Find the Roots/Zeros Using the Rational Roots Test h(x)=2x^3+23x^2+58x-35

$h (x) = 2 x^{3} + 23 x^{2} + 58 x - 35$

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.

$p = \pm 1, \pm 5, \pm 7, \pm 35$

$q = \pm 1, \pm 2$

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

$\pm 1, \pm \frac{1}{2}, \pm 5, \pm \frac{5}{2}, \pm 7, \pm \frac{7}{2}, \pm 35, \pm \frac{35}{2}$

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.

$2 {(\frac{1}{2})}^{3} + 23 {(\frac{1}{2})}^{2} + 58 (\frac{1}{2}) - 35$

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 $\frac{1}{2}$ .

$2 \frac{1^{3}}{2^{3}} + 23 {(\frac{1}{2})}^{2} + 58 (\frac{1}{2}) - 35$

One to any power is one.

$2 \frac{1}{2^{3}} + 23 {(\frac{1}{2})}^{2} + 58 (\frac{1}{2}) - 35$

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

$2 (\frac{1}{8}) + 23 {(\frac{1}{2})}^{2} + 58 (\frac{1}{2}) - 35$

Cancel the common factor of $2$ .

Factor $2$ out of $8$ .

$2 \frac{1}{2 (4)} + 23 {(\frac{1}{2})}^{2} + 58 (\frac{1}{2}) - 35$

Cancel the common factor.

$2 \frac{1}{2 \cdot 4} + 23 {(\frac{1}{2})}^{2} + 58 (\frac{1}{2}) - 35$

Rewrite the expression.

$\frac{1}{4} + 23 {(\frac{1}{2})}^{2} + 58 (\frac{1}{2}) - 35$

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

$\frac{1}{4} + 23 \frac{1^{2}}{2^{2}} + 58 (\frac{1}{2}) - 35$

One to any power is one.

$\frac{1}{4} + 23 \frac{1}{2^{2}} + 58 (\frac{1}{2}) - 35$

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

$\frac{1}{4} + 23 (\frac{1}{4}) + 58 (\frac{1}{2}) - 35$

Combine $23$ and $\frac{1}{4}$ .

$\frac{1}{4} + \frac{23}{4} + 58 (\frac{1}{2}) - 35$

Cancel the common factor of $2$ .

Factor $2$ out of $58$ .

$\frac{1}{4} + \frac{23}{4} + 2 (29) \frac{1}{2} - 35$

Cancel the common factor.

$\frac{1}{4} + \frac{23}{4} + 2 \cdot 29 \frac{1}{2} - 35$

Rewrite the expression.

$\frac{1}{4} + \frac{23}{4} + 29 - 35$

Combine fractions.

Combine the numerators over the common denominator.

$29 - 35 + \frac{1 + 23}{4}$

Add $1$ and $23$ .

$29 - 35 + \frac{24}{4}$

Find the common denominator.

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

$\frac{29}{1} - 35 + \frac{24}{4}$

Multiply $\frac{29}{1}$ by $\frac{4}{4}$ .

$\frac{29}{1} \cdot \frac{4}{4} - 35 + \frac{24}{4}$

Multiply $\frac{29}{1}$ by $\frac{4}{4}$ .

$\frac{29 \cdot 4}{4} - 35 + \frac{24}{4}$

Write $- 35$ as a fraction with denominator $1$ .

$\frac{29 \cdot 4}{4} + \frac{- 35}{1} + \frac{24}{4}$

Multiply $\frac{- 35}{1}$ by $\frac{4}{4}$ .

$\frac{29 \cdot 4}{4} + \frac{- 35}{1} \cdot \frac{4}{4} + \frac{24}{4}$

Multiply $\frac{- 35}{1}$ by $\frac{4}{4}$ .

$\frac{29 \cdot 4}{4} + \frac{- 35 \cdot 4}{4} + \frac{24}{4}$

Combine the numerators over the common denominator.

$\frac{29 \cdot 4 - 35 \cdot 4 + 24}{4}$

Simplify each term.

Multiply $29$ by $4$ .

$\frac{116 - 35 \cdot 4 + 24}{4}$

Multiply $- 35$ by $4$ .

$\frac{116 - 140 + 24}{4}$

Simplify the expression.

Subtract $140$ from $116$ .

$\frac{- 24 + 24}{4}$

Add $- 24$ and $24$ .

$\frac{0}{4}$

Divide $0$ by $4$ .

$0$

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.

$\frac{2 x^{3} + 23 x^{2} + 58 x - 35}{x - \frac{1}{2}}$

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.

$\frac{1}{2}$	$2$	$23$	$58$	$- 35$
	\xa0			\xa0

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

$\frac{1}{2}$	$2$	$23$	$58$	$- 35$
	\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 $(23)$ .

$\frac{1}{2}$	$2$	$23$	$58$	$- 35$
	\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.

$\frac{1}{2}$	$2$	$23$	$58$	$- 35$
	\xa0	$1$		\xa0
	$2$	$24$

Multiply the newest entry in the result $(24)$ by the divisor $(\frac{1}{2})$ and place the result of $(12)$ under the next term in the dividend $(58)$ .

$\frac{1}{2}$	$2$	$23$	$58$	$- 35$
	\xa0	$1$	$12$	\xa0
	$2$	$24$

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

$\frac{1}{2}$	$2$	$23$	$58$	$- 35$
	\xa0	$1$	$12$	\xa0
	$2$	$24$	$70$

Multiply the newest entry in the result $(70)$ by the divisor $(\frac{1}{2})$ and place the result of $(35)$ under the next term in the dividend $(- 35)$ .

$\frac{1}{2}$	$2$	$23$	$58$	$- 35$
	\xa0	$1$	$12$	$35$
	$2$	$24$	$70$

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

$\frac{1}{2}$	$2$	$23$	$58$	$- 35$
	\xa0	$1$	$12$	$35$
	$2$	$24$	$70$	$0$

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

$2 x^{2} + 24 x + 70$

Factor $2$ out of $2 x^{2} + 24 x + 70$ .

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

$(x - \frac{1}{2}) (2 (x^{2}) + 24 x + 70)$

Factor $2$ out of $24 x$ .

$(x - \frac{1}{2}) (2 (x^{2}) + 2 (12 x) + 70)$

Factor $2$ out of $70$ .

$(x - \frac{1}{2}) (2 x^{2} + 2 (12 x) + 2 \cdot 35)$

Factor $2$ out of $2 x^{2} + 2 (12 x)$ .

$(x - \frac{1}{2}) (2 (x^{2} + 12 x) + 2 \cdot 35)$

Factor $2$ out of $2 (x^{2} + 12 x) + 2 \cdot 35$ .

$(x - \frac{1}{2}) (2 (x^{2} + 12 x + 35))$

Factor.

Factor $x^{2} + 12 x + 35$ 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 $35$ and whose sum is $12$ .

$5, 7$

Write the factored form using these integers.

$(x - \frac{1}{2}) (2 ((x + 5) (x + 7)))$

Remove unnecessary parentheses.

$(x - \frac{1}{2}) (2 (x + 5) (x + 7))$

Factor the left side of the equation.

Factor $2 x^{3} + 23 x^{2} + 58 x - 35$ 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.

$p = \pm 1, \pm 35, \pm 5, \pm 7 q = \pm 1, \pm 2$

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

$\pm 1, \pm 0.5, \pm 35, \pm 17.5, \pm 5, \pm 2.5, \pm 7, \pm 3.5$

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.

$2 \cdot {0.5}^{3} + 23 \cdot {0.5}^{2} + 58 \cdot 0.5 - 35$

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

$2 \cdot 0.125 + 23 \cdot {0.5}^{2} + 58 \cdot 0.5 - 35$

Multiply $2$ by $0.125$ .

$0.25 + 23 \cdot {0.5}^{2} + 58 \cdot 0.5 - 35$

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

$0.25 + 23 \cdot 0.25 + 58 \cdot 0.5 - 35$

Multiply $23$ by $0.25$ .

$0.25 + 5.75 + 58 \cdot 0.5 - 35$

Add $0.25$ and $5.75$ .

$6 + 58 \cdot 0.5 - 35$

Multiply $58$ by $0.5$ .

$6 + 29 - 35$

Add $6$ and $29$ .

$35 - 35$

Subtract $35$ from $35$ .

$0$

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.

$\frac{2 x^{3} + 23 x^{2} + 58 x - 35}{2 x - 1}$

Divide $2 x^{3} + 23 x^{2} + 58 x - 35$ 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 $0$ .

$2 x$

$1$

$2 x^{3}$

$23 x^{2}$

$58 x$

$35$

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

					$x^{2}$
	$2 x$	-	$1$		$2 x^{3}$	+	$23 x^{2}$	+	$58 x$	-	$35$

Multiply the new quotient term by the divisor.

				$x^{2}$
$2 x$	-	$1$		$2 x^{3}$	+	$23 x^{2}$	+	$58 x$	-	$35$
			+	$2 x^{3}$	-	$x^{2}$

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

				$x^{2}$
$2 x$	-	$1$		$2 x^{3}$	+	$23 x^{2}$	+	$58 x$	-	$35$
			-	$2 x^{3}$	+	$x^{2}$

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

				$x^{2}$
$2 x$	-	$1$		$2 x^{3}$	+	$23 x^{2}$	+	$58 x$	-	$35$
			-	$2 x^{3}$	+	$x^{2}$
					+	$24 x^{2}$

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

				$x^{2}$
$2 x$	-	$1$		$2 x^{3}$	+	$23 x^{2}$	+	$58 x$	-	$35$
			-	$2 x^{3}$	+	$x^{2}$
					+	$24 x^{2}$	+	$58 x$

Divide the highest order term in the dividend $24 x^{2}$ by the highest order term in divisor $2 x$ .

				$x^{2}$	+	$12 x$
$2 x$	-	$1$		$2 x^{3}$	+	$23 x^{2}$	+	$58 x$	-	$35$
			-	$2 x^{3}$	+	$x^{2}$
					+	$24 x^{2}$	+	$58 x$

Multiply the new quotient term by the divisor.

				$x^{2}$	+	$12 x$
$2 x$	-	$1$		$2 x^{3}$	+	$23 x^{2}$	+	$58 x$	-	$35$
			-	$2 x^{3}$	+	$x^{2}$
					+	$24 x^{2}$	+	$58 x$
					+	$24 x^{2}$	-	$12 x$

The expression needs to be subtracted from the dividend, so change all the signs in $24 x^{2} - 12 x$

				$x^{2}$	+	$12 x$
$2 x$	-	$1$		$2 x^{3}$	+	$23 x^{2}$	+	$58 x$	-	$35$
			-	$2 x^{3}$	+	$x^{2}$
					+	$24 x^{2}$	+	$58 x$
					-	$24 x^{2}$	+	$12 x$

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

				$x^{2}$	+	$12 x$
$2 x$	-	$1$		$2 x^{3}$	+	$23 x^{2}$	+	$58 x$	-	$35$
			-	$2 x^{3}$	+	$x^{2}$
					+	$24 x^{2}$	+	$58 x$
					-	$24 x^{2}$	+	$12 x$
							+	$70 x$

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

				$x^{2}$	+	$12 x$
$2 x$	-	$1$		$2 x^{3}$	+	$23 x^{2}$	+	$58 x$	-	$35$
			-	$2 x^{3}$	+	$x^{2}$
					+	$24 x^{2}$	+	$58 x$
					-	$24 x^{2}$	+	$12 x$
							+	$70 x$	-	$35$

Divide the highest order term in the dividend $70 x$ by the highest order term in divisor $2 x$ .

				$x^{2}$	+	$12 x$	+	$35$
$2 x$	-	$1$		$2 x^{3}$	+	$23 x^{2}$	+	$58 x$	-	$35$
			-	$2 x^{3}$	+	$x^{2}$
					+	$24 x^{2}$	+	$58 x$
					-	$24 x^{2}$	+	$12 x$
							+	$70 x$	-	$35$

Multiply the new quotient term by the divisor.

				$x^{2}$	+	$12 x$	+	$35$
$2 x$	-	$1$		$2 x^{3}$	+	$23 x^{2}$	+	$58 x$	-	$35$
			-	$2 x^{3}$	+	$x^{2}$
					+	$24 x^{2}$	+	$58 x$
					-	$24 x^{2}$	+	$12 x$
							+	$70 x$	-	$35$
							+	$70 x$	-	$35$

The expression needs to be subtracted from the dividend, so change all the signs in $70 x - 35$

				$x^{2}$	+	$12 x$	+	$35$
$2 x$	-	$1$		$2 x^{3}$	+	$23 x^{2}$	+	$58 x$	-	$35$
			-	$2 x^{3}$	+	$x^{2}$
					+	$24 x^{2}$	+	$58 x$
					-	$24 x^{2}$	+	$12 x$
							+	$70 x$	-	$35$
							-	$70 x$	+	$35$

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

				$x^{2}$	+	$12 x$	+	$35$
$2 x$	-	$1$		$2 x^{3}$	+	$23 x^{2}$	+	$58 x$	-	$35$
			-	$2 x^{3}$	+	$x^{2}$
					+	$24 x^{2}$	+	$58 x$
					-	$24 x^{2}$	+	$12 x$
							+	$70 x$	-	$35$
							-	$70 x$	+	$35$
										$0$

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

$x^{2} + 12 x + 35$

Write $2 x^{3} + 23 x^{2} + 58 x - 35$ as a set of factors.

$(2 x - 1) (x^{2} + 12 x + 35) = 0$

Factor $x^{2} + 12 x + 35$ 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 $35$ and whose sum is $12$ .

$5, 7$

Write the factored form using these integers.

$(2 x - 1) ((x + 5) (x + 7)) = 0$

Remove unnecessary parentheses.

$(2 x - 1) (x + 5) (x + 7) = 0$

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

$2 x - 1 = 0$

$x + 5 = 0$

$x + 7 = 0$

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 $x + 5$ equal to $0$ and solve for $x$ .

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

$x + 5 = 0$

Subtract $5$ from both sides of the equation.

$x = - 5$

Set $x + 7$ equal to $0$ and solve for $x$ .

Set $x + 7$ equal to $0$ .

$x + 7 = 0$

Subtract $7$ from both sides of the equation.

$x = - 7$

The final solution is all the values that make $(2 x - 1) (x + 5) (x + 7) = 0$ true.

$x = \frac{1}{2}, - 5, - 7$

Find the Roots/Zeros Using the Rational Roots Test h(x)=2x^3+23x^2+58x-35

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