Find the Roots (Zeros) x^4-2x^2-16x-15=0

$x^{4} - 2 x^{2} - 16 x - 15 = 0$

Step 1

Factor the left side of the equation.

Regroup terms.

$- 16 x + x^{4} - 2 x^{2} - 15 = 0$

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

$- 16 x + {(x^{2})}^{2} - 2 x^{2} - 15 = 0$

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

$- 16 x + u^{2} - 2 u - 15 = 0$

Factor $u^{2} - 2 u - 15$ 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 $- 2$ .

$- 5, 3$

Write the factored form using these integers.

$- 16 x + (u - 5) (u + 3) = 0$

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

$- 16 x + (x^{2} - 5) (x^{2} + 3) = 0$

Factor $- 1$ out of $- 16 x + (x^{2} - 5) (x^{2} + 3)$ .

Reorder $- 16 x$ and $(x^{2} - 5) (x^{2} + 3)$ .

$(x^{2} - 5) (x^{2} + 3) - 16 x = 0$

Factor $- 1$ out of $(x^{2} - 5) (x^{2} + 3)$ .

$- ((- x^{2} + 5) (x^{2} + 3)) - 16 x = 0$

Factor $- 1$ out of $- 16 x$ .

$- ((- x^{2} + 5) (x^{2} + 3)) - (16 x) = 0$

Factor $- 1$ out of $- ((- x^{2} + 5) (x^{2} + 3)) - (16 x)$ .

$- ((- x^{2} + 5) (x^{2} + 3) + 16 x) = 0$

Expand $(- x^{2} + 5) (x^{2} + 3)$ using the FOIL Method.

Apply the distributive property.

$- (- x^{2} (x^{2} + 3) + 5 (x^{2} + 3) + 16 x) = 0$

Apply the distributive property.

$- (- x^{2} x^{2} - x^{2} \cdot 3 + 5 (x^{2} + 3) + 16 x) = 0$

Apply the distributive property.

$- (- x^{2} x^{2} - x^{2} \cdot 3 + 5 x^{2} + 5 \cdot 3 + 16 x) = 0$

Simplify and combine like terms.

Simplify each term.

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

Move $x^{2}$ .

$- (- (x^{2} x^{2}) - x^{2} \cdot 3 + 5 x^{2} + 5 \cdot 3 + 16 x) = 0$

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

$- (- x^{2 + 2} - x^{2} \cdot 3 + 5 x^{2} + 5 \cdot 3 + 16 x) = 0$

Add $2$ and $2$ .

$- (- x^{4} - x^{2} \cdot 3 + 5 x^{2} + 5 \cdot 3 + 16 x) = 0$

Multiply $3$ by $- 1$ .

$- (- x^{4} - 3 x^{2} + 5 x^{2} + 5 \cdot 3 + 16 x) = 0$

Multiply $5$ by $3$ .

$- (- x^{4} - 3 x^{2} + 5 x^{2} + 15 + 16 x) = 0$

Add $- 3 x^{2}$ and $5 x^{2}$ .

$- (- x^{4} + 2 x^{2} + 15 + 16 x) = 0$

Reorder terms.

$- (- x^{4} + 2 x^{2} + 16 x + 15) = 0$

Factor.

Rewrite $- x^{4} + 2 x^{2} + 16 x + 15$ in a factored form.

Factor $- x^{4} + 2 x^{2} + 16 x + 15$ 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.

$p = \pm 1, \pm 15, \pm 3, \pm 5 q = \pm 1$

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

$\pm 1, \pm 15, \pm 3, \pm 5$

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.

$- {(- 1)}^{4} + 2 {(- 1)}^{2} + 16 \cdot - 1 + 15$

Raise $- 1$ to the power of $4$ .

$- 1 \cdot 1 + 2 {(- 1)}^{2} + 16 \cdot - 1 + 15$

Multiply $- 1$ by $1$ .

$- 1 + 2 {(- 1)}^{2} + 16 \cdot - 1 + 15$

Raise $- 1$ to the power of $2$ .

$- 1 + 2 \cdot 1 + 16 \cdot - 1 + 15$

Multiply $2$ by $1$ .

$- 1 + 2 + 16 \cdot - 1 + 15$

Add $- 1$ and $2$ .

$1 + 16 \cdot - 1 + 15$

Multiply $16$ by $- 1$ .

$1 - 16 + 15$

Subtract $16$ from $1$ .

$- 15 + 15$

Add $- 15$ and $15$ .

$0$

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.

$\frac{- x^{4} + 2 x^{2} + 16 x + 15}{x + 1}$

Divide $- x^{4} + 2 x^{2} + 16 x + 15$ 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 $0$ .

$x$

$1$

$x^{4}$

$0 x^{3}$

$2 x^{2}$

$16 x$

$15$

Divide the highest order term in the dividend $- x^{4}$ by the highest order term in divisor $x$ .

$x^{3}$

$x$

$1$

$x^{4}$

$0 x^{3}$

$2 x^{2}$

$16 x$

$15$

Multiply the new quotient term by the divisor.

$x^{3}$

$x$

$1$

$x^{4}$

$0 x^{3}$

$2 x^{2}$

$16 x$

$15$

$x^{4}$

$x^{3}$

The expression needs to be subtracted from the dividend, so change all the signs in $- x^{4} - x^{3}$

$x^{3}$

$x$

$1$

$x^{4}$

$0 x^{3}$

$2 x^{2}$

$16 x$

$15$

$x^{4}$

$x^{3}$

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

$x^{3}$

$x$

$1$

$x^{4}$

$0 x^{3}$

$2 x^{2}$

$16 x$

$15$

$x^{4}$

$x^{3}$

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

$x^{3}$

$x$

$1$

$x^{4}$

$0 x^{3}$

$2 x^{2}$

$16 x$

$15$

$x^{4}$

$x^{3}$

$2 x^{2}$

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

$x^{3}$

$x^{2}$

$x$

$1$

$x^{4}$

$0 x^{3}$

$2 x^{2}$

$16 x$

$15$

$x^{4}$

$x^{3}$

$2 x^{2}$

Multiply the new quotient term by the divisor.

$x^{3}$

$x^{2}$

$x$

$1$

$x^{4}$

$0 x^{3}$

$2 x^{2}$

$16 x$

$15$

$x^{4}$

$x^{3}$

$2 x^{2}$

$x^{3}$

$x^{2}$

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

$x^{3}$

$x^{2}$

$x$

$1$

$x^{4}$

$0 x^{3}$

$2 x^{2}$

$16 x$

$15$

$x^{4}$

$x^{3}$

$2 x^{2}$

$x^{3}$

$x^{2}$

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

$x^{3}$

$x^{2}$

$x$

$1$

$x^{4}$

$0 x^{3}$

$2 x^{2}$

$16 x$

$15$

$x^{4}$

$x^{3}$

$2 x^{2}$

$x^{3}$

$x^{2}$

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

$x^{3}$

$x^{2}$

$x$

$1$

$x^{4}$

$0 x^{3}$

$2 x^{2}$

$16 x$

$15$

$x^{4}$

$x^{3}$

$2 x^{2}$

$x^{3}$

$x^{2}$

$16 x$

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

$x^{3}$

$x^{2}$

$x$

$1$

$x^{4}$

$0 x^{3}$

$2 x^{2}$

$16 x$

$15$

$x^{4}$

$x^{3}$

$2 x^{2}$

$x^{3}$

$x^{2}$

$16 x$

Multiply the new quotient term by the divisor.

$x^{3}$

$x^{2}$

$x$

$1$

$x^{4}$

$0 x^{3}$

$2 x^{2}$

$16 x$

$15$

$x^{4}$

$x^{3}$

$2 x^{2}$

$x^{3}$

$x^{2}$

$16 x$

$x^{2}$

$x$

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

$x^{3}$

$x^{2}$

$x$

$1$

$x^{4}$

$0 x^{3}$

$2 x^{2}$

$16 x$

$15$

$x^{4}$

$x^{3}$

$2 x^{2}$

$x^{3}$

$x^{2}$

$16 x$

$x^{2}$

$x$

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

$x^{3}$

$x^{2}$

$x$

$1$

$x^{4}$

$0 x^{3}$

$2 x^{2}$

$16 x$

$15$

$x^{4}$

$x^{3}$

$2 x^{2}$

$x^{3}$

$x^{2}$

$16 x$

$x^{2}$

$x$

$15 x$

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

$x^{3}$

$x^{2}$

$x$

$1$

$x^{4}$

$0 x^{3}$

$2 x^{2}$

$16 x$

$15$

$x^{4}$

$x^{3}$

$2 x^{2}$

$x^{3}$

$x^{2}$

$16 x$

$x^{2}$

$x$

$15 x$

$15$

Divide the highest order term in the dividend $15 x$ by the highest order term in divisor $x$ .

$x^{3}$

$x^{2}$

$x$

$15$

$x$

$1$

$x^{4}$

$0 x^{3}$

$2 x^{2}$

$16 x$

$15$

$x^{4}$

$x^{3}$

$2 x^{2}$

$x^{3}$

$x^{2}$

$16 x$

$x^{2}$

$x$

$15 x$

$15$

Multiply the new quotient term by the divisor.

$x^{3}$

$x^{2}$

$x$

$15$

$x$

$1$

$x^{4}$

$0 x^{3}$

$2 x^{2}$

$16 x$

$15$

$x^{4}$

$x^{3}$

$2 x^{2}$

$x^{3}$

$x^{2}$

$16 x$

$x^{2}$

$x$

$15 x$

$15$

$15 x$

$15$

The expression needs to be subtracted from the dividend, so change all the signs in $15 x + 15$

$x^{3}$

$x^{2}$

$x$

$15$

$x$

$1$

$x^{4}$

$0 x^{3}$

$2 x^{2}$

$16 x$

$15$

$x^{4}$

$x^{3}$

$2 x^{2}$

$x^{3}$

$x^{2}$

$16 x$

$x^{2}$

$x$

$15 x$

$15$

$15 x$

$15$

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

$x^{3}$

$x^{2}$

$x$

$15$

$x$

$1$

$x^{4}$

$0 x^{3}$

$2 x^{2}$

$16 x$

$15$

$x^{4}$

$x^{3}$

$2 x^{2}$

$x^{3}$

$x^{2}$

$16 x$

$x^{2}$

$x$

$15 x$

$15$

$15 x$

$15$

$0$

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

$- x^{3} + x^{2} + x + 15$

Write $- x^{4} + 2 x^{2} + 16 x + 15$ as a set of factors.

$- ((x + 1) (- x^{3} + x^{2} + x + 15)) = 0$

Factor $- x^{3} + x^{2} + x + 15$ 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.

$p = \pm 1, \pm 15, \pm 3, \pm 5 q = \pm 1$

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

$\pm 1, \pm 15, \pm 3, \pm 5$

Substitute $3$ and simplify the expression. In this case, the expression is equal to $0$ so $3$ is a root of the polynomial.

Substitute $3$ into the polynomial.

$- 3^{3} + 3^{2} + 3 + 15$

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

$- 1 \cdot 27 + 3^{2} + 3 + 15$

Multiply $- 1$ by $27$ .

$- 27 + 3^{2} + 3 + 15$

Raise $3$ to the power of $2$ .

$- 27 + 9 + 3 + 15$

Add $- 27$ and $9$ .

$- 18 + 3 + 15$

Add $- 18$ and $3$ .

$- 15 + 15$

Add $- 15$ and $15$ .

$0$

Since $3$ is a known root, divide the polynomial by $x - 3$ to find the quotient polynomial. This polynomial can then be used to find the remaining roots.

$\frac{- x^{3} + x^{2} + x + 15}{x - 3}$

Divide $- x^{3} + x^{2} + x + 15$ by $x - 3$ .

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

$x$

$3$

$x^{3}$

$x^{2}$

$x$

$15$

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

				-	$x^{2}$
	$x$	-	$3$	-	$x^{3}$	+	$x^{2}$	+	$x$	+	$15$

Multiply the new quotient term by the divisor.

			-	$x^{2}$
$x$	-	$3$	-	$x^{3}$	+	$x^{2}$	+	$x$	+	$15$
			-	$x^{3}$	+	$3 x^{2}$

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

			-	$x^{2}$
$x$	-	$3$	-	$x^{3}$	+	$x^{2}$	+	$x$	+	$15$
			+	$x^{3}$	-	$3 x^{2}$

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

			-	$x^{2}$
$x$	-	$3$	-	$x^{3}$	+	$x^{2}$	+	$x$	+	$15$
			+	$x^{3}$	-	$3 x^{2}$
					-	$2 x^{2}$

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

			-	$x^{2}$
$x$	-	$3$	-	$x^{3}$	+	$x^{2}$	+	$x$	+	$15$
			+	$x^{3}$	-	$3 x^{2}$
					-	$2 x^{2}$	+	$x$

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

			-	$x^{2}$	-	$2 x$
$x$	-	$3$	-	$x^{3}$	+	$x^{2}$	+	$x$	+	$15$
			+	$x^{3}$	-	$3 x^{2}$
					-	$2 x^{2}$	+	$x$

Multiply the new quotient term by the divisor.

			-	$x^{2}$	-	$2 x$
$x$	-	$3$	-	$x^{3}$	+	$x^{2}$	+	$x$	+	$15$
			+	$x^{3}$	-	$3 x^{2}$
					-	$2 x^{2}$	+	$x$
					-	$2 x^{2}$	+	$6 x$

The expression needs to be subtracted from the dividend, so change all the signs in $- 2 x^{2} + 6 x$

			-	$x^{2}$	-	$2 x$
$x$	-	$3$	-	$x^{3}$	+	$x^{2}$	+	$x$	+	$15$
			+	$x^{3}$	-	$3 x^{2}$
					-	$2 x^{2}$	+	$x$
					+	$2 x^{2}$	-	$6 x$

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

			-	$x^{2}$	-	$2 x$
$x$	-	$3$	-	$x^{3}$	+	$x^{2}$	+	$x$	+	$15$
			+	$x^{3}$	-	$3 x^{2}$
					-	$2 x^{2}$	+	$x$
					+	$2 x^{2}$	-	$6 x$
							-	$5 x$

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

			-	$x^{2}$	-	$2 x$
$x$	-	$3$	-	$x^{3}$	+	$x^{2}$	+	$x$	+	$15$
			+	$x^{3}$	-	$3 x^{2}$
					-	$2 x^{2}$	+	$x$
					+	$2 x^{2}$	-	$6 x$
							-	$5 x$	+	$15$

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

			-	$x^{2}$	-	$2 x$	-	$5$
$x$	-	$3$	-	$x^{3}$	+	$x^{2}$	+	$x$	+	$15$
			+	$x^{3}$	-	$3 x^{2}$
					-	$2 x^{2}$	+	$x$
					+	$2 x^{2}$	-	$6 x$
							-	$5 x$	+	$15$

Multiply the new quotient term by the divisor.

			-	$x^{2}$	-	$2 x$	-	$5$
$x$	-	$3$	-	$x^{3}$	+	$x^{2}$	+	$x$	+	$15$
			+	$x^{3}$	-	$3 x^{2}$
					-	$2 x^{2}$	+	$x$
					+	$2 x^{2}$	-	$6 x$
							-	$5 x$	+	$15$
							-	$5 x$	+	$15$

The expression needs to be subtracted from the dividend, so change all the signs in $- 5 x + 15$

			-	$x^{2}$	-	$2 x$	-	$5$
$x$	-	$3$	-	$x^{3}$	+	$x^{2}$	+	$x$	+	$15$
			+	$x^{3}$	-	$3 x^{2}$
					-	$2 x^{2}$	+	$x$
					+	$2 x^{2}$	-	$6 x$
							-	$5 x$	+	$15$
							+	$5 x$	-	$15$

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

			-	$x^{2}$	-	$2 x$	-	$5$
$x$	-	$3$	-	$x^{3}$	+	$x^{2}$	+	$x$	+	$15$
			+	$x^{3}$	-	$3 x^{2}$
					-	$2 x^{2}$	+	$x$
					+	$2 x^{2}$	-	$6 x$
							-	$5 x$	+	$15$
							+	$5 x$	-	$15$
										$0$

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

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

Write $- x^{3} + x^{2} + x + 15$ as a set of factors.

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

Remove unnecessary parentheses.

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

Remove unnecessary parentheses.

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

Step 2

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

$x + 1 = 0$

$x - 3 = 0$

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

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

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

$x - 3 = 0$

Add $3$ to both sides of the equation.

$x = 3$

Step 5

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

$x = \frac{1 \pm 2 i}{- 1}$

Move the negative one from the denominator of $\frac{1 \pm 2 i}{- 1}$ .

$x = - 1 \cdot (1 \pm 2 i)$

Rewrite $- 1 \cdot (1 \pm 2 i)$ as $- (1 \pm 2 i)$ .

$x = - (1 \pm 2 i)$

The final answer is the combination of both solutions.

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

Step 6

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

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

Step 7

Find the Roots (Zeros) x^4-2x^2-16x-15=0

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