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

$x^{4} - 4 x^{3} + x^{2} + 12 x - 12$

Step 1

Set $x^{4} - 4 x^{3} + x^{2} + 12 x - 12$ equal to $0$ .

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

Step 2

Solve for $x$ .

Factor the left side of the equation.

Regroup terms.

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

Factor $- 4 x$ out of $- 4 x^{3} + 12 x$ .

Factor $- 4 x$ out of $- 4 x^{3}$ .

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

Factor $- 4 x$ out of $12 x$ .

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

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

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

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

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

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

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

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.

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

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

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

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

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

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

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

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

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

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

Factor using the perfect square rule.

Rearrange terms.

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

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

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

Check that the middle term is two times the product of the numbers being squared in the first term and third term.

$4 u = 2 \cdot u \cdot 2$

Rewrite the polynomial.

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

Factor using the perfect square trinomial rule $a^{2} - 2 a b + b^{2} = {(a - b)}^{2}$ , where $a = u$ and $b = 2$ .

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

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

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

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

$x^{2} - 3 = 0$

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

Set $x^{2} - 3$ equal to $0$ and solve for $x$ .

Set $x^{2} - 3$ equal to $0$ .

$x^{2} - 3 = 0$

Solve $x^{2} - 3 = 0$ for $x$ .

Add $3$ to both sides of the equation.

$x^{2} = 3$

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

$x = \pm \sqrt{3}$

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.

$x = \sqrt{3}$

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

$x = - \sqrt{3}$

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

$x = \sqrt{3}, - \sqrt{3}$

Set ${(x - 2)}^{2}$ equal to $0$ and solve for $x$ .

Set ${(x - 2)}^{2}$ equal to $0$ .

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

Solve ${(x - 2)}^{2} = 0$ for $x$ .

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

$x - 2 = 0$

Add $2$ to both sides of the equation.

$x = 2$

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

$x = \sqrt{3}, - \sqrt{3}, 2$

Step 3

The result can be shown in multiple forms.

Exact Form:

$x = \sqrt{3}, - \sqrt{3}, 2$

Decimal Form:

$x = 1.73205080 \dots, - 1.73205080 \dots, 2$

Step 4

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

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