Find the Roots (Zeros) x^4-4x^3-x^2-16x-20=0

$x^{4} - 4 x^{3} - x^{2} - 16 x - 20 = 0$

Step 1

Factor the left side of the equation.

Regroup terms.

$x^{4} - x^{2} - 4 x^{3} - 16 x - 20 = 0$

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

Factor $x^{2}$ out of $x^{4}$ .

$x^{2} x^{2} - x^{2} - 4 x^{3} - 16 x - 20 = 0$

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

$x^{2} x^{2} + x^{2} \cdot - 1 - 4 x^{3} - 16 x - 20 = 0$

Factor $x^{2}$ out of $x^{2} x^{2} + x^{2} \cdot - 1$ .

$x^{2} (x^{2} - 1) - 4 x^{3} - 16 x - 20 = 0$

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

$x^{2} (x^{2} - 1^{2}) - 4 x^{3} - 16 x - 20 = 0$

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 $b = 1$ .

$x^{2} ((x + 1) (x - 1)) - 4 x^{3} - 16 x - 20 = 0$

Remove unnecessary parentheses.

$x^{2} (x + 1) (x - 1) - 4 x^{3} - 16 x - 20 = 0$

Factor $4$ out of $- 4 x^{3} - 16 x - 20$ .

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

$x^{2} (x + 1) (x - 1) + 4 (- x^{3}) - 16 x - 20 = 0$

Factor $4$ out of $- 16 x$ .

$x^{2} (x + 1) (x - 1) + 4 (- x^{3}) + 4 (- 4 x) - 20 = 0$

Factor $4$ out of $- 20$ .

$x^{2} (x + 1) (x - 1) + 4 (- x^{3}) + 4 (- 4 x) + 4 (- 5) = 0$

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

$x^{2} (x + 1) (x - 1) + 4 (- x^{3} - 4 x) + 4 (- 5) = 0$

Factor $4$ out of $4 (- x^{3} - 4 x) + 4 (- 5)$ .

$x^{2} (x + 1) (x - 1) + 4 (- x^{3} - 4 x - 5) = 0$

Factor.

Factor $- x^{3} - 4 x - 5$ using the rational roots test.

If a polynomial function has integer coefficients, then every rational zero will have the form $\frac{p}{q}$ where $p$ is a factor of the constant and $q$ is a factor of the leading coefficient.

$p = \pm 1, \pm 5 q = \pm 1$

Find every combination of $\pm \frac{p}{q}$ . These are the possible roots of the polynomial function.

$\pm 1, \pm 5$

Substitute $- 1$ and simplify the expression. In this case, the expression is equal to $0$ so $- 1$ is a root of the polynomial.

Substitute $- 1$ into the polynomial.

$- {(- 1)}^{3} - 4 \cdot - 1 - 5$

Raise $- 1$ to the power of $3$ .

$- - 1 - 4 \cdot - 1 - 5$

Multiply $- 1$ by $- 1$ .

$1 - 4 \cdot - 1 - 5$

Multiply $- 4$ by $- 1$ .

$1 + 4 - 5$

Add $1$ and $4$ .

$5 - 5$

Subtract $5$ from $5$ .

$0$

Since $- 1$ is a known root, divide the polynomial by $x + 1$ to find the quotient polynomial. This polynomial can then be used to find the remaining roots.

$\frac{- x^{3} - 4 x - 5}{x + 1}$

Divide $- x^{3} - 4 x - 5$ by $x + 1$ .

Set up the polynomials to be divided. If there is not a term for every exponent, insert one with a value of $0$ .

$x$

$1$

$x^{3}$

$0 x^{2}$

$4 x$

$5$

Divide the highest order term in the dividend $- x^{3}$ by the highest order term in divisor $x$ .

				-	$x^{2}$
	$x$	+	$1$	-	$x^{3}$	+	$0 x^{2}$	-	$4 x$	-	$5$

Multiply the new quotient term by the divisor.

			-	$x^{2}$
$x$	+	$1$	-	$x^{3}$	+	$0 x^{2}$	-	$4 x$	-	$5$
			-	$x^{3}$	-	$x^{2}$

The expression needs to be subtracted from the dividend, so change all the signs in $- x^{3} - x^{2}$

			-	$x^{2}$
$x$	+	$1$	-	$x^{3}$	+	$0 x^{2}$	-	$4 x$	-	$5$
			+	$x^{3}$	+	$x^{2}$

After changing the signs, add the last dividend from the multiplied polynomial to find the new dividend.

			-	$x^{2}$
$x$	+	$1$	-	$x^{3}$	+	$0 x^{2}$	-	$4 x$	-	$5$
			+	$x^{3}$	+	$x^{2}$
					+	$x^{2}$

Pull the next terms from the original dividend down into the current dividend.

			-	$x^{2}$
$x$	+	$1$	-	$x^{3}$	+	$0 x^{2}$	-	$4 x$	-	$5$
			+	$x^{3}$	+	$x^{2}$
					+	$x^{2}$	-	$4 x$

Divide the highest order term in the dividend $x^{2}$ by the highest order term in divisor $x$ .

			-	$x^{2}$	+	$x$
$x$	+	$1$	-	$x^{3}$	+	$0 x^{2}$	-	$4 x$	-	$5$
			+	$x^{3}$	+	$x^{2}$
					+	$x^{2}$	-	$4 x$

Multiply the new quotient term by the divisor.

			-	$x^{2}$	+	$x$
$x$	+	$1$	-	$x^{3}$	+	$0 x^{2}$	-	$4 x$	-	$5$
			+	$x^{3}$	+	$x^{2}$
					+	$x^{2}$	-	$4 x$
					+	$x^{2}$	+	$x$

The expression needs to be subtracted from the dividend, so change all the signs in $x^{2} + x$

			-	$x^{2}$	+	$x$
$x$	+	$1$	-	$x^{3}$	+	$0 x^{2}$	-	$4 x$	-	$5$
			+	$x^{3}$	+	$x^{2}$
					+	$x^{2}$	-	$4 x$
					-	$x^{2}$	-	$x$

After changing the signs, add the last dividend from the multiplied polynomial to find the new dividend.

			-	$x^{2}$	+	$x$
$x$	+	$1$	-	$x^{3}$	+	$0 x^{2}$	-	$4 x$	-	$5$
			+	$x^{3}$	+	$x^{2}$
					+	$x^{2}$	-	$4 x$
					-	$x^{2}$	-	$x$
							-	$5 x$

Pull the next terms from the original dividend down into the current dividend.

			-	$x^{2}$	+	$x$
$x$	+	$1$	-	$x^{3}$	+	$0 x^{2}$	-	$4 x$	-	$5$
			+	$x^{3}$	+	$x^{2}$
					+	$x^{2}$	-	$4 x$
					-	$x^{2}$	-	$x$
							-	$5 x$	-	$5$

Divide the highest order term in the dividend $- 5 x$ by the highest order term in divisor $x$ .

			-	$x^{2}$	+	$x$	-	$5$
$x$	+	$1$	-	$x^{3}$	+	$0 x^{2}$	-	$4 x$	-	$5$
			+	$x^{3}$	+	$x^{2}$
					+	$x^{2}$	-	$4 x$
					-	$x^{2}$	-	$x$
							-	$5 x$	-	$5$

Multiply the new quotient term by the divisor.

			-	$x^{2}$	+	$x$	-	$5$
$x$	+	$1$	-	$x^{3}$	+	$0 x^{2}$	-	$4 x$	-	$5$
			+	$x^{3}$	+	$x^{2}$
					+	$x^{2}$	-	$4 x$
					-	$x^{2}$	-	$x$
							-	$5 x$	-	$5$
							-	$5 x$	-	$5$

The expression needs to be subtracted from the dividend, so change all the signs in $- 5 x - 5$

			-	$x^{2}$	+	$x$	-	$5$
$x$	+	$1$	-	$x^{3}$	+	$0 x^{2}$	-	$4 x$	-	$5$
			+	$x^{3}$	+	$x^{2}$
					+	$x^{2}$	-	$4 x$
					-	$x^{2}$	-	$x$
							-	$5 x$	-	$5$
							+	$5 x$	+	$5$

After changing the signs, add the last dividend from the multiplied polynomial to find the new dividend.

			-	$x^{2}$	+	$x$	-	$5$
$x$	+	$1$	-	$x^{3}$	+	$0 x^{2}$	-	$4 x$	-	$5$
			+	$x^{3}$	+	$x^{2}$
					+	$x^{2}$	-	$4 x$
					-	$x^{2}$	-	$x$
							-	$5 x$	-	$5$
							+	$5 x$	+	$5$
										$0$

Since the remander is $0$ , the final answer is the quotient.

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

Write $- x^{3} - 4 x - 5$ as a set of factors.

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

Remove unnecessary parentheses.

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

Factor $x + 1$ out of $x^{2} (x + 1) (x - 1) + 4 (x + 1) (- x^{2} + x - 5)$ .

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

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

Factor $x + 1$ out of $4 (x + 1) (- x^{2} + x - 5)$ .

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

Factor $x + 1$ out of $(x + 1) (x^{2} (x - 1)) + (x + 1) (4 (- x^{2} + x - 5))$ .

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

Apply the distributive property.

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

Multiply $x^{2}$ by $x$ by adding the exponents.

Multiply $x^{2}$ by $x$ .

Raise $x$ to the power of $1$ .

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

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

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

Add $2$ and $1$ .

$(x + 1) (x^{3} + x^{2} \cdot - 1 + 4 (- x^{2} + x - 5)) = 0$

Move $- 1$ to the left of $x^{2}$ .

$(x + 1) (x^{3} - 1 x^{2} + 4 (- x^{2} + x - 5)) = 0$

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

$(x + 1) (x^{3} - x^{2} + 4 (- x^{2} + x - 5)) = 0$

Apply the distributive property.

$(x + 1) (x^{3} - x^{2} + 4 (- x^{2}) + 4 x + 4 \cdot - 5) = 0$

Simplify.

Multiply $- 1$ by $4$ .

$(x + 1) (x^{3} - x^{2} - 4 x^{2} + 4 x + 4 \cdot - 5) = 0$

Multiply $4$ by $- 5$ .

$(x + 1) (x^{3} - x^{2} - 4 x^{2} + 4 x - 20) = 0$

Subtract $4 x^{2}$ from $- x^{2}$ .

$(x + 1) (x^{3} - 5 x^{2} + 4 x - 20) = 0$

Factor.

Rewrite $x^{3} - 5 x^{2} + 4 x - 20$ in a factored form.

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

$(x + 1) ((x^{3} - 5 x^{2}) + 4 x - 20) = 0$

Factor out the greatest common factor (GCF) from each group.

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

Factor the polynomial by factoring out the greatest common factor, $x - 5$ .

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

Remove unnecessary parentheses.

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

Step 2

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

$x + 1 = 0$

$x - 5 = 0$

$x^{2} + 4 = 0$

Step 3

Set $x + 1$ equal to $0$ and solve for $x$ .

Set $x + 1$ equal to $0$ .

$x + 1 = 0$

Subtract $1$ from both sides of the equation.

$x = - 1$

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

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

$x^{2} + 4 = 0$

Solve $x^{2} + 4 = 0$ for $x$ .

Subtract $4$ from both sides of the equation.

$x^{2} = - 4$

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

$x = \pm \sqrt{- 4}$

Simplify $\pm \sqrt{- 4}$ .

Rewrite $- 4$ as $- 1 (4)$ .

$x = \pm \sqrt{- 1 (4)}$

Rewrite $\sqrt{- 1 (4)}$ as $\sqrt{- 1} \cdot \sqrt{4}$ .

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

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

$x = \pm i \cdot \sqrt{4}$

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

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

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

$x = \pm i \cdot 2$

Move $2$ to the left of $i$ .

$x = \pm 2 i$

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.

$x = 2 i$

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

$x = - 2 i$

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

$x = 2 i, - 2 i$

Step 6

The final solution is all the values that make $(x + 1) (x - 5) (x^{2} + 4) = 0$ true.

$x = - 1, 5, 2 i, - 2 i$

Step 7

Find the Roots (Zeros) x^4-4x^3-x^2-16x-20=0

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