Find the Roots/Zeros Using the Rational Roots Test 2x^4+7x^3+x^2-7x-3

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

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.

$p = \pm 1, \pm 3$

$q = \pm 1, \pm 2$

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

$\pm 1, \pm \frac{1}{2}, \pm 3, \pm \frac{3}{2}$

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.

$2 {(1)}^{4} + 7 {(1)}^{3} + {(1)}^{2} - 7 \cdot 1 - 3$

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.

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

Multiply $2$ by $1$ .

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

One to any power is one.

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

Multiply $7$ by $1$ .

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

One to any power is one.

$2 + 7 + 1 - 7 \cdot 1 - 3$

Multiply $- 7$ by $1$ .

$2 + 7 + 1 - 7 - 3$

Simplify by adding and subtracting.

Add $2$ and $7$ .

$9 + 1 - 7 - 3$

Add $9$ and $1$ .

$10 - 7 - 3$

Subtract $7$ from $10$ .

$3 - 3$

Subtract $3$ from $3$ .

$0$

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.

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

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.

$1$	$2$	$7$	$1$	$- 7$	$- 3$
	\xa0				\xa0

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

$1$	$2$	$7$	$1$	$- 7$	$- 3$
	\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 $(7)$ .

$1$	$2$	$7$	$1$	$- 7$	$- 3$
	\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.

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

Multiply the newest entry in the result $(9)$ by the divisor $(1)$ and place the result of $(9)$ under the next term in the dividend $(1)$ .

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

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

$1$	$2$	$7$	$1$	$- 7$	$- 3$
	\xa0	$2$	$9$		\xa0
	$2$	$9$	$10$

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 $(- 7)$ .

$1$	$2$	$7$	$1$	$- 7$	$- 3$
	\xa0	$2$	$9$	$10$	\xa0
	$2$	$9$	$10$

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

$1$	$2$	$7$	$1$	$- 7$	$- 3$
	\xa0	$2$	$9$	$10$	\xa0
	$2$	$9$	$10$	$3$

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

$1$	$2$	$7$	$1$	$- 7$	$- 3$
	\xa0	$2$	$9$	$10$	$3$
	$2$	$9$	$10$	$3$

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

$1$	$2$	$7$	$1$	$- 7$	$- 3$
	\xa0	$2$	$9$	$10$	$3$
	$2$	$9$	$10$	$3$	$0$

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

$2 x^{3} + 9 x^{2} + (10) x + 3$

Simplify the quotient polynomial.

$2 x^{3} + 9 x^{2} + 10 x + 3$

Solve the equation to find any remaining roots.

Factor the left side of the equation.

Factor $2 x^{3} + 9 x^{2} + 10 x + 3$ 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.

$p = \pm 1, \pm 3 q = \pm 1, \pm 2$

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

$\pm 1, \pm 0.5, \pm 3, \pm 1.5$

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.

$2 {(- 1)}^{3} + 9 {(- 1)}^{2} + 10 \cdot - 1 + 3$

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

$2 \cdot - 1 + 9 {(- 1)}^{2} + 10 \cdot - 1 + 3$

Multiply $2$ by $- 1$ .

$- 2 + 9 {(- 1)}^{2} + 10 \cdot - 1 + 3$

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

$- 2 + 9 \cdot 1 + 10 \cdot - 1 + 3$

Multiply $9$ by $1$ .

$- 2 + 9 + 10 \cdot - 1 + 3$

Add $- 2$ and $9$ .

$7 + 10 \cdot - 1 + 3$

Multiply $10$ by $- 1$ .

$7 - 10 + 3$

Subtract $10$ from $7$ .

$- 3 + 3$

Add $- 3$ and $3$ .

$0$

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.

$\frac{2 x^{3} + 9 x^{2} + 10 x + 3}{x + 1}$

Divide $2 x^{3} + 9 x^{2} + 10 x + 3$ 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 $0$ .

$x$

$1$

$2 x^{3}$

$9 x^{2}$

$10 x$

$3$

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

					$2 x^{2}$
	$x$	+	$1$		$2 x^{3}$	+	$9 x^{2}$	+	$10 x$	+	$3$

Multiply the new quotient term by the divisor.

				$2 x^{2}$
$x$	+	$1$		$2 x^{3}$	+	$9 x^{2}$	+	$10 x$	+	$3$
			+	$2 x^{3}$	+	$2 x^{2}$

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

				$2 x^{2}$
$x$	+	$1$		$2 x^{3}$	+	$9 x^{2}$	+	$10 x$	+	$3$
			-	$2 x^{3}$	-	$2 x^{2}$

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

				$2 x^{2}$
$x$	+	$1$		$2 x^{3}$	+	$9 x^{2}$	+	$10 x$	+	$3$
			-	$2 x^{3}$	-	$2 x^{2}$
					+	$7 x^{2}$

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

				$2 x^{2}$
$x$	+	$1$		$2 x^{3}$	+	$9 x^{2}$	+	$10 x$	+	$3$
			-	$2 x^{3}$	-	$2 x^{2}$
					+	$7 x^{2}$	+	$10 x$

Divide the highest order term in the dividend $7 x^{2}$ by the highest order term in divisor $x$ .

				$2 x^{2}$	+	$7 x$
$x$	+	$1$		$2 x^{3}$	+	$9 x^{2}$	+	$10 x$	+	$3$
			-	$2 x^{3}$	-	$2 x^{2}$
					+	$7 x^{2}$	+	$10 x$

Multiply the new quotient term by the divisor.

				$2 x^{2}$	+	$7 x$
$x$	+	$1$		$2 x^{3}$	+	$9 x^{2}$	+	$10 x$	+	$3$
			-	$2 x^{3}$	-	$2 x^{2}$
					+	$7 x^{2}$	+	$10 x$
					+	$7 x^{2}$	+	$7 x$

The expression needs to be subtracted from the dividend, so change all the signs in $7 x^{2} + 7 x$

				$2 x^{2}$	+	$7 x$
$x$	+	$1$		$2 x^{3}$	+	$9 x^{2}$	+	$10 x$	+	$3$
			-	$2 x^{3}$	-	$2 x^{2}$
					+	$7 x^{2}$	+	$10 x$
					-	$7 x^{2}$	-	$7 x$

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

				$2 x^{2}$	+	$7 x$
$x$	+	$1$		$2 x^{3}$	+	$9 x^{2}$	+	$10 x$	+	$3$
			-	$2 x^{3}$	-	$2 x^{2}$
					+	$7 x^{2}$	+	$10 x$
					-	$7 x^{2}$	-	$7 x$
							+	$3 x$

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

				$2 x^{2}$	+	$7 x$
$x$	+	$1$		$2 x^{3}$	+	$9 x^{2}$	+	$10 x$	+	$3$
			-	$2 x^{3}$	-	$2 x^{2}$
					+	$7 x^{2}$	+	$10 x$
					-	$7 x^{2}$	-	$7 x$
							+	$3 x$	+	$3$

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

				$2 x^{2}$	+	$7 x$	+	$3$
$x$	+	$1$		$2 x^{3}$	+	$9 x^{2}$	+	$10 x$	+	$3$
			-	$2 x^{3}$	-	$2 x^{2}$
					+	$7 x^{2}$	+	$10 x$
					-	$7 x^{2}$	-	$7 x$
							+	$3 x$	+	$3$

Multiply the new quotient term by the divisor.

				$2 x^{2}$	+	$7 x$	+	$3$
$x$	+	$1$		$2 x^{3}$	+	$9 x^{2}$	+	$10 x$	+	$3$
			-	$2 x^{3}$	-	$2 x^{2}$
					+	$7 x^{2}$	+	$10 x$
					-	$7 x^{2}$	-	$7 x$
							+	$3 x$	+	$3$
							+	$3 x$	+	$3$

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

				$2 x^{2}$	+	$7 x$	+	$3$
$x$	+	$1$		$2 x^{3}$	+	$9 x^{2}$	+	$10 x$	+	$3$
			-	$2 x^{3}$	-	$2 x^{2}$
					+	$7 x^{2}$	+	$10 x$
					-	$7 x^{2}$	-	$7 x$
							+	$3 x$	+	$3$
							-	$3 x$	-	$3$

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

				$2 x^{2}$	+	$7 x$	+	$3$
$x$	+	$1$		$2 x^{3}$	+	$9 x^{2}$	+	$10 x$	+	$3$
			-	$2 x^{3}$	-	$2 x^{2}$
					+	$7 x^{2}$	+	$10 x$
					-	$7 x^{2}$	-	$7 x$
							+	$3 x$	+	$3$
							-	$3 x$	-	$3$
										$0$

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

$2 x^{2} + 7 x + 3$

Write $2 x^{3} + 9 x^{2} + 10 x + 3$ as a set of factors.

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

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 $a \cdot c = 2 \cdot 3 = 6$ and whose sum is $b = 7$ .

Factor $7$ out of $7 x$ .

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

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

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

Apply the distributive property.

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

Multiply $x$ by $1$ .

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

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

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

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

$(x + 1) (x (2 x + 1) + 3 (2 x + 1)) = 0$

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

$(x + 1) ((2 x + 1) (x + 3)) = 0$

Remove unnecessary parentheses.

$(x + 1) (2 x + 1) (x + 3) = 0$

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

$x + 1 = 0$

$2 x + 1 = 0$

$x + 3 = 0$

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.

$x = - 1$

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

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

$x + 3 = 0$

Subtract $3$ from both sides of the equation.

$x = - 3$

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

$x = - 1, - \frac{1}{2}, - 3$

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

$(x - 1) (x + 1) (x + \frac{1}{2}) (x + 3)$

These are the roots (zeros) of the polynomial $2 x^{4} + 7 x^{3} + x^{2} - 7 x - 3$ .

$x = 1, - 1, - \frac{1}{2}, - 3$

Find the Roots/Zeros Using the Rational Roots Test 2x^4+7x^3+x^2-7x-3

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