Find the Roots/Zeros Using the Rational Roots Test h(x)=2x^3-7x^2-17x+10

$h (x) = 2 x^{3} - 7 x^{2} - 17 x + 10$

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.

$p = \pm 1, \pm 2, \pm 5, \pm 10$

$q = \pm 1, \pm 2$

Step 2

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

$\pm 1, \pm \frac{1}{2}, \pm 2, \pm 5, \pm \frac{5}{2}, \pm 10$

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.

$2 {(\frac{1}{2})}^{3} - 7 {(\frac{1}{2})}^{2} - 17 (\frac{1}{2}) + 10$

Step 4

Simplify the expression. In this case, the expression is equal to $0$ so $x = \frac{1}{2}$ is a root of the polynomial.

Simplify each term.

Apply the product rule to $\frac{1}{2}$ .

$2 \frac{1^{3}}{2^{3}} - 7 {(\frac{1}{2})}^{2} - 17 (\frac{1}{2}) + 10$

One to any power is one.

$2 \frac{1}{2^{3}} - 7 {(\frac{1}{2})}^{2} - 17 (\frac{1}{2}) + 10$

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

$2 (\frac{1}{8}) - 7 {(\frac{1}{2})}^{2} - 17 (\frac{1}{2}) + 10$

Cancel the common factor of $2$ .

Factor $2$ out of $8$ .

$2 \frac{1}{2 (4)} - 7 {(\frac{1}{2})}^{2} - 17 (\frac{1}{2}) + 10$

Cancel the common factor.

$2 \frac{1}{2 \cdot 4} - 7 {(\frac{1}{2})}^{2} - 17 (\frac{1}{2}) + 10$

Rewrite the expression.

$\frac{1}{4} - 7 {(\frac{1}{2})}^{2} - 17 (\frac{1}{2}) + 10$

Apply the product rule to $\frac{1}{2}$ .

$\frac{1}{4} - 7 \frac{1^{2}}{2^{2}} - 17 (\frac{1}{2}) + 10$

One to any power is one.

$\frac{1}{4} - 7 \frac{1}{2^{2}} - 17 (\frac{1}{2}) + 10$

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

$\frac{1}{4} - 7 (\frac{1}{4}) - 17 (\frac{1}{2}) + 10$

Combine $- 7$ and $\frac{1}{4}$ .

$\frac{1}{4} + \frac{- 7}{4} - 17 (\frac{1}{2}) + 10$

Move the negative in front of the fraction.

$\frac{1}{4} - \frac{7}{4} - 17 (\frac{1}{2}) + 10$

Combine $- 17$ and $\frac{1}{2}$ .

$\frac{1}{4} - \frac{7}{4} + \frac{- 17}{2} + 10$

Move the negative in front of the fraction.

$\frac{1}{4} - \frac{7}{4} - \frac{17}{2} + 10$

Combine fractions.

Combine the numerators over the common denominator.

$10 + \frac{1 - 7}{4} + \frac{- 17}{2}$

Subtract $7$ from $1$ .

$10 + \frac{- 6}{4} + \frac{- 17}{2}$

Find the common denominator.

Write $10$ as a fraction with denominator $1$ .

$\frac{10}{1} + \frac{- 6}{4} + \frac{- 17}{2}$

Multiply $\frac{10}{1}$ by $\frac{4}{4}$ .

$\frac{10}{1} \cdot \frac{4}{4} + \frac{- 6}{4} + \frac{- 17}{2}$

Multiply $\frac{10}{1}$ by $\frac{4}{4}$ .

$\frac{10 \cdot 4}{4} + \frac{- 6}{4} + \frac{- 17}{2}$

Multiply $\frac{- 17}{2}$ by $\frac{2}{2}$ .

$\frac{10 \cdot 4}{4} + \frac{- 6}{4} + \frac{- 17}{2} \cdot \frac{2}{2}$

Multiply $\frac{- 17}{2}$ by $\frac{2}{2}$ .

$\frac{10 \cdot 4}{4} + \frac{- 6}{4} + \frac{- 17 \cdot 2}{2 \cdot 2}$

Multiply $2$ by $2$ .

$\frac{10 \cdot 4}{4} + \frac{- 6}{4} + \frac{- 17 \cdot 2}{4}$

Combine the numerators over the common denominator.

$\frac{10 \cdot 4 - 6 - 17 \cdot 2}{4}$

Simplify each term.

Multiply $10$ by $4$ .

$\frac{40 - 6 - 17 \cdot 2}{4}$

Multiply $- 17$ by $2$ .

$\frac{40 - 6 - 34}{4}$

Simplify the expression.

Subtract $6$ from $40$ .

$\frac{34 - 34}{4}$

Subtract $34$ from $34$ .

$\frac{0}{4}$

Divide $0$ by $4$ .

$0$

Step 5

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

$\frac{2 x^{3} - 7 x^{2} - 17 x + 10}{x - \frac{1}{2}}$

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.

$\frac{1}{2}$	$2$	$- 7$	$- 17$	$10$
	\xa0			\xa0

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

$\frac{1}{2}$	$2$	$- 7$	$- 17$	$10$
	\xa0			\xa0
	$2$

Multiply the newest entry in the result $(2)$ by the divisor $(\frac{1}{2})$ and place the result of $(1)$ under the next term in the dividend $(- 7)$ .

$\frac{1}{2}$	$2$	$- 7$	$- 17$	$10$
	\xa0	$1$		\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.

$\frac{1}{2}$	$2$	$- 7$	$- 17$	$10$
	\xa0	$1$		\xa0
	$2$	$- 6$

Multiply the newest entry in the result $(- 6)$ by the divisor $(\frac{1}{2})$ and place the result of $(- 3)$ under the next term in the dividend $(- 17)$ .

$\frac{1}{2}$	$2$	$- 7$	$- 17$	$10$
	\xa0	$1$	$- 3$	\xa0
	$2$	$- 6$

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

$\frac{1}{2}$	$2$	$- 7$	$- 17$	$10$
	\xa0	$1$	$- 3$	\xa0
	$2$	$- 6$	$- 20$

Multiply the newest entry in the result $(- 20)$ by the divisor $(\frac{1}{2})$ and place the result of $(- 10)$ under the next term in the dividend $(10)$ .

$\frac{1}{2}$	$2$	$- 7$	$- 17$	$10$
	\xa0	$1$	$- 3$	$- 10$
	$2$	$- 6$	$- 20$

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

$\frac{1}{2}$	$2$	$- 7$	$- 17$	$10$
	\xa0	$1$	$- 3$	$- 10$
	$2$	$- 6$	$- 20$	$0$

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

$2 x^{2} + - 6 x - 20$

Simplify the quotient polynomial.

$2 x^{2} - 6 x - 20$

Step 7

Factor $2$ out of $2 x^{2} - 6 x - 20$ .

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

$(x - \frac{1}{2}) (2 (x^{2}) - 6 x - 20)$

Factor $2$ out of $- 6 x$ .

$(x - \frac{1}{2}) (2 (x^{2}) + 2 (- 3 x) - 20)$

Factor $2$ out of $- 20$ .

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

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

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

Factor $2$ out of $2 (x^{2} - 3 x) + 2 \cdot - 10$ .

$(x - \frac{1}{2}) (2 (x^{2} - 3 x - 10))$

Step 8

Factor.

Factor $x^{2} - 3 x - 10$ 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 $- 10$ and whose sum is $- 3$ .

$- 5, 2$

Write the factored form using these integers.

$(x - \frac{1}{2}) (2 ((x - 5) (x + 2)))$

Remove unnecessary parentheses.

$(x - \frac{1}{2}) (2 (x - 5) (x + 2))$

Step 9

Factor the left side of the equation.

Factor $2 x^{3} - 7 x^{2} - 17 x + 10$ 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.

$p = \pm 1, \pm 10, \pm 2, \pm 5 q = \pm 1, \pm 2$

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

$\pm 1, \pm 0.5, \pm 10, \pm 5, \pm 2, \pm 2.5$

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.

$2 \cdot {0.5}^{3} - 7 \cdot {0.5}^{2} - 17 \cdot 0.5 + 10$

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

$2 \cdot 0.125 - 7 \cdot {0.5}^{2} - 17 \cdot 0.5 + 10$

Multiply $2$ by $0.125$ .

$0.25 - 7 \cdot {0.5}^{2} - 17 \cdot 0.5 + 10$

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

$0.25 - 7 \cdot 0.25 - 17 \cdot 0.5 + 10$

Multiply $- 7$ by $0.25$ .

$0.25 - 1.75 - 17 \cdot 0.5 + 10$

Subtract $1.75$ from $0.25$ .

$- 1.5 - 17 \cdot 0.5 + 10$

Multiply $- 17$ by $0.5$ .

$- 1.5 - 8.5 + 10$

Subtract $8.5$ from $- 1.5$ .

$- 10 + 10$

Add $- 10$ and $10$ .

$0$

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.

$\frac{2 x^{3} - 7 x^{2} - 17 x + 10}{2 x - 1}$

Divide $2 x^{3} - 7 x^{2} - 17 x + 10$ 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 $0$ .

$2 x$

$1$

$2 x^{3}$

$7 x^{2}$

$17 x$

$10$

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

					$x^{2}$
	$2 x$	-	$1$		$2 x^{3}$	-	$7 x^{2}$	-	$17 x$	+	$10$

Multiply the new quotient term by the divisor.

				$x^{2}$
$2 x$	-	$1$		$2 x^{3}$	-	$7 x^{2}$	-	$17 x$	+	$10$
			+	$2 x^{3}$	-	$x^{2}$

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

				$x^{2}$
$2 x$	-	$1$		$2 x^{3}$	-	$7 x^{2}$	-	$17 x$	+	$10$
			-	$2 x^{3}$	+	$x^{2}$

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

				$x^{2}$
$2 x$	-	$1$		$2 x^{3}$	-	$7 x^{2}$	-	$17 x$	+	$10$
			-	$2 x^{3}$	+	$x^{2}$
					-	$6 x^{2}$

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

				$x^{2}$
$2 x$	-	$1$		$2 x^{3}$	-	$7 x^{2}$	-	$17 x$	+	$10$
			-	$2 x^{3}$	+	$x^{2}$
					-	$6 x^{2}$	-	$17 x$

Divide the highest order term in the dividend $- 6 x^{2}$ by the highest order term in divisor $2 x$ .

				$x^{2}$	-	$3 x$
$2 x$	-	$1$		$2 x^{3}$	-	$7 x^{2}$	-	$17 x$	+	$10$
			-	$2 x^{3}$	+	$x^{2}$
					-	$6 x^{2}$	-	$17 x$

Multiply the new quotient term by the divisor.

				$x^{2}$	-	$3 x$
$2 x$	-	$1$		$2 x^{3}$	-	$7 x^{2}$	-	$17 x$	+	$10$
			-	$2 x^{3}$	+	$x^{2}$
					-	$6 x^{2}$	-	$17 x$
					-	$6 x^{2}$	+	$3 x$

The expression needs to be subtracted from the dividend, so change all the signs in $- 6 x^{2} + 3 x$

				$x^{2}$	-	$3 x$
$2 x$	-	$1$		$2 x^{3}$	-	$7 x^{2}$	-	$17 x$	+	$10$
			-	$2 x^{3}$	+	$x^{2}$
					-	$6 x^{2}$	-	$17 x$
					+	$6 x^{2}$	-	$3 x$

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

				$x^{2}$	-	$3 x$
$2 x$	-	$1$		$2 x^{3}$	-	$7 x^{2}$	-	$17 x$	+	$10$
			-	$2 x^{3}$	+	$x^{2}$
					-	$6 x^{2}$	-	$17 x$
					+	$6 x^{2}$	-	$3 x$
							-	$20 x$

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

				$x^{2}$	-	$3 x$
$2 x$	-	$1$		$2 x^{3}$	-	$7 x^{2}$	-	$17 x$	+	$10$
			-	$2 x^{3}$	+	$x^{2}$
					-	$6 x^{2}$	-	$17 x$
					+	$6 x^{2}$	-	$3 x$
							-	$20 x$	+	$10$

Divide the highest order term in the dividend $- 20 x$ by the highest order term in divisor $2 x$ .

				$x^{2}$	-	$3 x$	-	$10$
$2 x$	-	$1$		$2 x^{3}$	-	$7 x^{2}$	-	$17 x$	+	$10$
			-	$2 x^{3}$	+	$x^{2}$
					-	$6 x^{2}$	-	$17 x$
					+	$6 x^{2}$	-	$3 x$
							-	$20 x$	+	$10$

Multiply the new quotient term by the divisor.

				$x^{2}$	-	$3 x$	-	$10$
$2 x$	-	$1$		$2 x^{3}$	-	$7 x^{2}$	-	$17 x$	+	$10$
			-	$2 x^{3}$	+	$x^{2}$
					-	$6 x^{2}$	-	$17 x$
					+	$6 x^{2}$	-	$3 x$
							-	$20 x$	+	$10$
							-	$20 x$	+	$10$

The expression needs to be subtracted from the dividend, so change all the signs in $- 20 x + 10$

				$x^{2}$	-	$3 x$	-	$10$
$2 x$	-	$1$		$2 x^{3}$	-	$7 x^{2}$	-	$17 x$	+	$10$
			-	$2 x^{3}$	+	$x^{2}$
					-	$6 x^{2}$	-	$17 x$
					+	$6 x^{2}$	-	$3 x$
							-	$20 x$	+	$10$
							+	$20 x$	-	$10$

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

				$x^{2}$	-	$3 x$	-	$10$
$2 x$	-	$1$		$2 x^{3}$	-	$7 x^{2}$	-	$17 x$	+	$10$
			-	$2 x^{3}$	+	$x^{2}$
					-	$6 x^{2}$	-	$17 x$
					+	$6 x^{2}$	-	$3 x$
							-	$20 x$	+	$10$
							+	$20 x$	-	$10$
										$0$

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

$x^{2} - 3 x - 10$

Write $2 x^{3} - 7 x^{2} - 17 x + 10$ as a set of factors.

$(2 x - 1) (x^{2} - 3 x - 10) = 0$

Factor $x^{2} - 3 x - 10$ 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 $- 10$ and whose sum is $- 3$ .

$- 5, 2$

Write the factored form using these integers.

$(2 x - 1) ((x - 5) (x + 2)) = 0$

Remove unnecessary parentheses.

$(2 x - 1) (x - 5) (x + 2) = 0$

Step 10

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

$2 x - 1 = 0$

$x - 5 = 0$

$x + 2 = 0$

Step 11

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 $1$ .

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

Step 12

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

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

$x - 5 = 0$

Add $5$ to both sides of the equation.

$x = 5$

Step 13

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 14

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

$x = \frac{1}{2}, 5, - 2$

Step 15

Find the Roots/Zeros Using the Rational Roots Test h(x)=2x^3-7x^2-17x+10

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