Identify the Zeros and Their Multiplicities x^4-19x^2+48

$x^{4} - 19 x^{2} + 48$

Step 1

Set $x^{4} - 19 x^{2} + 48$ equal to $0$ .

$x^{4} - 19 x^{2} + 48 = 0$

Step 2

Solve for $x$ .

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

$u^{2} - 19 u + 48 = 0$

$u = x^{2}$

Factor $u^{2} - 19 u + 48$ 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 $48$ and whose sum is $- 19$ .

$- 16, - 3$

Write the factored form using these integers.

$(u - 16) (u - 3) = 0$

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

$u - 16 = 0$

$u - 3 = 0$

Set $u - 16$ equal to $0$ and solve for $u$ .

Set $u - 16$ equal to $0$ .

$u - 16 = 0$

Add $16$ to both sides of the equation.

$u = 16$

Set $u - 3$ equal to $0$ and solve for $u$ .

Set $u - 3$ equal to $0$ .

$u - 3 = 0$

Add $3$ to both sides of the equation.

$u = 3$

The final solution is all the values that make $(u - 16) (u - 3) = 0$ true. The multiplicity of a root is the number of times the root appears.

$u = 16$ (Multiplicity of $1$ )

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

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

$x^{2} = 16$

${(x^{2})}^{1} = 3$

Solve the first equation for $x$ .

$x^{2} = 16$

Solve the equation for $x$ .

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

$x = \pm \sqrt{16}$

Simplify $\pm \sqrt{16}$ .

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

$x = \pm \sqrt{4^{2}}$

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

$x = \pm 4$

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.

$x = 4$

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

$x = - 4$

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

$x = 4, - 4$

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 $3$ .

$x = 4$ (Multiplicity of $1$ )

$x = - 4$ (Multiplicity of $1$ )

$x = 4$ (Multiplicity of $1$ )

$x = - 4$ (Multiplicity of $1$ )

Solve the second equation for $x$ .

${(x^{2})}^{1} = 3$

Solve the equation for $x$ .

Remove parentheses.

$x^{2} = 3$

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

$x = \pm \sqrt{3}$

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.

$x = \sqrt{3}$

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

$x = - \sqrt{3}$

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

$x = \sqrt{3}, - \sqrt{3}$

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 $3$ .

$x = \sqrt{3}$ (Multiplicity of $1$ )

$x = - \sqrt{3}$ (Multiplicity of $1$ )

$x = \sqrt{3}$ (Multiplicity of $1$ )

$x = - \sqrt{3}$ (Multiplicity of $1$ )

The solution to $x^{4} - 19 x^{2} + 48 = 0$ is $x = 4, - 4, \sqrt{3}, - \sqrt{3}$ .

$x = 4, - 4, \sqrt{3}, - \sqrt{3}$

Step 3

The result can be shown in multiple forms.

Exact Form:

$x = 4, - 4, \sqrt{3}, - \sqrt{3}$

Decimal Form:

$x = 4, - 4, 1.73205080 \dots, - 1.73205080 \dots$

Step 4

Identify the Zeros and Their Multiplicities x^4-19x^2+48

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