Find the Roots/Zeros Using the Rational Roots Test -2x^4+13x^3-21x^2+2x+8

$- 2 x^{4} + 13 x^{3} - 21 x^{2} + 2 x + 8$

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.

$p = \pm 1, \pm 2, \pm 4, \pm 8$

$q = \pm 1, \pm 2$

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

$\pm 1, \pm \frac{1}{2}, \pm 2, \pm 4, \pm 8$

Substitute the possible roots one by one into the polynomial to find the actual roots. Simplify to check if the value is $0$ , which means it is a root.

$- 2 {(1)}^{4} + 13 {(1)}^{3} - 21 {(1)}^{2} + 2 (1) + 8$

Simplify the expression. In this case, the expression is equal to $0$ so $x = 1$ is a root of the polynomial.

Simplify each term.

One to any power is one.

$- 2 \cdot 1 + 13 {(1)}^{3} - 21 {(1)}^{2} + 2 (1) + 8$

Multiply $- 2$ by $1$ .

$- 2 + 13 {(1)}^{3} - 21 {(1)}^{2} + 2 (1) + 8$

One to any power is one.

$- 2 + 13 \cdot 1 - 21 {(1)}^{2} + 2 (1) + 8$

Multiply $13$ by $1$ .

$- 2 + 13 - 21 {(1)}^{2} + 2 (1) + 8$

One to any power is one.

$- 2 + 13 - 21 \cdot 1 + 2 (1) + 8$

Multiply $- 21$ by $1$ .

$- 2 + 13 - 21 + 2 (1) + 8$

Multiply $2$ by $1$ .

$- 2 + 13 - 21 + 2 + 8$

Simplify by adding and subtracting.

Add $- 2$ and $13$ .

$11 - 21 + 2 + 8$

Subtract $21$ from $11$ .

$- 10 + 2 + 8$

Add $- 10$ and $2$ .

$- 8 + 8$

Add $- 8$ and $8$ .

$0$

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.

$\frac{- 2 x^{4} + 13 x^{3} - 21 x^{2} + 2 x + 8}{x - 1}$

Next, find the roots of the remaining polynomial. The order of the polynomial has been reduced by $1$ .

Place the numbers representing the divisor and the dividend into a division-like configuration.

$1$	$- 2$	$13$	$- 21$	$2$	$8$
	\xa0				\xa0

The first number in the dividend $(- 2)$ is put into the first position of the result area (below the horizontal line).

$1$	$- 2$	$13$	$- 21$	$2$	$8$
	\xa0				\xa0
	$- 2$

Multiply the newest entry in the result $(- 2)$ by the divisor $(1)$ and place the result of $(- 2)$ under the next term in the dividend $(13)$ .

$1$	$- 2$	$13$	$- 21$	$2$	$8$
	\xa0	$- 2$			\xa0
	$- 2$

Add the product of the multiplication and the number from the dividend and put the result in the next position on the result line.

$1$	$- 2$	$13$	$- 21$	$2$	$8$
	\xa0	$- 2$			\xa0
	$- 2$	$11$

Multiply the newest entry in the result $(11)$ by the divisor $(1)$ and place the result of $(11)$ under the next term in the dividend $(- 21)$ .

$1$	$- 2$	$13$	$- 21$	$2$	$8$
	\xa0	$- 2$	$11$		\xa0
	$- 2$	$11$

Add the product of the multiplication and the number from the dividend and put the result in the next position on the result line.

$1$	$- 2$	$13$	$- 21$	$2$	$8$
	\xa0	$- 2$	$11$		\xa0
	$- 2$	$11$	$- 10$

Multiply the newest entry in the result $(- 10)$ by the divisor $(1)$ and place the result of $(- 10)$ under the next term in the dividend $(2)$ .

$1$	$- 2$	$13$	$- 21$	$2$	$8$
	\xa0	$- 2$	$11$	$- 10$	\xa0
	$- 2$	$11$	$- 10$

Add the product of the multiplication and the number from the dividend and put the result in the next position on the result line.

$1$	$- 2$	$13$	$- 21$	$2$	$8$
	\xa0	$- 2$	$11$	$- 10$	\xa0
	$- 2$	$11$	$- 10$	$- 8$

Multiply the newest entry in the result $(- 8)$ by the divisor $(1)$ and place the result of $(- 8)$ under the next term in the dividend $(8)$ .

$1$	$- 2$	$13$	$- 21$	$2$	$8$
	\xa0	$- 2$	$11$	$- 10$	$- 8$
	$- 2$	$11$	$- 10$	$- 8$

Add the product of the multiplication and the number from the dividend and put the result in the next position on the result line.

$1$	$- 2$	$13$	$- 21$	$2$	$8$
	\xa0	$- 2$	$11$	$- 10$	$- 8$
	$- 2$	$11$	$- 10$	$- 8$	$0$

All numbers except the last become the coefficients of the quotient polynomial. The last value in the result line is the remainder.

$- 2 x^{3} + 11 x^{2} + (- 10) x - 8$

Simplify the quotient polynomial.

$- 2 x^{3} + 11 x^{2} - 10 x - 8$

Solve the equation to find any remaining roots.

Factor the left side of the equation.

Factor $- 2 x^{3} + 11 x^{2} - 10 x - 8$ 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.

$p = \pm 1, \pm 8, \pm 2, \pm 4 q = \pm 1, \pm 2$

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

$\pm 1, \pm 0.5, \pm 8, \pm 4, \pm 2$

Substitute $- 0.5$ and simplify the expression. In this case, the expression is equal to $0$ so $- 0.5$ is a root of the polynomial.

Substitute $- 0.5$ into the polynomial.

$- 2 {(- 0.5)}^{3} + 11 {(- 0.5)}^{2} - 10 \cdot - 0.5 - 8$

Raise $- 0.5$ to the power of $3$ .

$- 2 \cdot - 0.125 + 11 {(- 0.5)}^{2} - 10 \cdot - 0.5 - 8$

Multiply $- 2$ by $- 0.125$ .

$0.25 + 11 {(- 0.5)}^{2} - 10 \cdot - 0.5 - 8$

Raise $- 0.5$ to the power of $2$ .

$0.25 + 11 \cdot 0.25 - 10 \cdot - 0.5 - 8$

Multiply $11$ by $0.25$ .

$0.25 + 2.75 - 10 \cdot - 0.5 - 8$

Add $0.25$ and $2.75$ .

$3 - 10 \cdot - 0.5 - 8$

Multiply $- 10$ by $- 0.5$ .

$3 + 5 - 8$

Add $3$ and $5$ .

$8 - 8$

Subtract $8$ from $8$ .

$0$

Since $- 0.5$ is a known root, divide the polynomial by $2 x + 1$ to find the quotient polynomial. This polynomial can then be used to find the remaining roots.

$\frac{- 2 x^{3} + 11 x^{2} - 10 x - 8}{2 x + 1}$

Divide $- 2 x^{3} + 11 x^{2} - 10 x - 8$ by $2 x + 1$ .

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

$2 x$

$1$

$2 x^{3}$

$11 x^{2}$

$10 x$

$8$

Divide the highest order term in the dividend $- 2 x^{3}$ by the highest order term in divisor $2 x$ .

				-	$x^{2}$
	$2 x$	+	$1$	-	$2 x^{3}$	+	$11 x^{2}$	-	$10 x$	-	$8$

Multiply the new quotient term by the divisor.

			-	$x^{2}$
$2 x$	+	$1$	-	$2 x^{3}$	+	$11 x^{2}$	-	$10 x$	-	$8$
			-	$2 x^{3}$	-	$x^{2}$

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

			-	$x^{2}$
$2 x$	+	$1$	-	$2 x^{3}$	+	$11 x^{2}$	-	$10 x$	-	$8$
			+	$2 x^{3}$	+	$x^{2}$

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

			-	$x^{2}$
$2 x$	+	$1$	-	$2 x^{3}$	+	$11 x^{2}$	-	$10 x$	-	$8$
			+	$2 x^{3}$	+	$x^{2}$
					+	$12 x^{2}$

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

			-	$x^{2}$
$2 x$	+	$1$	-	$2 x^{3}$	+	$11 x^{2}$	-	$10 x$	-	$8$
			+	$2 x^{3}$	+	$x^{2}$
					+	$12 x^{2}$	-	$10 x$

Divide the highest order term in the dividend $12 x^{2}$ by the highest order term in divisor $2 x$ .

			-	$x^{2}$	+	$6 x$
$2 x$	+	$1$	-	$2 x^{3}$	+	$11 x^{2}$	-	$10 x$	-	$8$
			+	$2 x^{3}$	+	$x^{2}$
					+	$12 x^{2}$	-	$10 x$

Multiply the new quotient term by the divisor.

			-	$x^{2}$	+	$6 x$
$2 x$	+	$1$	-	$2 x^{3}$	+	$11 x^{2}$	-	$10 x$	-	$8$
			+	$2 x^{3}$	+	$x^{2}$
					+	$12 x^{2}$	-	$10 x$
					+	$12 x^{2}$	+	$6 x$

The expression needs to be subtracted from the dividend, so change all the signs in $12 x^{2} + 6 x$

			-	$x^{2}$	+	$6 x$
$2 x$	+	$1$	-	$2 x^{3}$	+	$11 x^{2}$	-	$10 x$	-	$8$
			+	$2 x^{3}$	+	$x^{2}$
					+	$12 x^{2}$	-	$10 x$
					-	$12 x^{2}$	-	$6 x$

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

			-	$x^{2}$	+	$6 x$
$2 x$	+	$1$	-	$2 x^{3}$	+	$11 x^{2}$	-	$10 x$	-	$8$
			+	$2 x^{3}$	+	$x^{2}$
					+	$12 x^{2}$	-	$10 x$
					-	$12 x^{2}$	-	$6 x$
							-	$16 x$

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

			-	$x^{2}$	+	$6 x$
$2 x$	+	$1$	-	$2 x^{3}$	+	$11 x^{2}$	-	$10 x$	-	$8$
			+	$2 x^{3}$	+	$x^{2}$
					+	$12 x^{2}$	-	$10 x$
					-	$12 x^{2}$	-	$6 x$
							-	$16 x$	-	$8$

Divide the highest order term in the dividend $- 16 x$ by the highest order term in divisor $2 x$ .

			-	$x^{2}$	+	$6 x$	-	$8$
$2 x$	+	$1$	-	$2 x^{3}$	+	$11 x^{2}$	-	$10 x$	-	$8$
			+	$2 x^{3}$	+	$x^{2}$
					+	$12 x^{2}$	-	$10 x$
					-	$12 x^{2}$	-	$6 x$
							-	$16 x$	-	$8$

Multiply the new quotient term by the divisor.

			-	$x^{2}$	+	$6 x$	-	$8$
$2 x$	+	$1$	-	$2 x^{3}$	+	$11 x^{2}$	-	$10 x$	-	$8$
			+	$2 x^{3}$	+	$x^{2}$
					+	$12 x^{2}$	-	$10 x$
					-	$12 x^{2}$	-	$6 x$
							-	$16 x$	-	$8$
							-	$16 x$	-	$8$

The expression needs to be subtracted from the dividend, so change all the signs in $- 16 x - 8$

			-	$x^{2}$	+	$6 x$	-	$8$
$2 x$	+	$1$	-	$2 x^{3}$	+	$11 x^{2}$	-	$10 x$	-	$8$
			+	$2 x^{3}$	+	$x^{2}$
					+	$12 x^{2}$	-	$10 x$
					-	$12 x^{2}$	-	$6 x$
							-	$16 x$	-	$8$
							+	$16 x$	+	$8$

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

			-	$x^{2}$	+	$6 x$	-	$8$
$2 x$	+	$1$	-	$2 x^{3}$	+	$11 x^{2}$	-	$10 x$	-	$8$
			+	$2 x^{3}$	+	$x^{2}$
					+	$12 x^{2}$	-	$10 x$
					-	$12 x^{2}$	-	$6 x$
							-	$16 x$	-	$8$
							+	$16 x$	+	$8$
										$0$

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

$- x^{2} + 6 x - 8$

Write $- 2 x^{3} + 11 x^{2} - 10 x - 8$ as a set of factors.

$(2 x + 1) (- x^{2} + 6 x - 8) = 0$

Factor by grouping.

For a polynomial of the form $a x^{2} + b x + c$ , rewrite the middle term as a sum of two terms whose product is $a \cdot c = - 1 \cdot - 8 = 8$ and whose sum is $b = 6$ .

Factor $6$ out of $6 x$ .

$(2 x + 1) (- x^{2} + 6 (x) - 8) = 0$

Rewrite $6$ as $2$ plus $4$

$(2 x + 1) (- x^{2} + (2 + 4) x - 8) = 0$

Apply the distributive property.

$(2 x + 1) (- x^{2} + 2 x + 4 x - 8) = 0$

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

$(2 x + 1) ((- x^{2} + 2 x) + 4 x - 8) = 0$

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

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

Factor the polynomial by factoring out the greatest common factor, $- x + 2$ .

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

Remove unnecessary parentheses.

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

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

$2 x + 1 = 0$

$- x + 2 = 0$

$x - 4 = 0$

Set $2 x + 1$ equal to $0$ and solve for $x$ .

Set $2 x + 1$ equal to $0$ .

$2 x + 1 = 0$

Solve $2 x + 1 = 0$ for $x$ .

Subtract $1$ from both sides of the equation.

$2 x = - 1$

Divide each term in $2 x = - 1$ by $2$ and simplify.

Divide each term in $2 x = - 1$ by $2$ .

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

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

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

Divide $x$ by $1$ .

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

Simplify the right side.

Move the negative in front of the fraction.

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

Set $- x + 2$ equal to $0$ and solve for $x$ .

Set $- x + 2$ equal to $0$ .

$- x + 2 = 0$

Solve $- x + 2 = 0$ for $x$ .

Subtract $2$ from both sides of the equation.

$- x = - 2$

Divide each term in $- x = - 2$ by $- 1$ and simplify.

Divide each term in $- x = - 2$ by $- 1$ .

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

Simplify the left side.

Dividing two negative values results in a positive value.

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

Divide $x$ by $1$ .

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

Simplify the right side.

Divide $- 2$ by $- 1$ .

$x = 2$

Set $x - 4$ equal to $0$ and solve for $x$ .

Set $x - 4$ equal to $0$ .

$x - 4 = 0$

Add $4$ to both sides of the equation.

$x = 4$

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

$x = - \frac{1}{2}, 2, 4$

The polynomial can be written as a set of linear factors.

$(x - 1) (x + \frac{1}{2}) (x - 2) (x - 4)$

These are the roots (zeros) of the polynomial $- 2 x^{4} + 13 x^{3} - 21 x^{2} + 2 x + 8$ .

$x = 1, - \frac{1}{2}, 2, 4$

Find the Roots/Zeros Using the Rational Roots Test -2x^4+13x^3-21x^2+2x+8

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