Find the Roots/Zeros Using the Rational Roots Test 2x^4-7x^3-17x^2+58x-24

$2 x^{4} - 7 x^{3} - 17 x^{2} + 58 x - 24$

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.

$p = \pm 1, \pm 2, \pm 3, \pm 4, \pm 6, \pm 8, \pm 12, \pm 24$

$q = \pm 1, \pm 2$

Step 2

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

$\pm 1, \pm \frac{1}{2}, \pm 2, \pm 3, \pm \frac{3}{2}, \pm 4, \pm 6, \pm 8, \pm 12, \pm 24$

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.

$2 {(\frac{1}{2})}^{4} - 7 {(\frac{1}{2})}^{3} - 17 {(\frac{1}{2})}^{2} + 58 (\frac{1}{2}) - 24$

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 $\frac{1}{2}$ .

$2 \frac{1^{4}}{2^{4}} - 7 {(\frac{1}{2})}^{3} - 17 {(\frac{1}{2})}^{2} + 58 (\frac{1}{2}) - 24$

One to any power is one.

$2 \frac{1}{2^{4}} - 7 {(\frac{1}{2})}^{3} - 17 {(\frac{1}{2})}^{2} + 58 (\frac{1}{2}) - 24$

Raise $2$ to the power of $4$ .

$2 (\frac{1}{16}) - 7 {(\frac{1}{2})}^{3} - 17 {(\frac{1}{2})}^{2} + 58 (\frac{1}{2}) - 24$

Cancel the common factor of $2$ .

Factor $2$ out of $16$ .

$2 \frac{1}{2 (8)} - 7 {(\frac{1}{2})}^{3} - 17 {(\frac{1}{2})}^{2} + 58 (\frac{1}{2}) - 24$

Cancel the common factor.

$2 \frac{1}{2 \cdot 8} - 7 {(\frac{1}{2})}^{3} - 17 {(\frac{1}{2})}^{2} + 58 (\frac{1}{2}) - 24$

Rewrite the expression.

$\frac{1}{8} - 7 {(\frac{1}{2})}^{3} - 17 {(\frac{1}{2})}^{2} + 58 (\frac{1}{2}) - 24$

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

$\frac{1}{8} - 7 \frac{1^{3}}{2^{3}} - 17 {(\frac{1}{2})}^{2} + 58 (\frac{1}{2}) - 24$

One to any power is one.

$\frac{1}{8} - 7 \frac{1}{2^{3}} - 17 {(\frac{1}{2})}^{2} + 58 (\frac{1}{2}) - 24$

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

$\frac{1}{8} - 7 (\frac{1}{8}) - 17 {(\frac{1}{2})}^{2} + 58 (\frac{1}{2}) - 24$

Combine $- 7$ and $\frac{1}{8}$ .

$\frac{1}{8} + \frac{- 7}{8} - 17 {(\frac{1}{2})}^{2} + 58 (\frac{1}{2}) - 24$

Move the negative in front of the fraction.

$\frac{1}{8} - \frac{7}{8} - 17 {(\frac{1}{2})}^{2} + 58 (\frac{1}{2}) - 24$

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

$\frac{1}{8} - \frac{7}{8} - 17 \frac{1^{2}}{2^{2}} + 58 (\frac{1}{2}) - 24$

One to any power is one.

$\frac{1}{8} - \frac{7}{8} - 17 \frac{1}{2^{2}} + 58 (\frac{1}{2}) - 24$

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

$\frac{1}{8} - \frac{7}{8} - 17 (\frac{1}{4}) + 58 (\frac{1}{2}) - 24$

Combine $- 17$ and $\frac{1}{4}$ .

$\frac{1}{8} - \frac{7}{8} + \frac{- 17}{4} + 58 (\frac{1}{2}) - 24$

Move the negative in front of the fraction.

$\frac{1}{8} - \frac{7}{8} - \frac{17}{4} + 58 (\frac{1}{2}) - 24$

Cancel the common factor of $2$ .

Factor $2$ out of $58$ .

$\frac{1}{8} - \frac{7}{8} - \frac{17}{4} + 2 (29) \frac{1}{2} - 24$

Cancel the common factor.

$\frac{1}{8} - \frac{7}{8} - \frac{17}{4} + 2 \cdot 29 \frac{1}{2} - 24$

Rewrite the expression.

$\frac{1}{8} - \frac{7}{8} - \frac{17}{4} + 29 - 24$

Combine fractions.

Combine the numerators over the common denominator.

$29 - 24 + \frac{1 - 7}{8} + \frac{- 17}{4}$

Subtract $7$ from $1$ .

$29 - 24 + \frac{- 6}{8} + \frac{- 17}{4}$

Find the common denominator.

Write $29$ as a fraction with denominator $1$ .

$\frac{29}{1} - 24 + \frac{- 6}{8} + \frac{- 17}{4}$

Multiply $\frac{29}{1}$ by $\frac{8}{8}$ .

$\frac{29}{1} \cdot \frac{8}{8} - 24 + \frac{- 6}{8} + \frac{- 17}{4}$

Multiply $\frac{29}{1}$ by $\frac{8}{8}$ .

$\frac{29 \cdot 8}{8} - 24 + \frac{- 6}{8} + \frac{- 17}{4}$

Write $- 24$ as a fraction with denominator $1$ .

$\frac{29 \cdot 8}{8} + \frac{- 24}{1} + \frac{- 6}{8} + \frac{- 17}{4}$

Multiply $\frac{- 24}{1}$ by $\frac{8}{8}$ .

$\frac{29 \cdot 8}{8} + \frac{- 24}{1} \cdot \frac{8}{8} + \frac{- 6}{8} + \frac{- 17}{4}$

Multiply $\frac{- 24}{1}$ by $\frac{8}{8}$ .

$\frac{29 \cdot 8}{8} + \frac{- 24 \cdot 8}{8} + \frac{- 6}{8} + \frac{- 17}{4}$

Multiply $\frac{- 17}{4}$ by $\frac{2}{2}$ .

$\frac{29 \cdot 8}{8} + \frac{- 24 \cdot 8}{8} + \frac{- 6}{8} + \frac{- 17}{4} \cdot \frac{2}{2}$

Multiply $\frac{- 17}{4}$ by $\frac{2}{2}$ .

$\frac{29 \cdot 8}{8} + \frac{- 24 \cdot 8}{8} + \frac{- 6}{8} + \frac{- 17 \cdot 2}{4 \cdot 2}$

Reorder the factors of $4 \cdot 2$ .

$\frac{29 \cdot 8}{8} + \frac{- 24 \cdot 8}{8} + \frac{- 6}{8} + \frac{- 17 \cdot 2}{2 \cdot 4}$

Multiply $2$ by $4$ .

$\frac{29 \cdot 8}{8} + \frac{- 24 \cdot 8}{8} + \frac{- 6}{8} + \frac{- 17 \cdot 2}{8}$

Combine the numerators over the common denominator.

$\frac{29 \cdot 8 - 24 \cdot 8 - 6 - 17 \cdot 2}{8}$

Simplify each term.

Multiply $29$ by $8$ .

$\frac{232 - 24 \cdot 8 - 6 - 17 \cdot 2}{8}$

Multiply $- 24$ by $8$ .

$\frac{232 - 192 - 6 - 17 \cdot 2}{8}$

Multiply $- 17$ by $2$ .

$\frac{232 - 192 - 6 - 34}{8}$

Simplify the expression.

Subtract $192$ from $232$ .

$\frac{40 - 6 - 34}{8}$

Subtract $6$ from $40$ .

$\frac{34 - 34}{8}$

Subtract $34$ from $34$ .

$\frac{0}{8}$

Divide $0$ by $8$ .

$0$

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.

$\frac{2 x^{4} - 7 x^{3} - 17 x^{2} + 58 x - 24}{x - \frac{1}{2}}$

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.

$\frac{1}{2}$	$2$	$- 7$	$- 17$	$58$	$- 24$
	\xa0				\xa0

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

$\frac{1}{2}$	$2$	$- 7$	$- 17$	$58$	$- 24$
	\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 $(- 7)$ .

$\frac{1}{2}$	$2$	$- 7$	$- 17$	$58$	$- 24$
	\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.

$\frac{1}{2}$	$2$	$- 7$	$- 17$	$58$	$- 24$
	\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 $(- 17)$ .

$\frac{1}{2}$	$2$	$- 7$	$- 17$	$58$	$- 24$
	\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.

$\frac{1}{2}$	$2$	$- 7$	$- 17$	$58$	$- 24$
	\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 $(58)$ .

$\frac{1}{2}$	$2$	$- 7$	$- 17$	$58$	$- 24$
	\xa0	$1$	$- 3$	$- 10$	\xa0
	$2$	$- 6$	$- 20$

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

$\frac{1}{2}$	$2$	$- 7$	$- 17$	$58$	$- 24$
	\xa0	$1$	$- 3$	$- 10$	\xa0
	$2$	$- 6$	$- 20$	$48$

Multiply the newest entry in the result $(48)$ by the divisor $(\frac{1}{2})$ and place the result of $(24)$ under the next term in the dividend $(- 24)$ .

$\frac{1}{2}$	$2$	$- 7$	$- 17$	$58$	$- 24$
	\xa0	$1$	$- 3$	$- 10$	$24$
	$2$	$- 6$	$- 20$	$48$

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

$\frac{1}{2}$	$2$	$- 7$	$- 17$	$58$	$- 24$
	\xa0	$1$	$- 3$	$- 10$	$24$
	$2$	$- 6$	$- 20$	$48$	$0$

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

$2 x^{3} + - 6 x^{2} + (- 20) x + 48$

Simplify the quotient polynomial.

$2 x^{3} - 6 x^{2} - 20 x + 48$

Step 7

Factor $2$ out of $2 x^{3} - 6 x^{2} - 20 x + 48$ .

Factor $2$ out of $2 x^{3}$ .

$(x - \frac{1}{2}) (2 (x^{3}) - 6 x^{2} - 20 x + 48)$

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

$(x - \frac{1}{2}) (2 (x^{3}) + 2 (- 3 x^{2}) - 20 x + 48)$

Factor $2$ out of $- 20 x$ .

$(x - \frac{1}{2}) (2 (x^{3}) + 2 (- 3 x^{2}) + 2 (- 10 x) + 48)$

Factor $2$ out of $48$ .

$(x - \frac{1}{2}) (2 x^{3} + 2 (- 3 x^{2}) + 2 (- 10 x) + 2 \cdot 24)$

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

$(x - \frac{1}{2}) (2 (x^{3} - 3 x^{2}) + 2 (- 10 x) + 2 \cdot 24)$

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

$(x - \frac{1}{2}) (2 (x^{3} - 3 x^{2} - 10 x) + 2 \cdot 24)$

Factor $2$ out of $2 (x^{3} - 3 x^{2} - 10 x) + 2 \cdot 24$ .

$(x - \frac{1}{2}) (2 (x^{3} - 3 x^{2} - 10 x + 24))$

Step 8

Factor the left side of the equation.

Factor $2 x^{4} - 7 x^{3} - 17 x^{2} + 58 x - 24$ 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.

$p = \pm 1, \pm 24, \pm 2, \pm 12, \pm 3, \pm 8, \pm 4, \pm 6 q = \pm 1, \pm 2$

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

$\pm 1, \pm 0.5, \pm 24, \pm 12, \pm 2, \pm 6, \pm 3, \pm 1.5, \pm 8, \pm 4$

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.

$2 \cdot {0.5}^{4} - 7 \cdot {0.5}^{3} - 17 \cdot {0.5}^{2} + 58 \cdot 0.5 - 24$

Raise $0.5$ to the power of $4$ .

$2 \cdot 0.0625 - 7 \cdot {0.5}^{3} - 17 \cdot {0.5}^{2} + 58 \cdot 0.5 - 24$

Multiply $2$ by $0.0625$ .

$0.125 - 7 \cdot {0.5}^{3} - 17 \cdot {0.5}^{2} + 58 \cdot 0.5 - 24$

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

$0.125 - 7 \cdot 0.125 - 17 \cdot {0.5}^{2} + 58 \cdot 0.5 - 24$

Multiply $- 7$ by $0.125$ .

$0.125 - 0.875 - 17 \cdot {0.5}^{2} + 58 \cdot 0.5 - 24$

Subtract $0.875$ from $0.125$ .

$- 0.75 - 17 \cdot {0.5}^{2} + 58 \cdot 0.5 - 24$

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

$- 0.75 - 17 \cdot 0.25 + 58 \cdot 0.5 - 24$

Multiply $- 17$ by $0.25$ .

$- 0.75 - 4.25 + 58 \cdot 0.5 - 24$

Subtract $4.25$ from $- 0.75$ .

$- 5 + 58 \cdot 0.5 - 24$

Multiply $58$ by $0.5$ .

$- 5 + 29 - 24$

Add $- 5$ and $29$ .

$24 - 24$

Subtract $24$ from $24$ .

$0$

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.

$\frac{2 x^{4} - 7 x^{3} - 17 x^{2} + 58 x - 24}{2 x - 1}$

Divide $2 x^{4} - 7 x^{3} - 17 x^{2} + 58 x - 24$ 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 $0$ .

$2 x$

$1$

$2 x^{4}$

$7 x^{3}$

$17 x^{2}$

$58 x$

$24$

Divide the highest order term in the dividend $2 x^{4}$ by the highest order term in divisor $2 x$ .

					$x^{3}$
	$2 x$	-	$1$		$2 x^{4}$	-	$7 x^{3}$	-	$17 x^{2}$	+	$58 x$	-	$24$

Multiply the new quotient term by the divisor.

				$x^{3}$
$2 x$	-	$1$		$2 x^{4}$	-	$7 x^{3}$	-	$17 x^{2}$	+	$58 x$	-	$24$
			+	$2 x^{4}$	-	$x^{3}$

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

				$x^{3}$
$2 x$	-	$1$		$2 x^{4}$	-	$7 x^{3}$	-	$17 x^{2}$	+	$58 x$	-	$24$
			-	$2 x^{4}$	+	$x^{3}$

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

				$x^{3}$
$2 x$	-	$1$		$2 x^{4}$	-	$7 x^{3}$	-	$17 x^{2}$	+	$58 x$	-	$24$
			-	$2 x^{4}$	+	$x^{3}$
					-	$6 x^{3}$

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

				$x^{3}$
$2 x$	-	$1$		$2 x^{4}$	-	$7 x^{3}$	-	$17 x^{2}$	+	$58 x$	-	$24$
			-	$2 x^{4}$	+	$x^{3}$
					-	$6 x^{3}$	-	$17 x^{2}$

Divide the highest order term in the dividend $- 6 x^{3}$ by the highest order term in divisor $2 x$ .

				$x^{3}$	-	$3 x^{2}$
$2 x$	-	$1$		$2 x^{4}$	-	$7 x^{3}$	-	$17 x^{2}$	+	$58 x$	-	$24$
			-	$2 x^{4}$	+	$x^{3}$
					-	$6 x^{3}$	-	$17 x^{2}$

Multiply the new quotient term by the divisor.

				$x^{3}$	-	$3 x^{2}$
$2 x$	-	$1$		$2 x^{4}$	-	$7 x^{3}$	-	$17 x^{2}$	+	$58 x$	-	$24$
			-	$2 x^{4}$	+	$x^{3}$
					-	$6 x^{3}$	-	$17 x^{2}$
					-	$6 x^{3}$	+	$3 x^{2}$

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

				$x^{3}$	-	$3 x^{2}$
$2 x$	-	$1$		$2 x^{4}$	-	$7 x^{3}$	-	$17 x^{2}$	+	$58 x$	-	$24$
			-	$2 x^{4}$	+	$x^{3}$
					-	$6 x^{3}$	-	$17 x^{2}$
					+	$6 x^{3}$	-	$3 x^{2}$

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

				$x^{3}$	-	$3 x^{2}$
$2 x$	-	$1$		$2 x^{4}$	-	$7 x^{3}$	-	$17 x^{2}$	+	$58 x$	-	$24$
			-	$2 x^{4}$	+	$x^{3}$
					-	$6 x^{3}$	-	$17 x^{2}$
					+	$6 x^{3}$	-	$3 x^{2}$
							-	$20 x^{2}$

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

$x^{3}$

$3 x^{2}$

$2 x$

$1$

$2 x^{4}$

$7 x^{3}$

$17 x^{2}$

$58 x$

$24$

$2 x^{4}$

$x^{3}$

$6 x^{3}$

$17 x^{2}$

$6 x^{3}$

$3 x^{2}$

$20 x^{2}$

$58 x$

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

$x^{3}$

$3 x^{2}$

$10 x$

$2 x$

$1$

$2 x^{4}$

$7 x^{3}$

$17 x^{2}$

$58 x$

$24$

$2 x^{4}$

$x^{3}$

$6 x^{3}$

$17 x^{2}$

$6 x^{3}$

$3 x^{2}$

$20 x^{2}$

$58 x$

Multiply the new quotient term by the divisor.

$x^{3}$

$3 x^{2}$

$10 x$

$2 x$

$1$

$2 x^{4}$

$7 x^{3}$

$17 x^{2}$

$58 x$

$24$

$2 x^{4}$

$x^{3}$

$6 x^{3}$

$17 x^{2}$

$6 x^{3}$

$3 x^{2}$

$20 x^{2}$

$58 x$

$20 x^{2}$

$10 x$

The expression needs to be subtracted from the dividend, so change all the signs in $- 20 x^{2} + 10 x$

$x^{3}$

$3 x^{2}$

$10 x$

$2 x$

$1$

$2 x^{4}$

$7 x^{3}$

$17 x^{2}$

$58 x$

$24$

$2 x^{4}$

$x^{3}$

$6 x^{3}$

$17 x^{2}$

$6 x^{3}$

$3 x^{2}$

$20 x^{2}$

$58 x$

$20 x^{2}$

$10 x$

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

$x^{3}$

$3 x^{2}$

$10 x$

$2 x$

$1$

$2 x^{4}$

$7 x^{3}$

$17 x^{2}$

$58 x$

$24$

$2 x^{4}$

$x^{3}$

$6 x^{3}$

$17 x^{2}$

$6 x^{3}$

$3 x^{2}$

$20 x^{2}$

$58 x$

$20 x^{2}$

$10 x$

$48 x$

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

$x^{3}$

$3 x^{2}$

$10 x$

$2 x$

$1$

$2 x^{4}$

$7 x^{3}$

$17 x^{2}$

$58 x$

$24$

$2 x^{4}$

$x^{3}$

$6 x^{3}$

$17 x^{2}$

$6 x^{3}$

$3 x^{2}$

$20 x^{2}$

$58 x$

$20 x^{2}$

$10 x$

$48 x$

$24$

Divide the highest order term in the dividend $48 x$ by the highest order term in divisor $2 x$ .

$x^{3}$

$3 x^{2}$

$10 x$

$24$

$2 x$

$1$

$2 x^{4}$

$7 x^{3}$

$17 x^{2}$

$58 x$

$24$

$2 x^{4}$

$x^{3}$

$6 x^{3}$

$17 x^{2}$

$6 x^{3}$

$3 x^{2}$

$20 x^{2}$

$58 x$

$20 x^{2}$

$10 x$

$48 x$

$24$

Multiply the new quotient term by the divisor.

				$x^{3}$	-	$3 x^{2}$	-	$10 x$	+	$24$
$2 x$	-	$1$		$2 x^{4}$	-	$7 x^{3}$	-	$17 x^{2}$	+	$58 x$	-	$24$
			-	$2 x^{4}$	+	$x^{3}$
					-	$6 x^{3}$	-	$17 x^{2}$
					+	$6 x^{3}$	-	$3 x^{2}$
							-	$20 x^{2}$	+	$58 x$
							+	$20 x^{2}$	-	$10 x$
									+	$48 x$	-	$24$
									+	$48 x$	-	$24$

The expression needs to be subtracted from the dividend, so change all the signs in $48 x - 24$

				$x^{3}$	-	$3 x^{2}$	-	$10 x$	+	$24$
$2 x$	-	$1$		$2 x^{4}$	-	$7 x^{3}$	-	$17 x^{2}$	+	$58 x$	-	$24$
			-	$2 x^{4}$	+	$x^{3}$
					-	$6 x^{3}$	-	$17 x^{2}$
					+	$6 x^{3}$	-	$3 x^{2}$
							-	$20 x^{2}$	+	$58 x$
							+	$20 x^{2}$	-	$10 x$
									+	$48 x$	-	$24$
									-	$48 x$	+	$24$

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

				$x^{3}$	-	$3 x^{2}$	-	$10 x$	+	$24$
$2 x$	-	$1$		$2 x^{4}$	-	$7 x^{3}$	-	$17 x^{2}$	+	$58 x$	-	$24$
			-	$2 x^{4}$	+	$x^{3}$
					-	$6 x^{3}$	-	$17 x^{2}$
					+	$6 x^{3}$	-	$3 x^{2}$
							-	$20 x^{2}$	+	$58 x$
							+	$20 x^{2}$	-	$10 x$
									+	$48 x$	-	$24$
									-	$48 x$	+	$24$
												$0$

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

$x^{3} - 3 x^{2} - 10 x + 24$

Write $2 x^{4} - 7 x^{3} - 17 x^{2} + 58 x - 24$ as a set of factors.

$(2 x - 1) (x^{3} - 3 x^{2} - 10 x + 24) = 0$

Factor $x^{3} - 3 x^{2} - 10 x + 24$ 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.

$p = \pm 1, \pm 24, \pm 2, \pm 12, \pm 3, \pm 8, \pm 4, \pm 6 q = \pm 1$

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

$\pm 1, \pm 24, \pm 2, \pm 12, \pm 3, \pm 8, \pm 4, \pm 6$

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.

$2^{3} - 3 \cdot 2^{2} - 10 \cdot 2 + 24$

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

$8 - 3 \cdot 2^{2} - 10 \cdot 2 + 24$

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

$8 - 3 \cdot 4 - 10 \cdot 2 + 24$

Multiply $- 3$ by $4$ .

$8 - 12 - 10 \cdot 2 + 24$

Subtract $12$ from $8$ .

$- 4 - 10 \cdot 2 + 24$

Multiply $- 10$ by $2$ .

$- 4 - 20 + 24$

Subtract $20$ from $- 4$ .

$- 24 + 24$

Add $- 24$ and $24$ .

$0$

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.

$\frac{x^{3} - 3 x^{2} - 10 x + 24}{x - 2}$

Divide $x^{3} - 3 x^{2} - 10 x + 24$ 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 $0$ .

$x$

$2$

$x^{3}$

$3 x^{2}$

$10 x$

$24$

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

					$x^{2}$
	$x$	-	$2$		$x^{3}$	-	$3 x^{2}$	-	$10 x$	+	$24$

Multiply the new quotient term by the divisor.

				$x^{2}$
$x$	-	$2$		$x^{3}$	-	$3 x^{2}$	-	$10 x$	+	$24$
			+	$x^{3}$	-	$2 x^{2}$

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

				$x^{2}$
$x$	-	$2$		$x^{3}$	-	$3 x^{2}$	-	$10 x$	+	$24$
			-	$x^{3}$	+	$2 x^{2}$

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

				$x^{2}$
$x$	-	$2$		$x^{3}$	-	$3 x^{2}$	-	$10 x$	+	$24$
			-	$x^{3}$	+	$2 x^{2}$
					-	$x^{2}$

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

				$x^{2}$
$x$	-	$2$		$x^{3}$	-	$3 x^{2}$	-	$10 x$	+	$24$
			-	$x^{3}$	+	$2 x^{2}$
					-	$x^{2}$	-	$10 x$

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

				$x^{2}$	-	$x$
$x$	-	$2$		$x^{3}$	-	$3 x^{2}$	-	$10 x$	+	$24$
			-	$x^{3}$	+	$2 x^{2}$
					-	$x^{2}$	-	$10 x$

Multiply the new quotient term by the divisor.

				$x^{2}$	-	$x$
$x$	-	$2$		$x^{3}$	-	$3 x^{2}$	-	$10 x$	+	$24$
			-	$x^{3}$	+	$2 x^{2}$
					-	$x^{2}$	-	$10 x$
					-	$x^{2}$	+	$2 x$

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

				$x^{2}$	-	$x$
$x$	-	$2$		$x^{3}$	-	$3 x^{2}$	-	$10 x$	+	$24$
			-	$x^{3}$	+	$2 x^{2}$
					-	$x^{2}$	-	$10 x$
					+	$x^{2}$	-	$2 x$

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

				$x^{2}$	-	$x$
$x$	-	$2$		$x^{3}$	-	$3 x^{2}$	-	$10 x$	+	$24$
			-	$x^{3}$	+	$2 x^{2}$
					-	$x^{2}$	-	$10 x$
					+	$x^{2}$	-	$2 x$
							-	$12 x$

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

				$x^{2}$	-	$x$
$x$	-	$2$		$x^{3}$	-	$3 x^{2}$	-	$10 x$	+	$24$
			-	$x^{3}$	+	$2 x^{2}$
					-	$x^{2}$	-	$10 x$
					+	$x^{2}$	-	$2 x$
							-	$12 x$	+	$24$

Divide the highest order term in the dividend $- 12 x$ by the highest order term in divisor $x$ .

				$x^{2}$	-	$x$	-	$12$
$x$	-	$2$		$x^{3}$	-	$3 x^{2}$	-	$10 x$	+	$24$
			-	$x^{3}$	+	$2 x^{2}$
					-	$x^{2}$	-	$10 x$
					+	$x^{2}$	-	$2 x$
							-	$12 x$	+	$24$

Multiply the new quotient term by the divisor.

				$x^{2}$	-	$x$	-	$12$
$x$	-	$2$		$x^{3}$	-	$3 x^{2}$	-	$10 x$	+	$24$
			-	$x^{3}$	+	$2 x^{2}$
					-	$x^{2}$	-	$10 x$
					+	$x^{2}$	-	$2 x$
							-	$12 x$	+	$24$
							-	$12 x$	+	$24$

The expression needs to be subtracted from the dividend, so change all the signs in $- 12 x + 24$

				$x^{2}$	-	$x$	-	$12$
$x$	-	$2$		$x^{3}$	-	$3 x^{2}$	-	$10 x$	+	$24$
			-	$x^{3}$	+	$2 x^{2}$
					-	$x^{2}$	-	$10 x$
					+	$x^{2}$	-	$2 x$
							-	$12 x$	+	$24$
							+	$12 x$	-	$24$

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

				$x^{2}$	-	$x$	-	$12$
$x$	-	$2$		$x^{3}$	-	$3 x^{2}$	-	$10 x$	+	$24$
			-	$x^{3}$	+	$2 x^{2}$
					-	$x^{2}$	-	$10 x$
					+	$x^{2}$	-	$2 x$
							-	$12 x$	+	$24$
							+	$12 x$	-	$24$
										$0$

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

$x^{2} - x - 12$

Write $x^{3} - 3 x^{2} - 10 x + 24$ as a set of factors.

$(2 x - 1) ((x - 2) (x^{2} - x - 12)) = 0$

Factor $x^{2} - x - 12$ 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 $- 12$ and whose sum is $- 1$ .

$- 4, 3$

Write the factored form using these integers.

$(2 x - 1) ((x - 2) ((x - 4) (x + 3))) = 0$

Remove unnecessary parentheses.

$(2 x - 1) ((x - 2) (x - 4) (x + 3)) = 0$

Remove unnecessary parentheses.

$(2 x - 1) (x - 2) (x - 4) (x + 3) = 0$

Step 9

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

$2 x - 1 = 0$

$x - 2 = 0$

$x - 4 = 0$

$x + 3 = 0$

Step 10

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 11

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

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

$x - 2 = 0$

Add $2$ to both sides of the equation.

$x = 2$

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

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.

$x = - 3$

Step 14

The final solution is all the values that make $(2 x - 1) (x - 2) (x - 4) (x + 3) = 0$ true.

$x = \frac{1}{2}, 2, 4, - 3$

Step 15

Find the Roots/Zeros Using the Rational Roots Test 2x^4-7x^3-17x^2+58x-24

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