Find the Roots (Zeros) x^4+x^3-8x^2-9x-9=0

$x^{4} + x^{3} - 8 x^{2} - 9 x - 9 = 0$

Step 1

Factor the left side of the equation.

Regroup terms.

$x^{3} - 9 x + x^{4} - 8 x^{2} - 9 = 0$

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

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

$x \cdot x^{2} - 9 x + x^{4} - 8 x^{2} - 9 = 0$

Factor $x$ out of $- 9 x$ .

$x \cdot x^{2} + x \cdot - 9 + x^{4} - 8 x^{2} - 9 = 0$

Factor $x$ out of $x \cdot x^{2} + x \cdot - 9$ .

$x (x^{2} - 9) + x^{4} - 8 x^{2} - 9 = 0$

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

$x (x^{2} - 3^{2}) + x^{4} - 8 x^{2} - 9 = 0$

Factor.

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = x$ and $b = 3$ .

$x ((x + 3) (x - 3)) + x^{4} - 8 x^{2} - 9 = 0$

Remove unnecessary parentheses.

$x (x + 3) (x - 3) + x^{4} - 8 x^{2} - 9 = 0$

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

$x (x + 3) (x - 3) + {(x^{2})}^{2} - 8 x^{2} - 9 = 0$

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

$x (x + 3) (x - 3) + u^{2} - 8 u - 9 = 0$

Factor $u^{2} - 8 u - 9$ 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 $- 9$ and whose sum is $- 8$ .

$- 9, 1$

Write the factored form using these integers.

$x (x + 3) (x - 3) + (u - 9) (u + 1) = 0$

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

$x (x + 3) (x - 3) + (x^{2} - 9) (x^{2} + 1) = 0$

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

$x (x + 3) (x - 3) + (x^{2} - 3^{2}) (x^{2} + 1) = 0$

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = x$ and $b = 3$ .

$x (x + 3) (x - 3) + (x + 3) (x - 3) (x^{2} + 1) = 0$

Factor $(x + 3) (x - 3)$ out of $x (x + 3) (x - 3) + (x + 3) (x - 3) (x^{2} + 1)$ .

Factor $(x + 3) (x - 3)$ out of $x (x + 3) (x - 3)$ .

$(x + 3) (x - 3) (x) + (x + 3) (x - 3) (x^{2} + 1) = 0$

Factor $(x + 3) (x - 3)$ out of $(x + 3) (x - 3) (x) + (x + 3) (x - 3) (x^{2} + 1)$ .

$(x + 3) (x - 3) (x + x^{2} + 1) = 0$

Reorder terms.

$(x + 3) (x - 3) (x^{2} + x + 1) = 0$

Step 2

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

$x + 3 = 0$

$x - 3 = 0$

$x^{2} + x + 1 = 0$

Step 3

Set $x + 3$ equal to $0$ and solve for $x$ .

Set $x + 3$ equal to $0$ .

$x + 3 = 0$

Subtract $3$ from both sides of the equation.

$x = - 3$

Step 4

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

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

$x - 3 = 0$

Add $3$ to both sides of the equation.

$x = 3$

Step 5

Set $x^{2} + x + 1$ equal to $0$ and solve for $x$ .

Set $x^{2} + x + 1$ equal to $0$ .

$x^{2} + x + 1 = 0$

Solve $x^{2} + x + 1 = 0$ for $x$ .

Use the quadratic formula to find the solutions.

$\frac{- b \pm \sqrt{b^{2} - 4 (a c)}}{2 a}$

Substitute the values $a = 1$ , $b = 1$ , and $c = 1$ into the quadratic formula and solve for $x$ .

$\frac{- 1 \pm \sqrt{1^{2} - 4 \cdot (1 \cdot 1)}}{2 \cdot 1}$

Simplify.

Simplify the numerator.

One to any power is one.

$x = \frac{- 1 \pm \sqrt{1 - 4 \cdot 1 \cdot 1}}{2 \cdot 1}$

Multiply $- 4 \cdot 1 \cdot 1$ .

Multiply $- 4$ by $1$ .

$x = \frac{- 1 \pm \sqrt{1 - 4 \cdot 1}}{2 \cdot 1}$

Multiply $- 4$ by $1$ .

$x = \frac{- 1 \pm \sqrt{1 - 4}}{2 \cdot 1}$

Subtract $4$ from $1$ .

$x = \frac{- 1 \pm \sqrt{- 3}}{2 \cdot 1}$

Rewrite $- 3$ as $- 1 (3)$ .

$x = \frac{- 1 \pm \sqrt{- 1 \cdot 3}}{2 \cdot 1}$

Rewrite $\sqrt{- 1 (3)}$ as $\sqrt{- 1} \cdot \sqrt{3}$ .

$x = \frac{- 1 \pm \sqrt{- 1} \cdot \sqrt{3}}{2 \cdot 1}$

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

$x = \frac{- 1 \pm i \sqrt{3}}{2 \cdot 1}$

Multiply $2$ by $1$ .

$x = \frac{- 1 \pm i \sqrt{3}}{2}$

The final answer is the combination of both solutions.

$x = - \frac{1 - i \sqrt{3}}{2}, - \frac{1 + i \sqrt{3}}{2}$

Step 6

The final solution is all the values that make $(x + 3) (x - 3) (x^{2} + x + 1) = 0$ true.

$x = - 3, 3, - \frac{1 - i \sqrt{3}}{2}, - \frac{1 + i \sqrt{3}}{2}$

Step 7

Find the Roots (Zeros) x^4+x^3-8x^2-9x-9=0

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