Find the Roots (Zeros) x^4-2x^3-14x^2-2x-15=0

$x^{4} - 2 x^{3} - 14 x^{2} - 2 x - 15 = 0$

Step 1

Factor the left side of the equation.

Regroup terms.

$- 2 x^{3} - 2 x + x^{4} - 14 x^{2} - 15 = 0$

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

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

$- 2 x \cdot x^{2} - 2 x + x^{4} - 14 x^{2} - 15 = 0$

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

$- 2 x \cdot x^{2} - 2 x \cdot 1 + x^{4} - 14 x^{2} - 15 = 0$

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

$- 2 x (x^{2} + 1) + x^{4} - 14 x^{2} - 15 = 0$

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

$- 2 x (x^{2} + 1) + {(x^{2})}^{2} - 14 x^{2} - 15 = 0$

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

$- 2 x (x^{2} + 1) + u^{2} - 14 u - 15 = 0$

Factor $u^{2} - 14 u - 15$ 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 $- 15$ and whose sum is $- 14$ .

$- 15, 1$

Write the factored form using these integers.

$- 2 x (x^{2} + 1) + (u - 15) (u + 1) = 0$

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

$- 2 x (x^{2} + 1) + (x^{2} - 15) (x^{2} + 1) = 0$

Factor $x^{2} + 1$ out of $- 2 x (x^{2} + 1) + (x^{2} - 15) (x^{2} + 1)$ .

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

$(x^{2} + 1) (- 2 x) + (x^{2} - 15) (x^{2} + 1) = 0$

Factor $x^{2} + 1$ out of $(x^{2} - 15) (x^{2} + 1)$ .

$(x^{2} + 1) (- 2 x) + (x^{2} + 1) (x^{2} - 15) = 0$

Factor $x^{2} + 1$ out of $(x^{2} + 1) (- 2 x) + (x^{2} + 1) (x^{2} - 15)$ .

$(x^{2} + 1) (- 2 x + x^{2} - 15) = 0$

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

$(x^{2} + 1) (- 2 u + u^{2} - 15) = 0$

Factor $- 2 u + u^{2} - 15$ 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 $- 15$ and whose sum is $- 2$ .

$- 5, 3$

Write the factored form using these integers.

$(x^{2} + 1) ((u - 5) (u + 3)) = 0$

Factor.

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

$(x^{2} + 1) ((x - 5) (x + 3)) = 0$

Remove unnecessary parentheses.

$(x^{2} + 1) (x - 5) (x + 3) = 0$

Step 2

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

$x^{2} + 1 = 0$

$x - 5 = 0$

$x + 3 = 0$

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 $x - 5$ equal to $0$ and solve for $x$ .

Set $x - 5$ equal to $0$ .

$x - 5 = 0$

Add $5$ to both sides of the equation.

$x = 5$

Step 5

Set $x + 3$ equal to $0$ and solve for $x$ .

Set $x + 3$ equal to $0$ .

$x + 3 = 0$

Subtract $3$ from both sides of the equation.

$x = - 3$

Step 6

The final solution is all the values that make $(x^{2} + 1) (x - 5) (x + 3) = 0$ true.

$x = i, - i, 5, - 3$

Step 7

Find the Roots (Zeros) x^4-2x^3-14x^2-2x-15=0

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