Find the Roots/Zeros Using the Rational Roots Test f(x)=8x^3+34x^2-29x+5

$f (x) = 8 x^{3} + 34 x^{2} - 29 x + 5$

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.

$p = \pm 1, \pm 5$

$q = \pm 1, \pm 2, \pm 4, \pm 8$

Step 2

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

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

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.

$8 {(\frac{1}{2})}^{3} + 34 {(\frac{1}{2})}^{2} - 29 (\frac{1}{2}) + 5$

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 $\frac{1}{2}$ .

$8 \frac{1^{3}}{2^{3}} + 34 {(\frac{1}{2})}^{2} - 29 (\frac{1}{2}) + 5$

One to any power is one.

$8 \frac{1}{2^{3}} + 34 {(\frac{1}{2})}^{2} - 29 (\frac{1}{2}) + 5$

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

$8 (\frac{1}{8}) + 34 {(\frac{1}{2})}^{2} - 29 (\frac{1}{2}) + 5$

Cancel the common factor of $8$ .

Cancel the common factor.

$8 (\frac{1}{8}) + 34 {(\frac{1}{2})}^{2} - 29 (\frac{1}{2}) + 5$

Rewrite the expression.

$1 + 34 {(\frac{1}{2})}^{2} - 29 (\frac{1}{2}) + 5$

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

$1 + 34 \frac{1^{2}}{2^{2}} - 29 (\frac{1}{2}) + 5$

One to any power is one.

$1 + 34 \frac{1}{2^{2}} - 29 (\frac{1}{2}) + 5$

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

$1 + 34 (\frac{1}{4}) - 29 (\frac{1}{2}) + 5$

Cancel the common factor of $2$ .

Factor $2$ out of $34$ .

$1 + 2 (17) \frac{1}{4} - 29 (\frac{1}{2}) + 5$

Factor $2$ out of $4$ .

$1 + 2 \cdot 17 \frac{1}{2 \cdot 2} - 29 (\frac{1}{2}) + 5$

Cancel the common factor.

$1 + 2 \cdot 17 \frac{1}{2 \cdot 2} - 29 (\frac{1}{2}) + 5$

Rewrite the expression.

$1 + 17 (\frac{1}{2}) - 29 (\frac{1}{2}) + 5$

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

$1 + \frac{17}{2} - 29 (\frac{1}{2}) + 5$

Combine $- 29$ and $\frac{1}{2}$ .

$1 + \frac{17}{2} + \frac{- 29}{2} + 5$

Move the negative in front of the fraction.

$1 + \frac{17}{2} - \frac{29}{2} + 5$

Combine fractions.

Combine the numerators over the common denominator.

$1 + 5 + \frac{17 - 29}{2}$

Subtract $29$ from $17$ .

$1 + 5 + \frac{- 12}{2}$

Find the common denominator.

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

$\frac{1}{1} + 5 + \frac{- 12}{2}$

Multiply $\frac{1}{1}$ by $\frac{2}{2}$ .

$\frac{1}{1} \cdot \frac{2}{2} + 5 + \frac{- 12}{2}$

Multiply $\frac{1}{1}$ by $\frac{2}{2}$ .

$\frac{2}{2} + 5 + \frac{- 12}{2}$

Write $5$ as a fraction with denominator $1$ .

$\frac{2}{2} + \frac{5}{1} + \frac{- 12}{2}$

Multiply $\frac{5}{1}$ by $\frac{2}{2}$ .

$\frac{2}{2} + \frac{5}{1} \cdot \frac{2}{2} + \frac{- 12}{2}$

Multiply $\frac{5}{1}$ by $\frac{2}{2}$ .

$\frac{2}{2} + \frac{5 \cdot 2}{2} + \frac{- 12}{2}$

Combine the numerators over the common denominator.

$\frac{2 + 5 \cdot 2 - 12}{2}$

Simplify the expression.

Multiply $5$ by $2$ .

$\frac{2 + 10 - 12}{2}$

Add $2$ and $10$ .

$\frac{12 - 12}{2}$

Subtract $12$ from $12$ .

$\frac{0}{2}$

Divide $0$ by $2$ .

$0$

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.

$\frac{8 x^{3} + 34 x^{2} - 29 x + 5}{x - \frac{1}{2}}$

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.

$\frac{1}{2}$	$8$	$34$	$- 29$	$5$
	\xa0			\xa0

The first number in the dividend $(8)$ is put into the first position of the result area (below the horizontal line).

$\frac{1}{2}$	$8$	$34$	$- 29$	$5$
	\xa0			\xa0
	$8$

Multiply the newest entry in the result $(8)$ by the divisor $(\frac{1}{2})$ and place the result of $(4)$ under the next term in the dividend $(34)$ .

$\frac{1}{2}$	$8$	$34$	$- 29$	$5$
	\xa0	$4$		\xa0
	$8$

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

$\frac{1}{2}$	$8$	$34$	$- 29$	$5$
	\xa0	$4$		\xa0
	$8$	$38$

Multiply the newest entry in the result $(38)$ by the divisor $(\frac{1}{2})$ and place the result of $(19)$ under the next term in the dividend $(- 29)$ .

$\frac{1}{2}$	$8$	$34$	$- 29$	$5$
	\xa0	$4$	$19$	\xa0
	$8$	$38$

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

$\frac{1}{2}$	$8$	$34$	$- 29$	$5$
	\xa0	$4$	$19$	\xa0
	$8$	$38$	$- 10$

Multiply the newest entry in the result $(- 10)$ by the divisor $(\frac{1}{2})$ and place the result of $(- 5)$ under the next term in the dividend $(5)$ .

$\frac{1}{2}$	$8$	$34$	$- 29$	$5$
	\xa0	$4$	$19$	$- 5$
	$8$	$38$	$- 10$

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

$\frac{1}{2}$	$8$	$34$	$- 29$	$5$
	\xa0	$4$	$19$	$- 5$
	$8$	$38$	$- 10$	$0$

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

$8 x^{2} + 38 x - 10$

Step 7

Factor $2$ out of $8 x^{2} + 38 x - 10$ .

Factor $2$ out of $8 x^{2}$ .

$(x - \frac{1}{2}) (2 (4 x^{2}) + 38 x - 10)$

Factor $2$ out of $38 x$ .

$(x - \frac{1}{2}) (2 (4 x^{2}) + 2 (19 x) - 10)$

Factor $2$ out of $- 10$ .

$(x - \frac{1}{2}) (2 (4 x^{2}) + 2 (19 x) + 2 (- 5))$

Factor $2$ out of $2 (4 x^{2}) + 2 (19 x)$ .

$(x - \frac{1}{2}) (2 (4 x^{2} + 19 x) + 2 (- 5))$

Factor $2$ out of $2 (4 x^{2} + 19 x) + 2 (- 5)$ .

$(x - \frac{1}{2}) (2 (4 x^{2} + 19 x - 5))$

Step 8

Factor.

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 $a \cdot c = 4 \cdot - 5 = - 20$ and whose sum is $b = 19$ .

Factor $19$ out of $19 x$ .

$(x - \frac{1}{2}) (2 (4 x^{2} + 19 (x) - 5))$

Rewrite $19$ as $- 1$ plus $20$

$(x - \frac{1}{2}) (2 (4 x^{2} + (- 1 + 20) x - 5))$

Apply the distributive property.

$(x - \frac{1}{2}) (2 (4 x^{2} - 1 x + 20 x - 5))$

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

$(x - \frac{1}{2}) (2 ((4 x^{2} - 1 x) + 20 x - 5))$

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

$(x - \frac{1}{2}) (2 (x (4 x - 1) + 5 (4 x - 1)))$

Factor the polynomial by factoring out the greatest common factor, $4 x - 1$ .

$(x - \frac{1}{2}) (2 ((4 x - 1) (x + 5)))$

Remove unnecessary parentheses.

$(x - \frac{1}{2}) (2 (4 x - 1) (x + 5))$

Step 9

Factor the left side of the equation.

Factor $8 x^{3} + 34 x^{2} - 29 x + 5$ 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.

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

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

$\pm 1, \pm 0.125, \pm 0.5, \pm 0.25, \pm 5, \pm 0.625, \pm 2.5, \pm 1.25$

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.

$8 \cdot {0.5}^{3} + 34 \cdot {0.5}^{2} - 29 \cdot 0.5 + 5$

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

$8 \cdot 0.125 + 34 \cdot {0.5}^{2} - 29 \cdot 0.5 + 5$

Multiply $8$ by $0.125$ .

$1 + 34 \cdot {0.5}^{2} - 29 \cdot 0.5 + 5$

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

$1 + 34 \cdot 0.25 - 29 \cdot 0.5 + 5$

Multiply $34$ by $0.25$ .

$1 + 8.5 - 29 \cdot 0.5 + 5$

Add $1$ and $8.5$ .

$9.5 - 29 \cdot 0.5 + 5$

Multiply $- 29$ by $0.5$ .

$9.5 - 14.5 + 5$

Subtract $14.5$ from $9.5$ .

$- 5 + 5$

Add $- 5$ and $5$ .

$0$

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.

$\frac{8 x^{3} + 34 x^{2} - 29 x + 5}{2 x - 1}$

Divide $8 x^{3} + 34 x^{2} - 29 x + 5$ 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 $0$ .

$2 x$

$1$

$8 x^{3}$

$34 x^{2}$

$29 x$

$5$

Divide the highest order term in the dividend $8 x^{3}$ by the highest order term in divisor $2 x$ .

					$4 x^{2}$
	$2 x$	-	$1$		$8 x^{3}$	+	$34 x^{2}$	-	$29 x$	+	$5$

Multiply the new quotient term by the divisor.

				$4 x^{2}$
$2 x$	-	$1$		$8 x^{3}$	+	$34 x^{2}$	-	$29 x$	+	$5$
			+	$8 x^{3}$	-	$4 x^{2}$

The expression needs to be subtracted from the dividend, so change all the signs in $8 x^{3} - 4 x^{2}$

				$4 x^{2}$
$2 x$	-	$1$		$8 x^{3}$	+	$34 x^{2}$	-	$29 x$	+	$5$
			-	$8 x^{3}$	+	$4 x^{2}$

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

				$4 x^{2}$
$2 x$	-	$1$		$8 x^{3}$	+	$34 x^{2}$	-	$29 x$	+	$5$
			-	$8 x^{3}$	+	$4 x^{2}$
					+	$38 x^{2}$

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

				$4 x^{2}$
$2 x$	-	$1$		$8 x^{3}$	+	$34 x^{2}$	-	$29 x$	+	$5$
			-	$8 x^{3}$	+	$4 x^{2}$
					+	$38 x^{2}$	-	$29 x$

Divide the highest order term in the dividend $38 x^{2}$ by the highest order term in divisor $2 x$ .

				$4 x^{2}$	+	$19 x$
$2 x$	-	$1$		$8 x^{3}$	+	$34 x^{2}$	-	$29 x$	+	$5$
			-	$8 x^{3}$	+	$4 x^{2}$
					+	$38 x^{2}$	-	$29 x$

Multiply the new quotient term by the divisor.

				$4 x^{2}$	+	$19 x$
$2 x$	-	$1$		$8 x^{3}$	+	$34 x^{2}$	-	$29 x$	+	$5$
			-	$8 x^{3}$	+	$4 x^{2}$
					+	$38 x^{2}$	-	$29 x$
					+	$38 x^{2}$	-	$19 x$

The expression needs to be subtracted from the dividend, so change all the signs in $38 x^{2} - 19 x$

				$4 x^{2}$	+	$19 x$
$2 x$	-	$1$		$8 x^{3}$	+	$34 x^{2}$	-	$29 x$	+	$5$
			-	$8 x^{3}$	+	$4 x^{2}$
					+	$38 x^{2}$	-	$29 x$
					-	$38 x^{2}$	+	$19 x$

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

				$4 x^{2}$	+	$19 x$
$2 x$	-	$1$		$8 x^{3}$	+	$34 x^{2}$	-	$29 x$	+	$5$
			-	$8 x^{3}$	+	$4 x^{2}$
					+	$38 x^{2}$	-	$29 x$
					-	$38 x^{2}$	+	$19 x$
							-	$10 x$

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

				$4 x^{2}$	+	$19 x$
$2 x$	-	$1$		$8 x^{3}$	+	$34 x^{2}$	-	$29 x$	+	$5$
			-	$8 x^{3}$	+	$4 x^{2}$
					+	$38 x^{2}$	-	$29 x$
					-	$38 x^{2}$	+	$19 x$
							-	$10 x$	+	$5$

Divide the highest order term in the dividend $- 10 x$ by the highest order term in divisor $2 x$ .

				$4 x^{2}$	+	$19 x$	-	$5$
$2 x$	-	$1$		$8 x^{3}$	+	$34 x^{2}$	-	$29 x$	+	$5$
			-	$8 x^{3}$	+	$4 x^{2}$
					+	$38 x^{2}$	-	$29 x$
					-	$38 x^{2}$	+	$19 x$
							-	$10 x$	+	$5$

Multiply the new quotient term by the divisor.

				$4 x^{2}$	+	$19 x$	-	$5$
$2 x$	-	$1$		$8 x^{3}$	+	$34 x^{2}$	-	$29 x$	+	$5$
			-	$8 x^{3}$	+	$4 x^{2}$
					+	$38 x^{2}$	-	$29 x$
					-	$38 x^{2}$	+	$19 x$
							-	$10 x$	+	$5$
							-	$10 x$	+	$5$

The expression needs to be subtracted from the dividend, so change all the signs in $- 10 x + 5$

				$4 x^{2}$	+	$19 x$	-	$5$
$2 x$	-	$1$		$8 x^{3}$	+	$34 x^{2}$	-	$29 x$	+	$5$
			-	$8 x^{3}$	+	$4 x^{2}$
					+	$38 x^{2}$	-	$29 x$
					-	$38 x^{2}$	+	$19 x$
							-	$10 x$	+	$5$
							+	$10 x$	-	$5$

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

				$4 x^{2}$	+	$19 x$	-	$5$
$2 x$	-	$1$		$8 x^{3}$	+	$34 x^{2}$	-	$29 x$	+	$5$
			-	$8 x^{3}$	+	$4 x^{2}$
					+	$38 x^{2}$	-	$29 x$
					-	$38 x^{2}$	+	$19 x$
							-	$10 x$	+	$5$
							+	$10 x$	-	$5$
										$0$

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

$4 x^{2} + 19 x - 5$

Write $8 x^{3} + 34 x^{2} - 29 x + 5$ as a set of factors.

$(2 x - 1) (4 x^{2} + 19 x - 5) = 0$

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 $a \cdot c = 4 \cdot - 5 = - 20$ and whose sum is $b = 19$ .

Factor $19$ out of $19 x$ .

$(2 x - 1) (4 x^{2} + 19 (x) - 5) = 0$

Rewrite $19$ as $- 1$ plus $20$

$(2 x - 1) (4 x^{2} + (- 1 + 20) x - 5) = 0$

Apply the distributive property.

$(2 x - 1) (4 x^{2} - 1 x + 20 x - 5) = 0$

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

$(2 x - 1) ((4 x^{2} - 1 x) + 20 x - 5) = 0$

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

$(2 x - 1) (x (4 x - 1) + 5 (4 x - 1)) = 0$

Factor the polynomial by factoring out the greatest common factor, $4 x - 1$ .

$(2 x - 1) ((4 x - 1) (x + 5)) = 0$

Remove unnecessary parentheses.

$(2 x - 1) (4 x - 1) (x + 5) = 0$

Step 10

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

$2 x - 1 = 0$

$4 x - 1 = 0$

$x + 5 = 0$

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 $4 x - 1$ equal to $0$ and solve for $x$ .

Set $4 x - 1$ equal to $0$ .

$4 x - 1 = 0$

Solve $4 x - 1 = 0$ for $x$ .

Add $1$ to both sides of the equation.

$4 x = 1$

Divide each term in $4 x = 1$ by $4$ and simplify.

Divide each term in $4 x = 1$ by $4$ .

$\frac{4 x}{4} = \frac{1}{4}$

Simplify the left side.

Cancel the common factor of $4$ .

Cancel the common factor.

$\frac{4 x}{4} = \frac{1}{4}$

Divide $x$ by $1$ .

$x = \frac{1}{4}$

Step 13

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.

$x = - 5$

Step 14

The final solution is all the values that make $(2 x - 1) (4 x - 1) (x + 5) = 0$ true.

$x = \frac{1}{2}, \frac{1}{4}, - 5$

Step 15

Find the Roots/Zeros Using the Rational Roots Test f(x)=8x^3+34x^2-29x+5

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