Find the Roots (Zeros) x^4-6x^3-5x^2+24x+4=0

$x^{4} - 6 x^{3} - 5 x^{2} + 24 x + 4 = 0$

Step 1

Factor the left side of the equation.

Regroup terms.

$- 6 x^{3} + 24 x + x^{4} - 5 x^{2} + 4 = 0$

Factor $- 6 x$ out of $- 6 x^{3} + 24 x$ .

Factor $- 6 x$ out of $- 6 x^{3}$ .

$- 6 x \cdot x^{2} + 24 x + x^{4} - 5 x^{2} + 4 = 0$

Factor $- 6 x$ out of $24 x$ .

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

Factor $- 6 x$ out of $- 6 x (x^{2}) - 6 x (- 4)$ .

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

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

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

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 $b = 2$ .

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

Remove unnecessary parentheses.

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

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

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

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

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

Factor $u^{2} - 5 u + 4$ 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 $- 5$ .

$- 4, - 1$

Write the factored form using these integers.

$- 6 x (x + 2) (x - 2) + (u - 4) (u - 1) = 0$

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

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

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

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

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 $b = 2$ .

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

Rewrite $1$ as $1^{2}$ .

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

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 $b = 1$ .

$- 6 x (x + 2) (x - 2) + (x + 2) (x - 2) ((x + 1) (x - 1)) = 0$

Remove unnecessary parentheses.

$- 6 x (x + 2) (x - 2) + (x + 2) (x - 2) (x + 1) (x - 1) = 0$

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

Factor $(x + 2) (x - 2)$ out of $- 6 x (x + 2) (x - 2)$ .

$(x + 2) (x - 2) (- 6 x) + (x + 2) (x - 2) (x + 1) (x - 1) = 0$

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

$(x + 2) (x - 2) (- 6 x) + (x + 2) (x - 2) ((x + 1) (x - 1)) = 0$

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

$(x + 2) (x - 2) (- 6 x + (x + 1) (x - 1)) = 0$

Expand $(x + 1) (x - 1)$ using the FOIL Method.

Apply the distributive property.

$(x + 2) (x - 2) (- 6 x + x (x - 1) + 1 (x - 1)) = 0$

Apply the distributive property.

$(x + 2) (x - 2) (- 6 x + x \cdot x + x \cdot - 1 + 1 (x - 1)) = 0$

Apply the distributive property.

$(x + 2) (x - 2) (- 6 x + x \cdot x + x \cdot - 1 + 1 x + 1 \cdot - 1) = 0$

Simplify and combine like terms.

Simplify each term.

Multiply $x$ by $x$ .

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

Move $- 1$ to the left of $x$ .

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

Rewrite $- 1 x$ as $- x$ .

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

Multiply $x$ by $1$ .

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

Multiply $- 1$ by $1$ .

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

Add $- x$ and $x$ .

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

Add $x^{2}$ and $0$ .

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

Reorder terms.

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

Step 2

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

$x + 2 = 0$

$x - 2 = 0$

$x^{2} - 6 x - 1 = 0$

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 $x^{2} - 6 x - 1$ equal to $0$ and solve for $x$ .

Set $x^{2} - 6 x - 1$ equal to $0$ .

$x^{2} - 6 x - 1 = 0$

Solve $x^{2} - 6 x - 1 = 0$ for $x$ .

Use the quadratic formula to find the solutions.

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

Substitute the values $a = 1$ , $b = - 6$ , and $c = - 1$ into the quadratic formula and solve for $x$ .

$\frac{6 \pm \sqrt{{(- 6)}^{2} - 4 \cdot (1 \cdot - 1)}}{2 \cdot 1}$

Simplify.

Simplify the numerator.

Raise $- 6$ to the power of $2$ .

$x = \frac{6 \pm \sqrt{36 - 4 \cdot 1 \cdot - 1}}{2 \cdot 1}$

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

Multiply $- 4$ by $1$ .

$x = \frac{6 \pm \sqrt{36 - 4 \cdot - 1}}{2 \cdot 1}$

Multiply $- 4$ by $- 1$ .

$x = \frac{6 \pm \sqrt{36 + 4}}{2 \cdot 1}$

Add $36$ and $4$ .

$x = \frac{6 \pm \sqrt{40}}{2 \cdot 1}$

Rewrite $40$ as $2^{2} \cdot 10$ .

Factor $4$ out of $40$ .

$x = \frac{6 \pm \sqrt{4 (10)}}{2 \cdot 1}$

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

$x = \frac{6 \pm \sqrt{2^{2} \cdot 10}}{2 \cdot 1}$

Pull terms out from under the radical.

$x = \frac{6 \pm 2 \sqrt{10}}{2 \cdot 1}$

Multiply $2$ by $1$ .

$x = \frac{6 \pm 2 \sqrt{10}}{2}$

Simplify $\frac{6 \pm 2 \sqrt{10}}{2}$ .

$x = 3 \pm \sqrt{10}$

The final answer is the combination of both solutions.

$x = 3 + \sqrt{10}, 3 - \sqrt{10}$

Step 6

The final solution is all the values that make $(x + 2) (x - 2) (x^{2} - 6 x - 1) = 0$ true.

$x = - 2, 2, 3 + \sqrt{10}, 3 - \sqrt{10}$

Step 7

The result can be shown in multiple forms.

Exact Form:

$x = - 2, 2, 3 + \sqrt{10}, 3 - \sqrt{10}$

Decimal Form:

$x = - 2, 2, 6.16227766 \dots, - 0.16227766 \dots$

Step 8

Find the Roots (Zeros) x^4-6x^3-5x^2+24x+4=0

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