Find the Roots (Zeros) x^4-2x^3+x^2-8x-12=0

$x^{4} - 2 x^{3} + x^{2} - 8 x - 12 = 0$

Step 1

Factor the left side of the equation.

Regroup terms.

$- 2 x^{3} - 8 x + x^{4} + x^{2} - 12 = 0$

Factor $- 2 x$ out of $- 2 x^{3} - 8 x$ .

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

$- 2 x \cdot x^{2} - 8 x + x^{4} + x^{2} - 12 = 0$

Factor $- 2 x$ out of $- 8 x$ .

$- 2 x \cdot x^{2} - 2 x \cdot 4 + x^{4} + x^{2} - 12 = 0$

Factor $- 2 x$ out of $- 2 x (x^{2}) - 2 x (4)$ .

$- 2 x (x^{2} + 4) + x^{4} + x^{2} - 12 = 0$

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

$- 2 x (x^{2} + 4) + {(x^{2})}^{2} + x^{2} - 12 = 0$

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

$- 2 x (x^{2} + 4) + u^{2} + u - 12 = 0$

Factor $u^{2} + u - 12$ 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 $- 12$ and whose sum is $1$ .

$- 3, 4$

Write the factored form using these integers.

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

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

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

Factor $x^{2} + 4$ out of $- 2 x (x^{2} + 4) + (x^{2} - 3) (x^{2} + 4)$ .

Factor $x^{2} + 4$ out of $- 2 x (x^{2} + 4)$ .

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

Factor $x^{2} + 4$ out of $(x^{2} - 3) (x^{2} + 4)$ .

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

Factor $x^{2} + 4$ out of $(x^{2} + 4) (- 2 x) + (x^{2} + 4) (x^{2} - 3)$ .

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

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

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

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

$- 3, 1$

Write the factored form using these integers.

$(x^{2} + 4) ((u - 3) (u + 1)) = 0$

Factor.

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

$(x^{2} + 4) ((x - 3) (x + 1)) = 0$

Remove unnecessary parentheses.

$(x^{2} + 4) (x - 3) (x + 1) = 0$

Step 2

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

$x^{2} + 4 = 0$

$x - 3 = 0$

$x + 1 = 0$

Step 3

Set $x^{2} + 4$ equal to $0$ and solve for $x$ .

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

$x^{2} + 4 = 0$

Solve $x^{2} + 4 = 0$ for $x$ .

Subtract $4$ from both sides of the equation.

$x^{2} = - 4$

Take the specified 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 $- 1 (4)$ .

$x = \pm \sqrt{- 1 (4)}$

Rewrite $\sqrt{- 1 (4)}$ as $\sqrt{- 1} \cdot \sqrt{4}$ .

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

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

$x = \pm i \cdot \sqrt{4}$

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

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

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

$x = \pm i \cdot 2$

Move $2$ to the left of $i$ .

$x = \pm 2 i$

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.

$x = 2 i$

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

$x = - 2 i$

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

$x = 2 i, - 2 i$

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 $x + 1$ equal to $0$ and solve for $x$ .

Set $x + 1$ equal to $0$ .

$x + 1 = 0$

Subtract $1$ from both sides of the equation.

$x = - 1$

Step 6

The final solution is all the values that make $(x^{2} + 4) (x - 3) (x + 1) = 0$ true.

$x = 2 i, - 2 i, 3, - 1$

Step 7

Find the Roots (Zeros) x^4-2x^3+x^2-8x-12=0

Related Questions

Support LCMGCF.com by disabling your adblocker.