Find the Roots (Zeros) y=x^4-x^3-5x^2-x-6

$y = x^{4} - x^{3} - 5 x^{2} - x - 6$

Step 1

Set $x^{4} - x^{3} - 5 x^{2} - x - 6$ equal to $0$ .

$x^{4} - x^{3} - 5 x^{2} - x - 6 = 0$

Step 2

Solve for $x$ .

Factor the left side of the equation.

Regroup terms.

$- x^{3} - x + x^{4} - 5 x^{2} - 6 = 0$

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

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

$- x \cdot x^{2} - x + x^{4} - 5 x^{2} - 6 = 0$

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

$- x \cdot x^{2} - x \cdot 1 + x^{4} - 5 x^{2} - 6 = 0$

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

$- x (x^{2} + 1) + x^{4} - 5 x^{2} - 6 = 0$

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

$- x (x^{2} + 1) + {(x^{2})}^{2} - 5 x^{2} - 6 = 0$

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

$- x (x^{2} + 1) + u^{2} - 5 u - 6 = 0$

Factor $u^{2} - 5 u - 6$ 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 $- 6$ and whose sum is $- 5$ .

$- 6, 1$

Write the factored form using these integers.

$- x (x^{2} + 1) + (u - 6) (u + 1) = 0$

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

$- x (x^{2} + 1) + (x^{2} - 6) (x^{2} + 1) = 0$

Factor $x^{2} + 1$ out of $- x (x^{2} + 1) + (x^{2} - 6) (x^{2} + 1)$ .

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

$(x^{2} + 1) (- x) + (x^{2} - 6) (x^{2} + 1) = 0$

Factor $x^{2} + 1$ out of $(x^{2} - 6) (x^{2} + 1)$ .

$(x^{2} + 1) (- x) + (x^{2} + 1) (x^{2} - 6) = 0$

Factor $x^{2} + 1$ out of $(x^{2} + 1) (- x) + (x^{2} + 1) (x^{2} - 6)$ .

$(x^{2} + 1) (- x + x^{2} - 6) = 0$

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

$(x^{2} + 1) (- u + u^{2} - 6) = 0$

Factor $- u + u^{2} - 6$ 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 $- 6$ and whose sum is $- 1$ .

$- 3, 2$

Write the factored form using these integers.

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

Factor.

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

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

Remove unnecessary parentheses.

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

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

$x^{2} + 1 = 0$

$x - 3 = 0$

$x + 2 = 0$

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.

$x = i, - i$

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

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

$x - 3 = 0$

Add $3$ to both sides of the equation.

$x = 3$

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.

$x = - 2$

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

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

Step 3

Find the Roots (Zeros) y=x^4-x^3-5x^2-x-6

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