Find the Roots (Zeros) x^4-4x^3-3x^2+16x-4=0

$x^{4} - 4 x^{3} - 3 x^{2} + 16 x - 4 = 0$

Step 1

Factor the left side of the equation.

Regroup terms.

$- 4 x^{3} + 16 x + x^{4} - 3 x^{2} - 4 = 0$

Factor $- 4 x$ out of $- 4 x^{3} + 16 x$ .

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

$- 4 x \cdot x^{2} + 16 x + x^{4} - 3 x^{2} - 4 = 0$

Factor $- 4 x$ out of $16 x$ .

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

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

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

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

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

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 $b = 2$ .

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

Remove unnecessary parentheses.

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

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

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

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

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

Factor $u^{2} - 3 u - 4$ 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 $- 4$ and whose sum is $- 3$ .

$- 4, 1$

Write the factored form using these integers.

$- 4 x (x + 2) (x - 2) + (u - 4) (u + 1) = 0$

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

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

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

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

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 $b = 2$ .

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

Factor $(x + 2) (x - 2)$ out of $- 4 x (x + 2) (x - 2) + (x + 2) (x - 2) (x^{2} + 1)$ .

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

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

Factor $(x + 2) (x - 2)$ out of $(x + 2) (x - 2) (- 4 x) + (x + 2) (x - 2) (x^{2} + 1)$ .

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

Reorder terms.

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

Step 2

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

$x + 2 = 0$

$x - 2 = 0$

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

Step 3

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 4

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 5

Set $x^{2} - 4 x + 1$ equal to $0$ and solve for $x$ .

Set $x^{2} - 4 x + 1$ equal to $0$ .

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

Solve $x^{2} - 4 x + 1 = 0$ for $x$ .

Use the quadratic formula to find the solutions.

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

Substitute the values $a = 1$ , $b = - 4$ , and $c = 1$ into the quadratic formula and solve for $x$ .

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

Simplify.

Simplify the numerator.

Raise $- 4$ to the power of $2$ .

$x = \frac{4 \pm \sqrt{16 - 4 \cdot 1 \cdot 1}}{2 \cdot 1}$

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

Multiply $- 4$ by $1$ .

$x = \frac{4 \pm \sqrt{16 - 4 \cdot 1}}{2 \cdot 1}$

Multiply $- 4$ by $1$ .

$x = \frac{4 \pm \sqrt{16 - 4}}{2 \cdot 1}$

Subtract $4$ from $16$ .

$x = \frac{4 \pm \sqrt{12}}{2 \cdot 1}$

Rewrite $12$ as $2^{2} \cdot 3$ .

Factor $4$ out of $12$ .

$x = \frac{4 \pm \sqrt{4 (3)}}{2 \cdot 1}$

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

$x = \frac{4 \pm \sqrt{2^{2} \cdot 3}}{2 \cdot 1}$

Pull terms out from under the radical.

$x = \frac{4 \pm 2 \sqrt{3}}{2 \cdot 1}$

Multiply $2$ by $1$ .

$x = \frac{4 \pm 2 \sqrt{3}}{2}$

Simplify $\frac{4 \pm 2 \sqrt{3}}{2}$ .

$x = 2 \pm \sqrt{3}$

The final answer is the combination of both solutions.

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

Step 6

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

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

Step 7

The result can be shown in multiple forms.

Exact Form:

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

Decimal Form:

$x = - 2, 2, 3.73205080 \dots, 0.26794919 \dots$

Step 8

Find the Roots (Zeros) x^4-4x^3-3x^2+16x-4=0

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