Find the Roots/Zeros Using the Rational Roots Test f(x)=2x^3-9x^2-6x+40

$f (x) = 2 x^{3} - 9 x^{2} - 6 x + 40$

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.

$p = \pm 1, \pm 2, \pm 4, \pm 5, \pm 8, \pm 10, \pm 20, \pm 40$

$q = \pm 1, \pm 2$

Step 2

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

$\pm 1, \pm \frac{1}{2}, \pm 2, \pm 4, \pm 5, \pm \frac{5}{2}, \pm 8, \pm 10, \pm 20, \pm 40$

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.

$2 {(- 2)}^{3} - 9 {(- 2)}^{2} - 6 \cdot - 2 + 40$

Step 4

Simplify the expression. In this case, the expression is equal to $0$ so $x = - 2$ is a root of the polynomial.

Simplify each term.

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

$2 \cdot - 8 - 9 {(- 2)}^{2} - 6 \cdot - 2 + 40$

Multiply $2$ by $- 8$ .

$- 16 - 9 {(- 2)}^{2} - 6 \cdot - 2 + 40$

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

$- 16 - 9 \cdot 4 - 6 \cdot - 2 + 40$

Multiply $- 9$ by $4$ .

$- 16 - 36 - 6 \cdot - 2 + 40$

Multiply $- 6$ by $- 2$ .

$- 16 - 36 + 12 + 40$

Simplify by adding and subtracting.

Subtract $36$ from $- 16$ .

$- 52 + 12 + 40$

Add $- 52$ and $12$ .

$- 40 + 40$

Add $- 40$ and $40$ .

$0$

Step 5

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

$\frac{2 x^{3} - 9 x^{2} - 6 x + 40}{x + 2}$

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.

$- 2$	$2$	$- 9$	$- 6$	$40$
	\xa0			\xa0

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

$- 2$	$2$	$- 9$	$- 6$	$40$
	\xa0			\xa0
	$2$

Multiply the newest entry in the result $(2)$ by the divisor $(- 2)$ and place the result of $(- 4)$ under the next term in the dividend $(- 9)$ .

$- 2$	$2$	$- 9$	$- 6$	$40$
	\xa0	$- 4$		\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.

$- 2$	$2$	$- 9$	$- 6$	$40$
	\xa0	$- 4$		\xa0
	$2$	$- 13$

Multiply the newest entry in the result $(- 13)$ by the divisor $(- 2)$ and place the result of $(26)$ under the next term in the dividend $(- 6)$ .

$- 2$	$2$	$- 9$	$- 6$	$40$
	\xa0	$- 4$	$26$	\xa0
	$2$	$- 13$

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

$- 2$	$2$	$- 9$	$- 6$	$40$
	\xa0	$- 4$	$26$	\xa0
	$2$	$- 13$	$20$

Multiply the newest entry in the result $(20)$ by the divisor $(- 2)$ and place the result of $(- 40)$ under the next term in the dividend $(40)$ .

$- 2$	$2$	$- 9$	$- 6$	$40$
	\xa0	$- 4$	$26$	$- 40$
	$2$	$- 13$	$20$

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

$- 2$	$2$	$- 9$	$- 6$	$40$
	\xa0	$- 4$	$26$	$- 40$
	$2$	$- 13$	$20$	$0$

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

$2 x^{2} + - 13 x + 20$

Simplify the quotient polynomial.

$2 x^{2} - 13 x + 20$

Step 7

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 $a \cdot c = 2 \cdot 20 = 40$ and whose sum is $b = - 13$ .

Factor $- 13$ out of $- 13 x$ .

$(x + 2) + 2 x^{2} - 13 x + 20$

Rewrite $- 13$ as $- 5$ plus $- 8$

$(x + 2) + 2 x^{2} + (- 5 - 8) x + 20$

Apply the distributive property.

$(x + 2) + 2 x^{2} - 5 x - 8 x + 20$

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

$(x + 2) (2 x^{2} - 5 x) - 8 x + 20$

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

$(x + 2) + x (2 x - 5) - 4 (2 x - 5)$

Factor the polynomial by factoring out the greatest common factor, $2 x - 5$ .

$(x + 2) + (2 x - 5) (x - 4)$

$(x + 2) (2 x - 5) (x - 4)$

Step 8

Factor the left side of the equation.

Factor $2 x^{3} - 9 x^{2} - 6 x + 40$ 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.

$p = \pm 1, \pm 40, \pm 2, \pm 20, \pm 4, \pm 10, \pm 5, \pm 8 q = \pm 1, \pm 2$

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

$\pm 1, \pm 0.5, \pm 40, \pm 20, \pm 2, \pm 10, \pm 4, \pm 5, \pm 2.5, \pm 8$

Substitute $- 2$ and simplify the expression. In this case, the expression is equal to $0$ so $- 2$ is a root of the polynomial.

Substitute $- 2$ into the polynomial.

$2 {(- 2)}^{3} - 9 {(- 2)}^{2} - 6 \cdot - 2 + 40$

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

$2 \cdot - 8 - 9 {(- 2)}^{2} - 6 \cdot - 2 + 40$

Multiply $2$ by $- 8$ .

$- 16 - 9 {(- 2)}^{2} - 6 \cdot - 2 + 40$

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

$- 16 - 9 \cdot 4 - 6 \cdot - 2 + 40$

Multiply $- 9$ by $4$ .

$- 16 - 36 - 6 \cdot - 2 + 40$

Subtract $36$ from $- 16$ .

$- 52 - 6 \cdot - 2 + 40$

Multiply $- 6$ by $- 2$ .

$- 52 + 12 + 40$

Add $- 52$ and $12$ .

$- 40 + 40$

Add $- 40$ and $40$ .

$0$

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

$\frac{2 x^{3} - 9 x^{2} - 6 x + 40}{x + 2}$

Divide $2 x^{3} - 9 x^{2} - 6 x + 40$ by $x + 2$ .

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

$x$

$2$

$2 x^{3}$

$9 x^{2}$

$6 x$

$40$

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

					$2 x^{2}$
	$x$	+	$2$		$2 x^{3}$	-	$9 x^{2}$	-	$6 x$	+	$40$

Multiply the new quotient term by the divisor.

				$2 x^{2}$
$x$	+	$2$		$2 x^{3}$	-	$9 x^{2}$	-	$6 x$	+	$40$
			+	$2 x^{3}$	+	$4 x^{2}$

The expression needs to be subtracted from the dividend, so change all the signs in $2 x^{3} + 4 x^{2}$

				$2 x^{2}$
$x$	+	$2$		$2 x^{3}$	-	$9 x^{2}$	-	$6 x$	+	$40$
			-	$2 x^{3}$	-	$4 x^{2}$

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

				$2 x^{2}$
$x$	+	$2$		$2 x^{3}$	-	$9 x^{2}$	-	$6 x$	+	$40$
			-	$2 x^{3}$	-	$4 x^{2}$
					-	$13 x^{2}$

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

				$2 x^{2}$
$x$	+	$2$		$2 x^{3}$	-	$9 x^{2}$	-	$6 x$	+	$40$
			-	$2 x^{3}$	-	$4 x^{2}$
					-	$13 x^{2}$	-	$6 x$

Divide the highest order term in the dividend $- 13 x^{2}$ by the highest order term in divisor $x$ .

				$2 x^{2}$	-	$13 x$
$x$	+	$2$		$2 x^{3}$	-	$9 x^{2}$	-	$6 x$	+	$40$
			-	$2 x^{3}$	-	$4 x^{2}$
					-	$13 x^{2}$	-	$6 x$

Multiply the new quotient term by the divisor.

				$2 x^{2}$	-	$13 x$
$x$	+	$2$		$2 x^{3}$	-	$9 x^{2}$	-	$6 x$	+	$40$
			-	$2 x^{3}$	-	$4 x^{2}$
					-	$13 x^{2}$	-	$6 x$
					-	$13 x^{2}$	-	$26 x$

The expression needs to be subtracted from the dividend, so change all the signs in $- 13 x^{2} - 26 x$

				$2 x^{2}$	-	$13 x$
$x$	+	$2$		$2 x^{3}$	-	$9 x^{2}$	-	$6 x$	+	$40$
			-	$2 x^{3}$	-	$4 x^{2}$
					-	$13 x^{2}$	-	$6 x$
					+	$13 x^{2}$	+	$26 x$

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

				$2 x^{2}$	-	$13 x$
$x$	+	$2$		$2 x^{3}$	-	$9 x^{2}$	-	$6 x$	+	$40$
			-	$2 x^{3}$	-	$4 x^{2}$
					-	$13 x^{2}$	-	$6 x$
					+	$13 x^{2}$	+	$26 x$
							+	$20 x$

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

				$2 x^{2}$	-	$13 x$
$x$	+	$2$		$2 x^{3}$	-	$9 x^{2}$	-	$6 x$	+	$40$
			-	$2 x^{3}$	-	$4 x^{2}$
					-	$13 x^{2}$	-	$6 x$
					+	$13 x^{2}$	+	$26 x$
							+	$20 x$	+	$40$

Divide the highest order term in the dividend $20 x$ by the highest order term in divisor $x$ .

				$2 x^{2}$	-	$13 x$	+	$20$
$x$	+	$2$		$2 x^{3}$	-	$9 x^{2}$	-	$6 x$	+	$40$
			-	$2 x^{3}$	-	$4 x^{2}$
					-	$13 x^{2}$	-	$6 x$
					+	$13 x^{2}$	+	$26 x$
							+	$20 x$	+	$40$

Multiply the new quotient term by the divisor.

				$2 x^{2}$	-	$13 x$	+	$20$
$x$	+	$2$		$2 x^{3}$	-	$9 x^{2}$	-	$6 x$	+	$40$
			-	$2 x^{3}$	-	$4 x^{2}$
					-	$13 x^{2}$	-	$6 x$
					+	$13 x^{2}$	+	$26 x$
							+	$20 x$	+	$40$
							+	$20 x$	+	$40$

The expression needs to be subtracted from the dividend, so change all the signs in $20 x + 40$

				$2 x^{2}$	-	$13 x$	+	$20$
$x$	+	$2$		$2 x^{3}$	-	$9 x^{2}$	-	$6 x$	+	$40$
			-	$2 x^{3}$	-	$4 x^{2}$
					-	$13 x^{2}$	-	$6 x$
					+	$13 x^{2}$	+	$26 x$
							+	$20 x$	+	$40$
							-	$20 x$	-	$40$

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

				$2 x^{2}$	-	$13 x$	+	$20$
$x$	+	$2$		$2 x^{3}$	-	$9 x^{2}$	-	$6 x$	+	$40$
			-	$2 x^{3}$	-	$4 x^{2}$
					-	$13 x^{2}$	-	$6 x$
					+	$13 x^{2}$	+	$26 x$
							+	$20 x$	+	$40$
							-	$20 x$	-	$40$
										$0$

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

$2 x^{2} - 13 x + 20$

Write $2 x^{3} - 9 x^{2} - 6 x + 40$ as a set of factors.

$(x + 2) (2 x^{2} - 13 x + 20) = 0$

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 $a \cdot c = 2 \cdot 20 = 40$ and whose sum is $b = - 13$ .

Factor $- 13$ out of $- 13 x$ .

$(x + 2) (2 x^{2} - 13 x + 20) = 0$

Rewrite $- 13$ as $- 5$ plus $- 8$

$(x + 2) (2 x^{2} + (- 5 - 8) x + 20) = 0$

Apply the distributive property.

$(x + 2) (2 x^{2} - 5 x - 8 x + 20) = 0$

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

$(x + 2) ((2 x^{2} - 5 x) - 8 x + 20) = 0$

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

$(x + 2) (x (2 x - 5) - 4 (2 x - 5)) = 0$

Factor the polynomial by factoring out the greatest common factor, $2 x - 5$ .

$(x + 2) ((2 x - 5) (x - 4)) = 0$

Remove unnecessary parentheses.

$(x + 2) (2 x - 5) (x - 4) = 0$

Step 9

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

$x + 2 = 0$

$2 x - 5 = 0$

$x - 4 = 0$

Step 10

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.

$x = - 2$

Step 11

Set $2 x - 5$ equal to $0$ and solve for $x$ .

Set $2 x - 5$ equal to $0$ .

$2 x - 5 = 0$

Solve $2 x - 5 = 0$ for $x$ .

Add $5$ to both sides of the equation.

$2 x = 5$

Divide each term in $2 x = 5$ by $2$ and simplify.

Divide each term in $2 x = 5$ by $2$ .

$\frac{2 x}{2} = \frac{5}{2}$

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

$\frac{2 x}{2} = \frac{5}{2}$

Divide $x$ by $1$ .

$x = \frac{5}{2}$

Step 12

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.

$x = 4$

Step 13

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

$x = - 2, \frac{5}{2}, 4$

Step 14

Find the Roots/Zeros Using the Rational Roots Test f(x)=2x^3-9x^2-6x+40

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