Identify the Zeros and Their Multiplicities x^4+27x^2-324

$x^{4} + 27 x^{2} - 324$

Step 1

Set $x^{4} + 27 x^{2} - 324$ equal to $0$ .

$x^{4} + 27 x^{2} - 324 = 0$

Step 2

Solve for $x$ .

Substitute $u = x^{2}$ into the equation. This will make the quadratic formula easy to use.

$u^{2} + 27 u - 324 = 0$

$u = x^{2}$

Factor $u^{2} + 27 u - 324$ using the AC method.

Consider the form $x^{2} + b x + c$ . Find a pair of integers whose product is $c$ and whose sum is $b$ . In this case, whose product is $- 324$ and whose sum is $27$ .

$- 9, 36$

Write the factored form using these integers.

$(u - 9) (u + 36) = 0$

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

$u - 9 = 0$

$u + 36 = 0$

Set $u - 9$ equal to $0$ and solve for $u$ .

Set $u - 9$ equal to $0$ .

$u - 9 = 0$

Add $9$ to both sides of the equation.

$u = 9$

Set $u + 36$ equal to $0$ and solve for $u$ .

Set $u + 36$ equal to $0$ .

$u + 36 = 0$

Subtract $36$ from both sides of the equation.

$u = - 36$

The final solution is all the values that make $(u - 9) (u + 36) = 0$ true. The multiplicity of a root is the number of times the root appears.

$u = 9$ (Multiplicity of $1$ )

$u = - 36$ (Multiplicity of $1$ )

Substitute the real value of $u = x^{2}$ back into the solved equation.

$x^{2} = 9$

${(x^{2})}^{1} = - 36$

Solve the first equation for $x$ .

$x^{2} = 9$

Solve the equation for $x$ .

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

$x = \pm \sqrt{9}$

Simplify $\pm \sqrt{9}$ .

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

$x = \pm \sqrt{3^{2}}$

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

$x = \pm 3$

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

First, use the positive value of the $\pm$ to find the first solution.

$x = 3$

Next, use the negative value of the $\pm$ to find the second solution.

$x = - 3$

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

$x = 3, - 3$

The multiplicity of a root is the number of times the root appears. For example, a factor of ${(x + 5)}^{3}$ would have a root at $x = - 5$ with multiplicity of $3$ .

$x = 3$ (Multiplicity of $1$ )

$x = - 3$ (Multiplicity of $1$ )

$x = 3$ (Multiplicity of $1$ )

$x = - 3$ (Multiplicity of $1$ )

Solve the second equation for $x$ .

${(x^{2})}^{1} = - 36$

Solve the equation for $x$ .

Remove parentheses.

$x^{2} = - 36$

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

$x = \pm \sqrt{- 36}$

Simplify $\pm \sqrt{- 36}$ .

Rewrite $- 36$ as $- 1 (36)$ .

$x = \pm \sqrt{- 1 (36)}$

Rewrite $\sqrt{- 1 (36)}$ as $\sqrt{- 1} \cdot \sqrt{36}$ .

$x = \pm \sqrt{- 1} \cdot \sqrt{36}$

Rewrite $\sqrt{- 1}$ as $i$ .

$x = \pm i \cdot \sqrt{36}$

Rewrite $36$ as $6^{2}$ .

$x = \pm i \cdot \sqrt{6^{2}}$

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

$x = \pm i \cdot 6$

Move $6$ to the left of $i$ .

$x = \pm 6 i$

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

First, use the positive value of the $\pm$ to find the first solution.

$x = 6 i$

Next, use the negative value of the $\pm$ to find the second solution.

$x = - 6 i$

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

$x = 6 i, - 6 i$

The multiplicity of a root is the number of times the root appears. For example, a factor of ${(x + 5)}^{3}$ would have a root at $x = - 5$ with multiplicity of $3$ .

$x = 6 i$ (Multiplicity of $1$ )

$x = - 6 i$ (Multiplicity of $1$ )

$x = 6 i$ (Multiplicity of $1$ )

$x = - 6 i$ (Multiplicity of $1$ )

The solution to $x^{4} + 27 x^{2} - 324 = 0$ is $x = 3, - 3, 6 i, - 6 i$ .

$x = 3, - 3, 6 i, - 6 i$

Step 3

Identify the Zeros and Their Multiplicities x^4+27x^2-324

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