Find the Roots (Zeros) x^4-2x^3+14x^2-18x+45=0

$x^{4} - 2 x^{3} + 14 x^{2} - 18 x + 45 = 0$

Step 1

Factor the left side of the equation.

Regroup terms.

$- 2 x^{3} - 18 x + x^{4} + 14 x^{2} + 45 = 0$

Factor $- 2 x$ out of $- 2 x^{3} - 18 x$ .

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

$- 2 x \cdot x^{2} - 18 x + x^{4} + 14 x^{2} + 45 = 0$

Factor $- 2 x$ out of $- 18 x$ .

$- 2 x \cdot x^{2} - 2 x \cdot 9 + x^{4} + 14 x^{2} + 45 = 0$

Factor $- 2 x$ out of $- 2 x (x^{2}) - 2 x (9)$ .

$- 2 x (x^{2} + 9) + x^{4} + 14 x^{2} + 45 = 0$

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

$- 2 x (x^{2} + 9) + {(x^{2})}^{2} + 14 x^{2} + 45 = 0$

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

$- 2 x (x^{2} + 9) + u^{2} + 14 u + 45 = 0$

Factor $u^{2} + 14 u + 45$ 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 $45$ and whose sum is $14$ .

$5, 9$

Write the factored form using these integers.

$- 2 x (x^{2} + 9) + (u + 5) (u + 9) = 0$

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

$- 2 x (x^{2} + 9) + (x^{2} + 5) (x^{2} + 9) = 0$

Factor $x^{2} + 9$ out of $- 2 x (x^{2} + 9) + (x^{2} + 5) (x^{2} + 9)$ .

Factor $x^{2} + 9$ out of $- 2 x (x^{2} + 9)$ .

$(x^{2} + 9) (- 2 x) + (x^{2} + 5) (x^{2} + 9) = 0$

Factor $x^{2} + 9$ out of $(x^{2} + 5) (x^{2} + 9)$ .

$(x^{2} + 9) (- 2 x) + (x^{2} + 9) (x^{2} + 5) = 0$

Factor $x^{2} + 9$ out of $(x^{2} + 9) (- 2 x) + (x^{2} + 9) (x^{2} + 5)$ .

$(x^{2} + 9) (- 2 x + x^{2} + 5) = 0$

Reorder terms.

$(x^{2} + 9) (x^{2} - 2 x + 5) = 0$

Step 2

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

$x^{2} + 9 = 0$

$x^{2} - 2 x + 5 = 0$

Step 3

Set $x^{2} + 9$ equal to $0$ and solve for $x$ .

Set $x^{2} + 9$ equal to $0$ .

$x^{2} + 9 = 0$

Solve $x^{2} + 9 = 0$ for $x$ .

Subtract $9$ from both sides of the equation.

$x^{2} = - 9$

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

$x = \pm \sqrt{- 9}$

Simplify $\pm \sqrt{- 9}$ .

Rewrite $- 9$ as $- 1 (9)$ .

$x = \pm \sqrt{- 1 (9)}$

Rewrite $\sqrt{- 1 (9)}$ as $\sqrt{- 1} \cdot \sqrt{9}$ .

$x = \pm \sqrt{- 1} \cdot \sqrt{9}$

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

$x = \pm i \cdot \sqrt{9}$

Rewrite $9$ as $3^{2}$ .

$x = \pm i \cdot \sqrt{3^{2}}$

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

$x = \pm i \cdot 3$

Move $3$ to the left of $i$ .

$x = \pm 3 i$

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.

$x = 3 i$

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

$x = - 3 i$

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

$x = 3 i, - 3 i$

Step 4

Set $x^{2} - 2 x + 5$ equal to $0$ and solve for $x$ .

Set $x^{2} - 2 x + 5$ equal to $0$ .

$x^{2} - 2 x + 5 = 0$

Solve $x^{2} - 2 x + 5 = 0$ for $x$ .

Use the quadratic formula to find the solutions.

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

Substitute the values $a = 1$ , $b = - 2$ , and $c = 5$ into the quadratic formula and solve for $x$ .

$\frac{2 \pm \sqrt{{(- 2)}^{2} - 4 \cdot (1 \cdot 5)}}{2 \cdot 1}$

Simplify.

Simplify the numerator.

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

$x = \frac{2 \pm \sqrt{4 - 4 \cdot 1 \cdot 5}}{2 \cdot 1}$

Multiply $- 4 \cdot 1 \cdot 5$ .

Multiply $- 4$ by $1$ .

$x = \frac{2 \pm \sqrt{4 - 4 \cdot 5}}{2 \cdot 1}$

Multiply $- 4$ by $5$ .

$x = \frac{2 \pm \sqrt{4 - 20}}{2 \cdot 1}$

Subtract $20$ from $4$ .

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

Rewrite $- 16$ as $- 1 (16)$ .

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

Rewrite $\sqrt{- 1 (16)}$ as $\sqrt{- 1} \cdot \sqrt{16}$ .

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

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

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

Rewrite $16$ as $4^{2}$ .

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

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

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

Move $4$ to the left of $i$ .

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

Multiply $2$ by $1$ .

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

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

$x = 1 \pm 2 i$

The final answer is the combination of both solutions.

$x = 1 + 2 i, 1 - 2 i$

Step 5

The final solution is all the values that make $(x^{2} + 9) (x^{2} - 2 x + 5) = 0$ true.

$x = 3 i, - 3 i, 1 + 2 i, 1 - 2 i$

Step 6

Find the Roots (Zeros) x^4-2x^3+14x^2-18x+45=0

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