Find the Roots (Zeros) 0=x^4+x^3+4x^2+6x-12

$0 = x^{4} + x^{3} + 4 x^{2} + 6 x - 12$

Step 1

Rewrite the equation as $x^{4} + x^{3} + 4 x^{2} + 6 x - 12 = 0$ .

$x^{4} + x^{3} + 4 x^{2} + 6 x - 12 = 0$

Step 2

Factor the left side of the equation.

Regroup terms.

$x^{3} + 6 x + x^{4} + 4 x^{2} - 12 = 0$

Factor $x$ out of $x^{3} + 6 x$ .

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

$x \cdot x^{2} + 6 x + x^{4} + 4 x^{2} - 12 = 0$

Factor $x$ out of $6 x$ .

$x \cdot x^{2} + x \cdot 6 + x^{4} + 4 x^{2} - 12 = 0$

Factor $x$ out of $x \cdot x^{2} + x \cdot 6$ .

$x (x^{2} + 6) + x^{4} + 4 x^{2} - 12 = 0$

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

$x (x^{2} + 6) + {(x^{2})}^{2} + 4 x^{2} - 12 = 0$

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

$x (x^{2} + 6) + u^{2} + 4 u - 12 = 0$

Factor $u^{2} + 4 u - 12$ 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 $4$ .

$- 2, 6$

Write the factored form using these integers.

$x (x^{2} + 6) + (u - 2) (u + 6) = 0$

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

$x (x^{2} + 6) + (x^{2} - 2) (x^{2} + 6) = 0$

Factor $x^{2} + 6$ out of $x (x^{2} + 6) + (x^{2} - 2) (x^{2} + 6)$ .

Factor $x^{2} + 6$ out of $x (x^{2} + 6)$ .

$(x^{2} + 6) x + (x^{2} - 2) (x^{2} + 6) = 0$

Factor $x^{2} + 6$ out of $(x^{2} - 2) (x^{2} + 6)$ .

$(x^{2} + 6) x + (x^{2} + 6) (x^{2} - 2) = 0$

Factor $x^{2} + 6$ out of $(x^{2} + 6) x + (x^{2} + 6) (x^{2} - 2)$ .

$(x^{2} + 6) (x + x^{2} - 2) = 0$

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

$(x^{2} + 6) (u + u^{2} - 2) = 0$

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

$- 1, 2$

Write the factored form using these integers.

$(x^{2} + 6) ((u - 1) (u + 2)) = 0$

Factor.

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

$(x^{2} + 6) ((x - 1) (x + 2)) = 0$

Remove unnecessary parentheses.

$(x^{2} + 6) (x - 1) (x + 2) = 0$

Step 3

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

$x^{2} + 6 = 0$

$x - 1 = 0$

$x + 2 = 0$

Step 4

Set $x^{2} + 6$ equal to $0$ and solve for $x$ .

Set $x^{2} + 6$ equal to $0$ .

$x^{2} + 6 = 0$

Solve $x^{2} + 6 = 0$ for $x$ .

Subtract $6$ from both sides of the equation.

$x^{2} = - 6$

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

$x = \pm \sqrt{- 6}$

Simplify $\pm \sqrt{- 6}$ .

Rewrite $- 6$ as $- 1 (6)$ .

$x = \pm \sqrt{- 1 (6)}$

Rewrite $\sqrt{- 1 (6)}$ as $\sqrt{- 1} \cdot \sqrt{6}$ .

$x = \pm \sqrt{- 1} \cdot \sqrt{6}$

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

$x = \pm i \sqrt{6}$

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.

$x = i \sqrt{6}$

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

$x = - i \sqrt{6}$

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

$x = i \sqrt{6}, - i \sqrt{6}$

Step 5

Set $x - 1$ equal to $0$ and solve for $x$ .

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

$x - 1 = 0$

Add $1$ to both sides of the equation.

$x = 1$

Step 6

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 7

The final solution is all the values that make $(x^{2} + 6) (x - 1) (x + 2) = 0$ true.

$x = i \sqrt{6}, - i \sqrt{6}, 1, - 2$

Step 8

Find the Roots (Zeros) 0=x^4+x^3+4x^2+6x-12

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