Find the Roots (Zeros) P(x)=x^4+40x^2-441

$P (x) = x^{4} + 40 x^{2} - 441$

Step 1

Set $x^{4} + 40 x^{2} - 441$ equal to $0$ .

$x^{4} + 40 x^{2} - 441 = 0$

Step 2

Solve for $x$ .

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

$u^{2} + 40 u - 441 = 0$

$u = x^{2}$

Factor $u^{2} + 40 u - 441$ 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 $- 441$ and whose sum is $40$ .

$- 9, 49$

Write the factored form using these integers.

$(u - 9) (u + 49) = 0$

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

$u - 9 = 0$

$u + 49 = 0$

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 $u + 49$ equal to $0$ and solve for $u$ .

Set $u + 49$ equal to $0$ .

$u + 49 = 0$

Subtract $49$ from both sides of the equation.

$u = - 49$

The final solution is all the values that make $(u - 9) (u + 49) = 0$ true.

$u = 9, - 49$

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

$x^{2} = 9$

${(x^{2})}^{1} = - 49$

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 $x$ .

${(x^{2})}^{1} = - 49$

Solve the equation for $x$ .

Remove parentheses.

$x^{2} = - 49$

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

$x = \pm \sqrt{- 49}$

Simplify $\pm \sqrt{- 49}$ .

Rewrite $- 49$ as $- 1 (49)$ .

$x = \pm \sqrt{- 1 (49)}$

Rewrite $\sqrt{- 1 (49)}$ as $\sqrt{- 1} \cdot \sqrt{49}$ .

$x = \pm \sqrt{- 1} \cdot \sqrt{49}$

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

$x = \pm i \cdot \sqrt{49}$

Rewrite $49$ as $7^{2}$ .

$x = \pm i \cdot \sqrt{7^{2}}$

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

$x = \pm i \cdot 7$

Move $7$ to the left of $i$ .

$x = \pm 7 i$

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.

$x = 7 i$

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

$x = - 7 i$

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

$x = 7 i, - 7 i$

The solution to $x^{4} + 40 x^{2} - 441 = 0$ is $x = 3, - 3, 7 i, - 7 i$ .

$x = 3, - 3, 7 i, - 7 i$

Step 3

Find the Roots (Zeros) P(x)=x^4+40x^2-441

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