Find the Roots/Zeros Using the Rational Roots Test x^4-4x^3+14x^2-4x+13

$x^{4} - 4 x^{3} + 14 x^{2} - 4 x + 13$

Step 1

Factor the left side of the equation.

Regroup terms.

$- 4 x^{3} - 4 x + x^{4} + 14 x^{2} + 13 = 0$

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

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

$- 4 x \cdot x^{2} - 4 x + x^{4} + 14 x^{2} + 13 = 0$

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

$- 4 x \cdot x^{2} - 4 x \cdot 1 + x^{4} + 14 x^{2} + 13 = 0$

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

$- 4 x (x^{2} + 1) + x^{4} + 14 x^{2} + 13 = 0$

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

$- 4 x (x^{2} + 1) + {(x^{2})}^{2} + 14 x^{2} + 13 = 0$

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

$- 4 x (x^{2} + 1) + u^{2} + 14 u + 13 = 0$

Factor $u^{2} + 14 u + 13$ 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 $13$ and whose sum is $14$ .

$1, 13$

Write the factored form using these integers.

$- 4 x (x^{2} + 1) + (u + 1) (u + 13) = 0$

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

$- 4 x (x^{2} + 1) + (x^{2} + 1) (x^{2} + 13) = 0$

Factor $x^{2} + 1$ out of $- 4 x (x^{2} + 1) + (x^{2} + 1) (x^{2} + 13)$ .

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

$(x^{2} + 1) (- 4 x) + (x^{2} + 1) (x^{2} + 13) = 0$

Factor $x^{2} + 1$ out of $(x^{2} + 1) (- 4 x) + (x^{2} + 1) (x^{2} + 13)$ .

$(x^{2} + 1) (- 4 x + x^{2} + 13) = 0$

Reorder terms.

$(x^{2} + 1) (x^{2} - 4 x + 13) = 0$

Step 2

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

$x^{2} + 1 = 0$

$x^{2} - 4 x + 13 = 0$

Step 3

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

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

$x^{2} + 1 = 0$

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

Subtract $1$ from both sides of the equation.

$x^{2} = - 1$

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

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

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

$x = \pm i$

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.

$x = i$

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

$x = - i$

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

$x = i, - i$

Step 4

Set $x^{2} - 4 x + 13$ equal to $0$ and solve for $x$ .

Set $x^{2} - 4 x + 13$ equal to $0$ .

$x^{2} - 4 x + 13 = 0$

Solve $x^{2} - 4 x + 13 = 0$ 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 $c = 13$ into the quadratic formula and solve for $x$ .

$\frac{4 \pm \sqrt{{(- 4)}^{2} - 4 \cdot (1 \cdot 13)}}{2 \cdot 1}$

Simplify.

Simplify the numerator.

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

$x = \frac{4 \pm \sqrt{16 - 4 \cdot 1 \cdot 13}}{2 \cdot 1}$

Multiply $- 4 \cdot 1 \cdot 13$ .

Multiply $- 4$ by $1$ .

$x = \frac{4 \pm \sqrt{16 - 4 \cdot 13}}{2 \cdot 1}$

Multiply $- 4$ by $13$ .

$x = \frac{4 \pm \sqrt{16 - 52}}{2 \cdot 1}$

Subtract $52$ from $16$ .

$x = \frac{4 \pm \sqrt{- 36}}{2 \cdot 1}$

Rewrite $- 36$ as $- 1 (36)$ .

$x = \frac{4 \pm \sqrt{- 1 \cdot 36}}{2 \cdot 1}$

Rewrite $\sqrt{- 1 (36)}$ as $\sqrt{- 1} \cdot \sqrt{36}$ .

$x = \frac{4 \pm \sqrt{- 1} \cdot \sqrt{36}}{2 \cdot 1}$

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

$x = \frac{4 \pm i \cdot \sqrt{36}}{2 \cdot 1}$

Rewrite $36$ as $6^{2}$ .

$x = \frac{4 \pm i \cdot \sqrt{6^{2}}}{2 \cdot 1}$

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

$x = \frac{4 \pm i \cdot 6}{2 \cdot 1}$

Move $6$ to the left of $i$ .

$x = \frac{4 \pm 6 i}{2 \cdot 1}$

Multiply $2$ by $1$ .

$x = \frac{4 \pm 6 i}{2}$

Simplify $\frac{4 \pm 6 i}{2}$ .

$x = 2 \pm 3 i$

The final answer is the combination of both solutions.

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

Step 5

The final solution is all the values that make $(x^{2} + 1) (x^{2} - 4 x + 13) = 0$ true.

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

Step 6

Find the Roots/Zeros Using the Rational Roots Test x^4-4x^3+14x^2-4x+13

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