Find the Roots (Zeros) 0=x^4-3x^3+8x^2-18x+12

$0 = x^{4} - 3 x^{3} + 8 x^{2} - 18 x + 12$

Step 1

Rewrite the equation as $x^{4} - 3 x^{3} + 8 x^{2} - 18 x + 12 = 0$ .

$x^{4} - 3 x^{3} + 8 x^{2} - 18 x + 12 = 0$

Step 2

Factor the left side of the equation.

Regroup terms.

$- 3 x^{3} - 18 x + x^{4} + 8 x^{2} + 12 = 0$

Factor $- 3 x$ out of $- 3 x^{3} - 18 x$ .

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

$- 3 x \cdot x^{2} - 18 x + x^{4} + 8 x^{2} + 12 = 0$

Factor $- 3 x$ out of $- 18 x$ .

$- 3 x \cdot x^{2} - 3 x \cdot 6 + x^{4} + 8 x^{2} + 12 = 0$

Factor $- 3 x$ out of $- 3 x (x^{2}) - 3 x (6)$ .

$- 3 x (x^{2} + 6) + x^{4} + 8 x^{2} + 12 = 0$

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

$- 3 x (x^{2} + 6) + {(x^{2})}^{2} + 8 x^{2} + 12 = 0$

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

$- 3 x (x^{2} + 6) + u^{2} + 8 u + 12 = 0$

Factor $u^{2} + 8 u + 12$ 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 $8$ .

$2, 6$

Write the factored form using these integers.

$- 3 x (x^{2} + 6) + (u + 2) (u + 6) = 0$

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

$- 3 x (x^{2} + 6) + (x^{2} + 2) (x^{2} + 6) = 0$

Factor $x^{2} + 6$ out of $- 3 x (x^{2} + 6) + (x^{2} + 2) (x^{2} + 6)$ .

Factor $x^{2} + 6$ out of $- 3 x (x^{2} + 6)$ .

$(x^{2} + 6) (- 3 x) + (x^{2} + 2) (x^{2} + 6) = 0$

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

$(x^{2} + 6) (- 3 x) + (x^{2} + 6) (x^{2} + 2) = 0$

Factor $x^{2} + 6$ out of $(x^{2} + 6) (- 3 x) + (x^{2} + 6) (x^{2} + 2)$ .

$(x^{2} + 6) (- 3 x + x^{2} + 2) = 0$

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

$(x^{2} + 6) (- 3 u + u^{2} + 2) = 0$

Factor $- 3 u + u^{2} + 2$ 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 $- 3$ .

$- 2, - 1$

Write the factored form using these integers.

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

Factor.

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

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

Remove unnecessary parentheses.

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

Step 3

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

$x^{2} + 6 = 0$

$x - 2 = 0$

$x - 1 = 0$

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 $x - 2$ equal to $0$ and solve for $x$ .

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

$x - 2 = 0$

Add $2$ to both sides of the equation.

$x = 2$

Step 6

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 7

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

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

Step 8

Find the Roots (Zeros) 0=x^4-3x^3+8x^2-18x+12

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