Find the Roots (Zeros) x^4-3x^3-x^2-27x-90=0

$x^{4} - 3 x^{3} - x^{2} - 27 x - 90 = 0$

Step 1

Factor the left side of the equation.

Regroup terms.

$- 3 x^{3} - 27 x + x^{4} - x^{2} - 90 = 0$

Factor $- 3 x$ out of $- 3 x^{3} - 27 x$ .

Factor $- 3 x$ out of $- 3 x^{3}$ .

$- 3 x \cdot x^{2} - 27 x + x^{4} - x^{2} - 90 = 0$

Factor $- 3 x$ out of $- 27 x$ .

$- 3 x \cdot x^{2} - 3 x \cdot 9 + x^{4} - x^{2} - 90 = 0$

Factor $- 3 x$ out of $- 3 x (x^{2}) - 3 x (9)$ .

$- 3 x (x^{2} + 9) + x^{4} - x^{2} - 90 = 0$

Rewrite $x^{4}$ as ${(x^{2})}^{2}$ .

$- 3 x (x^{2} + 9) + {(x^{2})}^{2} - x^{2} - 90 = 0$

Let $u = x^{2}$ . Substitute $u$ for all occurrences of $x^{2}$ .

$- 3 x (x^{2} + 9) + u^{2} - u - 90 = 0$

Factor $u^{2} - u - 90$ 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 $- 90$ and whose sum is $- 1$ .

$- 10, 9$

Write the factored form using these integers.

$- 3 x (x^{2} + 9) + (u - 10) (u + 9) = 0$

Replace all occurrences of $u$ with $x^{2}$ .

$- 3 x (x^{2} + 9) + (x^{2} - 10) (x^{2} + 9) = 0$

Factor $x^{2} + 9$ out of $- 3 x (x^{2} + 9) + (x^{2} - 10) (x^{2} + 9)$ .

Factor $x^{2} + 9$ out of $- 3 x (x^{2} + 9)$ .

$(x^{2} + 9) (- 3 x) + (x^{2} - 10) (x^{2} + 9) = 0$

Factor $x^{2} + 9$ out of $(x^{2} - 10) (x^{2} + 9)$ .

$(x^{2} + 9) (- 3 x) + (x^{2} + 9) (x^{2} - 10) = 0$

Factor $x^{2} + 9$ out of $(x^{2} + 9) (- 3 x) + (x^{2} + 9) (x^{2} - 10)$ .

$(x^{2} + 9) (- 3 x + x^{2} - 10) = 0$

Let $u = x$ . Substitute $u$ for all occurrences of $x$ .

$(x^{2} + 9) (- 3 u + u^{2} - 10) = 0$

Factor $- 3 u + u^{2} - 10$ 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 $- 10$ and whose sum is $- 3$ .

$- 5, 2$

Write the factored form using these integers.

$(x^{2} + 9) ((u - 5) (u + 2)) = 0$

Factor.

Replace all occurrences of $u$ with $x$ .

$(x^{2} + 9) ((x - 5) (x + 2)) = 0$

Remove unnecessary parentheses.

$(x^{2} + 9) (x - 5) (x + 2) = 0$

Step 2

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

$x^{2} + 9 = 0$

$x - 5 = 0$

$x + 2 = 0$

Step 3

Set $x^{2} + 9$ equal to $0$ and solve for $x$ .

Set $x^{2} + 9$ equal to $0$ .

$x^{2} + 9 = 0$

Solve $x^{2} + 9 = 0$ for $x$ .

Subtract $9$ from both sides of the equation.

$x^{2} = - 9$

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 $- 1 (9)$ .

$x = \pm \sqrt{- 1 (9)}$

Rewrite $\sqrt{- 1 (9)}$ as $\sqrt{- 1} \cdot \sqrt{9}$ .

$x = \pm \sqrt{- 1} \cdot \sqrt{9}$

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

$x = \pm i \cdot \sqrt{9}$

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

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

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

$x = \pm i \cdot 3$

Move $3$ to the left of $i$ .

$x = \pm 3 i$

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.

$x = 3 i$

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

$x = - 3 i$

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

$x = 3 i, - 3 i$

Step 4

Set $x - 5$ equal to $0$ and solve for $x$ .

Set $x - 5$ equal to $0$ .

$x - 5 = 0$

Add $5$ to both sides of the equation.

$x = 5$

Step 5

Set $x + 2$ equal to $0$ and solve for $x$ .

Set $x + 2$ equal to $0$ .

$x + 2 = 0$

Subtract $2$ from both sides of the equation.

$x = - 2$

Step 6

The final solution is all the values that make $(x^{2} + 9) (x - 5) (x + 2) = 0$ true.

$x = 3 i, - 3 i, 5, - 2$

Step 7

Find the Roots (Zeros) x^4-3x^3-x^2-27x-90=0

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