Find the Roots (Zeros) (x^4-13x^2+36)÷(x+2)

$(x^{4} - 13 x^{2} + 36) \div (x + 2)$

Set $(x^{4} - 13 x^{2} + 36) \div (x + 2)$ equal to $0$ .

$(x^{4} - 13 x^{2} + 36) \div (x + 2) = 0$

Solve for $x$ .

Set the numerator equal to zero.

$x^{4} - 13 x^{2} + 36 = 0$

Solve the equation for $x$ .

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

$u^{2} - 13 u + 36 = 0$

$u = x^{2}$

Factor $u^{2} - 13 u + 36$ 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 $36$ and whose sum is $- 13$ .

$- 9, - 4$

Write the factored form using these integers.

$(u - 9) (u - 4) = 0$

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

$u - 9 = 0$

$u - 4 = 0$

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 $u - 4$ equal to $0$ and solve for $u$ .

Set $u - 4$ equal to $0$ .

$u - 4 = 0$

Add $4$ to both sides of the equation.

$u = 4$

The final solution is all the values that make $(u - 9) (u - 4) = 0$ true.

$u = 9, 4$

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

$x^{2} = 9$

${(x^{2})}^{1} = 4$

Solve the first equation for $x$ .

$x^{2} = 9$

Solve the equation for $x$ .

Take the square 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.

$x = 3, - 3$

Solve the second equation for $x$ .

${(x^{2})}^{1} = 4$

Solve the equation for $x$ .

Remove parentheses.

$x^{2} = 4$

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

$x = \pm \sqrt{4}$

Simplify $\pm \sqrt{4}$ .

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

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

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

$x = \pm 2$

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.

$x = 2$

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

$x = - 2$

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

$x = 2, - 2$

The solution to $x^{4} - 13 x^{2} + 36 = 0$ is $x = 3, - 3, 2, - 2$ .

$x = 3, - 3, 2, - 2$

Exclude the solutions that do not make $(x^{4} - 13 x^{2} + 36) \div (x + 2) = 0$ true.

$x = 3, - 3, 2$

Find the Roots (Zeros) (x^4-13x^2+36)÷(x+2)

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