Find the Roots/Zeros Using the Rational Roots Test f(x)=2x^5+9x^4+13x^3+5x^2-3x-2

$f (x) = 2 x^{5} + 9 x^{4} + 13 x^{3} + 5 x^{2} - 3 x - 2$

Step 1

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.

$p = \pm 1, \pm 2$

$q = \pm 1, \pm 2$

Step 2

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

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

Step 3

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.

$2 {(- 1)}^{5} + 9 {(- 1)}^{4} + 13 {(- 1)}^{3} + 5 {(- 1)}^{2} - 3 \cdot - 1 - 2$

Step 4

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

Simplify each term.

Raise $- 1$ to the power of $5$ .

$2 \cdot - 1 + 9 {(- 1)}^{4} + 13 {(- 1)}^{3} + 5 {(- 1)}^{2} - 3 \cdot - 1 - 2$

Multiply $2$ by $- 1$ .

$- 2 + 9 {(- 1)}^{4} + 13 {(- 1)}^{3} + 5 {(- 1)}^{2} - 3 \cdot - 1 - 2$

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

$- 2 + 9 \cdot 1 + 13 {(- 1)}^{3} + 5 {(- 1)}^{2} - 3 \cdot - 1 - 2$

Multiply $9$ by $1$ .

$- 2 + 9 + 13 {(- 1)}^{3} + 5 {(- 1)}^{2} - 3 \cdot - 1 - 2$

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

$- 2 + 9 + 13 \cdot - 1 + 5 {(- 1)}^{2} - 3 \cdot - 1 - 2$

Multiply $13$ by $- 1$ .

$- 2 + 9 - 13 + 5 {(- 1)}^{2} - 3 \cdot - 1 - 2$

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

$- 2 + 9 - 13 + 5 \cdot 1 - 3 \cdot - 1 - 2$

Multiply $5$ by $1$ .

$- 2 + 9 - 13 + 5 - 3 \cdot - 1 - 2$

Multiply $- 3$ by $- 1$ .

$- 2 + 9 - 13 + 5 + 3 - 2$

Simplify by adding and subtracting.

Add $- 2$ and $9$ .

$7 - 13 + 5 + 3 - 2$

Subtract $13$ from $7$ .

$- 6 + 5 + 3 - 2$

Add $- 6$ and $5$ .

$- 1 + 3 - 2$

Add $- 1$ and $3$ .

$2 - 2$

Subtract $2$ from $2$ .

$0$

Step 5

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.

$\frac{2 x^{5} + 9 x^{4} + 13 x^{3} + 5 x^{2} - 3 x - 2}{x + 1}$

Step 6

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.

$- 1$	$2$	$9$	$13$	$5$	$- 3$	$- 2$
	\xa0					\xa0

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

$- 1$	$2$	$9$	$13$	$5$	$- 3$	$- 2$
	\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 $(9)$ .

$- 1$	$2$	$9$	$13$	$5$	$- 3$	$- 2$
	\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.

$- 1$	$2$	$9$	$13$	$5$	$- 3$	$- 2$
	\xa0	$- 2$				\xa0
	$2$	$7$

Multiply the newest entry in the result $(7)$ by the divisor $(- 1)$ and place the result of $(- 7)$ under the next term in the dividend $(13)$ .

$- 1$	$2$	$9$	$13$	$5$	$- 3$	$- 2$
	\xa0	$- 2$	$- 7$			\xa0
	$2$	$7$

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

$- 1$	$2$	$9$	$13$	$5$	$- 3$	$- 2$
	\xa0	$- 2$	$- 7$			\xa0
	$2$	$7$	$6$

Multiply the newest entry in the result $(6)$ by the divisor $(- 1)$ and place the result of $(- 6)$ under the next term in the dividend $(5)$ .

$- 1$	$2$	$9$	$13$	$5$	$- 3$	$- 2$
	\xa0	$- 2$	$- 7$	$- 6$		\xa0
	$2$	$7$	$6$

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

$- 1$	$2$	$9$	$13$	$5$	$- 3$	$- 2$
	\xa0	$- 2$	$- 7$	$- 6$		\xa0
	$2$	$7$	$6$	$- 1$

Multiply the newest entry in the result $(- 1)$ by the divisor $(- 1)$ and place the result of $(1)$ under the next term in the dividend $(- 3)$ .

$- 1$	$2$	$9$	$13$	$5$	$- 3$	$- 2$
	\xa0	$- 2$	$- 7$	$- 6$	$1$	\xa0
	$2$	$7$	$6$	$- 1$

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

$- 1$	$2$	$9$	$13$	$5$	$- 3$	$- 2$
	\xa0	$- 2$	$- 7$	$- 6$	$1$	\xa0
	$2$	$7$	$6$	$- 1$	$- 2$

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 $(- 2)$ .

$- 1$	$2$	$9$	$13$	$5$	$- 3$	$- 2$
	\xa0	$- 2$	$- 7$	$- 6$	$1$	$2$
	$2$	$7$	$6$	$- 1$	$- 2$

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

$- 1$	$2$	$9$	$13$	$5$	$- 3$	$- 2$
	\xa0	$- 2$	$- 7$	$- 6$	$1$	$2$
	$2$	$7$	$6$	$- 1$	$- 2$	$0$

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

$2 x^{4} + 7 x^{3} + 6 x^{2} + - 1 x - 2$

Simplify the quotient polynomial.

$2 x^{4} + 7 x^{3} + 6 x^{2} - x - 2$

Step 7

Solve the equation to find any remaining roots.

Factor the left side of the equation.

Regroup terms.

$2 x^{4} - 2 + 7 x^{3} + 6 x^{2} - x = 0$

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

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

$2 (x^{4}) - 2 + 7 x^{3} + 6 x^{2} - x = 0$

Factor $2$ out of $- 2$ .

$2 (x^{4}) + 2 (- 1) + 7 x^{3} + 6 x^{2} - x = 0$

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

$2 (x^{4} - 1) + 7 x^{3} + 6 x^{2} - x = 0$

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

$2 ({(x^{2})}^{2} - 1) + 7 x^{3} + 6 x^{2} - x = 0$

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

$2 ({(x^{2})}^{2} - 1^{2}) + 7 x^{3} + 6 x^{2} - x = 0$

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = x^{2}$ and $b = 1$ .

$2 ((x^{2} + 1) (x^{2} - 1)) + 7 x^{3} + 6 x^{2} - x = 0$

Factor.

Simplify.

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

$2 ((x^{2} + 1) (x^{2} - 1^{2})) + 7 x^{3} + 6 x^{2} - x = 0$

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 $b = 1$ .

$2 ((x^{2} + 1) ((x + 1) (x - 1))) + 7 x^{3} + 6 x^{2} - x = 0$

Remove unnecessary parentheses.

$2 ((x^{2} + 1) (x + 1) (x - 1)) + 7 x^{3} + 6 x^{2} - x = 0$

Remove unnecessary parentheses.

$2 (x^{2} + 1) (x + 1) (x - 1) + 7 x^{3} + 6 x^{2} - x = 0$

Factor $x$ out of $7 x^{3} + 6 x^{2} - x$ .

Factor $x$ out of $7 x^{3}$ .

$2 (x^{2} + 1) (x + 1) (x - 1) + x (7 x^{2}) + 6 x^{2} - x = 0$

Factor $x$ out of $6 x^{2}$ .

$2 (x^{2} + 1) (x + 1) (x - 1) + x (7 x^{2}) + x (6 x) - x = 0$

Factor $x$ out of $- x$ .

$2 (x^{2} + 1) (x + 1) (x - 1) + x (7 x^{2}) + x (6 x) + x \cdot - 1 = 0$

Factor $x$ out of $x (7 x^{2}) + x (6 x)$ .

$2 (x^{2} + 1) (x + 1) (x - 1) + x (7 x^{2} + 6 x) + x \cdot - 1 = 0$

Factor $x$ out of $x (7 x^{2} + 6 x) + x \cdot - 1$ .

$2 (x^{2} + 1) (x + 1) (x - 1) + x (7 x^{2} + 6 x - 1) = 0$

Factor.

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 $a \cdot c = 7 \cdot - 1 = - 7$ and whose sum is $b = 6$ .

Factor $6$ out of $6 x$ .

$2 (x^{2} + 1) (x + 1) (x - 1) + x (7 x^{2} + 6 (x) - 1) = 0$

Rewrite $6$ as $- 1$ plus $7$

$2 (x^{2} + 1) (x + 1) (x - 1) + x (7 x^{2} + (- 1 + 7) x - 1) = 0$

Apply the distributive property.

$2 (x^{2} + 1) (x + 1) (x - 1) + x (7 x^{2} - 1 x + 7 x - 1) = 0$

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

$2 (x^{2} + 1) (x + 1) (x - 1) + x ((7 x^{2} - 1 x) + 7 x - 1) = 0$

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

$2 (x^{2} + 1) (x + 1) (x - 1) + x (x (7 x - 1) + 1 (7 x - 1)) = 0$

Factor the polynomial by factoring out the greatest common factor, $7 x - 1$ .

$2 (x^{2} + 1) (x + 1) (x - 1) + x ((7 x - 1) (x + 1)) = 0$

Remove unnecessary parentheses.

$2 (x^{2} + 1) (x + 1) (x - 1) + x (7 x - 1) (x + 1) = 0$

Factor $x + 1$ out of $2 (x^{2} + 1) (x + 1) (x - 1) + x (7 x - 1) (x + 1)$ .

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

$(x + 1) (2 (x^{2} + 1) (x - 1)) + x (7 x - 1) (x + 1) = 0$

Factor $x + 1$ out of $x (7 x - 1) (x + 1)$ .

$(x + 1) (2 (x^{2} + 1) (x - 1)) + (x + 1) (x (7 x - 1)) = 0$

Factor $x + 1$ out of $(x + 1) (2 (x^{2} + 1) (x - 1)) + (x + 1) (x (7 x - 1))$ .

$(x + 1) (2 (x^{2} + 1) (x - 1) + x (7 x - 1)) = 0$

Apply the distributive property.

$(x + 1) ((2 x^{2} + 2 \cdot 1) (x - 1) + x (7 x - 1)) = 0$

Multiply $2$ by $1$ .

$(x + 1) ((2 x^{2} + 2) (x - 1) + x (7 x - 1)) = 0$

Expand $(2 x^{2} + 2) (x - 1)$ using the FOIL Method.

Apply the distributive property.

$(x + 1) (2 x^{2} (x - 1) + 2 (x - 1) + x (7 x - 1)) = 0$

Apply the distributive property.

$(x + 1) (2 x^{2} x + 2 x^{2} \cdot - 1 + 2 (x - 1) + x (7 x - 1)) = 0$

Apply the distributive property.

$(x + 1) (2 x^{2} x + 2 x^{2} \cdot - 1 + 2 x + 2 \cdot - 1 + x (7 x - 1)) = 0$

Simplify each term.

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

Move $x$ .

$(x + 1) (2 (x \cdot x^{2}) + 2 x^{2} \cdot - 1 + 2 x + 2 \cdot - 1 + x (7 x - 1)) = 0$

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

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

$(x + 1) (2 (x \cdot x^{2}) + 2 x^{2} \cdot - 1 + 2 x + 2 \cdot - 1 + x (7 x - 1)) = 0$

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

$(x + 1) (2 x^{1 + 2} + 2 x^{2} \cdot - 1 + 2 x + 2 \cdot - 1 + x (7 x - 1)) = 0$

Add $1$ and $2$ .

$(x + 1) (2 x^{3} + 2 x^{2} \cdot - 1 + 2 x + 2 \cdot - 1 + x (7 x - 1)) = 0$

Multiply $- 1$ by $2$ .

$(x + 1) (2 x^{3} - 2 x^{2} + 2 x + 2 \cdot - 1 + x (7 x - 1)) = 0$

Multiply $2$ by $- 1$ .

$(x + 1) (2 x^{3} - 2 x^{2} + 2 x - 2 + x (7 x - 1)) = 0$

Apply the distributive property.

$(x + 1) (2 x^{3} - 2 x^{2} + 2 x - 2 + x (7 x) + x \cdot - 1) = 0$

Rewrite using the commutative property of multiplication.

$(x + 1) (2 x^{3} - 2 x^{2} + 2 x - 2 + 7 x \cdot x + x \cdot - 1) = 0$

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

$(x + 1) (2 x^{3} - 2 x^{2} + 2 x - 2 + 7 x \cdot x - 1 \cdot x) = 0$

Simplify each term.

Multiply $x$ by $x$ by adding the exponents.

Move $x$ .

$(x + 1) (2 x^{3} - 2 x^{2} + 2 x - 2 + 7 (x \cdot x) - 1 \cdot x) = 0$

Multiply $x$ by $x$ .

$(x + 1) (2 x^{3} - 2 x^{2} + 2 x - 2 + 7 x^{2} - 1 \cdot x) = 0$

$(x + 1) (2 x^{3} - 2 x^{2} + 2 x - 2 + 7 x^{2} - 1 x) = 0$

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

$(x + 1) (2 x^{3} - 2 x^{2} + 2 x - 2 + 7 x^{2} - x) = 0$

Add $- 2 x^{2}$ and $7 x^{2}$ .

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

Subtract $x$ from $2 x$ .

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

Factor.

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

Factor $2 x^{3} + 5 x^{2} + x - 2$ 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.

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

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

$\pm 1, \pm 0.5, \pm 2$

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.

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

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

$2 \cdot - 1 + 5 {(- 1)}^{2} - 1 - 2$

Multiply $2$ by $- 1$ .

$- 2 + 5 {(- 1)}^{2} - 1 - 2$

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

$- 2 + 5 \cdot 1 - 1 - 2$

Multiply $5$ by $1$ .

$- 2 + 5 - 1 - 2$

Add $- 2$ and $5$ .

$3 - 1 - 2$

Subtract $1$ from $3$ .

$2 - 2$

Subtract $2$ from $2$ .

$0$

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.

$\frac{2 x^{3} + 5 x^{2} + x - 2}{x + 1}$

Divide $2 x^{3} + 5 x^{2} + x - 2$ 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 $0$ .

$x$

$1$

$2 x^{3}$

$5 x^{2}$

$x$

$2$

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

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

Multiply the new quotient term by the divisor.

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

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

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

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

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

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

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

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

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

Multiply the new quotient term by the divisor.

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

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

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

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

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

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

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

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

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

Multiply the new quotient term by the divisor.

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

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

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

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

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

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

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

Write $2 x^{3} + 5 x^{2} + x - 2$ as a set of factors.

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

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 $a \cdot c = 2 \cdot - 2 = - 4$ and whose sum is $b = 3$ .

Factor $3$ out of $3 x$ .

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

Rewrite $3$ as $- 1$ plus $4$

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

Apply the distributive property.

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

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

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

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

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

Factor the polynomial by factoring out the greatest common factor, $2 x - 1$ .

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

Remove unnecessary parentheses.

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

Remove unnecessary parentheses.

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

Combine exponents.

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

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

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

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

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

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

Add $1$ and $1$ .

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

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

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

$2 x - 1 = 0$

$x + 2 = 0$

Set ${(x + 1)}^{2}$ equal to $0$ and solve for $x$ .

Set ${(x + 1)}^{2}$ equal to $0$ .

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

Solve ${(x + 1)}^{2} = 0$ for $x$ .

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

$x + 1 = 0$

Subtract $1$ from both sides of the equation.

$x = - 1$

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 $x$ .

Add $1$ to 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 $1$ .

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

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

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

Step 8

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

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

Step 9

These are the roots (zeros) of the polynomial $2 x^{5} + 9 x^{4} + 13 x^{3} + 5 x^{2} - 3 x - 2$ .

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

Step 10

Find the Roots/Zeros Using the Rational Roots Test f(x)=2x^5+9x^4+13x^3+5x^2-3x-2

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