x^{4}$ equal to $0$ and solve for $x$ .

Set $x^{4}$ equal to $0$ .

$x^{4} = 0$

Solve $x^{4} = 0$ for $x$ .

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

$x = \pm \sqrt[4]{0}$

Simplify $\pm \sqrt[4]{0}$ .

Rewrite $0$ as $0^{4}$ .

$x = \pm \sqrt[4]{0^{4}}$

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

$x = \pm 0$

Plus or minus $0$ is $0$ .

$x = 0$

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

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

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

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

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

$x + 4 = 0$

Subtract $4$ from both sides of the equation.

$x = - 4$

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

$x = 0$ (Multiplicity of $4$ )

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

$x = 0$ (Multiplicity of $4$ )

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

Step 3

Identify the Zeros and Their Multiplicities f(x)=x^6+8x^5+16x^4

Related Questions

Support LCMGCF.com by disabling your adblocker.