Find the Roots/Zeros Using the Rational Roots Test f(x)=6x^3+37x^2-34x+7

$f (x) = 6 x^{3} + 37 x^{2} - 34 x + 7$

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.

$p = \pm 1, \pm 7$

$q = \pm 1, \pm 2, \pm 3, \pm 6$

Step 2

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

$\pm 1, \pm \frac{1}{2}, \pm \frac{1}{3}, \pm \frac{1}{6}, \pm 7, \pm \frac{7}{2}, \pm \frac{7}{3}, \pm \frac{7}{6}$

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.

$6 {(\frac{1}{2})}^{3} + 37 {(\frac{1}{2})}^{2} - 34 (\frac{1}{2}) + 7$

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 $\frac{1}{2}$ .

$6 \frac{1^{3}}{2^{3}} + 37 {(\frac{1}{2})}^{2} - 34 (\frac{1}{2}) + 7$

One to any power is one.

$6 \frac{1}{2^{3}} + 37 {(\frac{1}{2})}^{2} - 34 (\frac{1}{2}) + 7$

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

$6 (\frac{1}{8}) + 37 {(\frac{1}{2})}^{2} - 34 (\frac{1}{2}) + 7$

Cancel the common factor of $2$ .

Factor $2$ out of $6$ .

$2 (3) \frac{1}{8} + 37 {(\frac{1}{2})}^{2} - 34 (\frac{1}{2}) + 7$

Factor $2$ out of $8$ .

$2 \cdot 3 \frac{1}{2 \cdot 4} + 37 {(\frac{1}{2})}^{2} - 34 (\frac{1}{2}) + 7$

Cancel the common factor.

$2 \cdot 3 \frac{1}{2 \cdot 4} + 37 {(\frac{1}{2})}^{2} - 34 (\frac{1}{2}) + 7$

Rewrite the expression.

$3 (\frac{1}{4}) + 37 {(\frac{1}{2})}^{2} - 34 (\frac{1}{2}) + 7$

Combine $3$ and $\frac{1}{4}$ .

$\frac{3}{4} + 37 {(\frac{1}{2})}^{2} - 34 (\frac{1}{2}) + 7$

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

$\frac{3}{4} + 37 \frac{1^{2}}{2^{2}} - 34 (\frac{1}{2}) + 7$

One to any power is one.

$\frac{3}{4} + 37 \frac{1}{2^{2}} - 34 (\frac{1}{2}) + 7$

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

$\frac{3}{4} + 37 (\frac{1}{4}) - 34 (\frac{1}{2}) + 7$

Combine $37$ and $\frac{1}{4}$ .

$\frac{3}{4} + \frac{37}{4} - 34 (\frac{1}{2}) + 7$

Cancel the common factor of $2$ .

Factor $2$ out of $- 34$ .

$\frac{3}{4} + \frac{37}{4} + 2 (- 17) \frac{1}{2} + 7$

Cancel the common factor.

$\frac{3}{4} + \frac{37}{4} + 2 \cdot - 17 \frac{1}{2} + 7$

Rewrite the expression.

$\frac{3}{4} + \frac{37}{4} - 17 + 7$

Combine fractions.

Combine the numerators over the common denominator.

$- 17 + 7 + \frac{3 + 37}{4}$

Add $3$ and $37$ .

$- 17 + 7 + \frac{40}{4}$

Find the common denominator.

Write $- 17$ as a fraction with denominator $1$ .

$\frac{- 17}{1} + 7 + \frac{40}{4}$

Multiply $\frac{- 17}{1}$ by $\frac{4}{4}$ .

$\frac{- 17}{1} \cdot \frac{4}{4} + 7 + \frac{40}{4}$

Multiply $\frac{- 17}{1}$ by $\frac{4}{4}$ .

$\frac{- 17 \cdot 4}{4} + 7 + \frac{40}{4}$

Write $7$ as a fraction with denominator $1$ .

$\frac{- 17 \cdot 4}{4} + \frac{7}{1} + \frac{40}{4}$

Multiply $\frac{7}{1}$ by $\frac{4}{4}$ .

$\frac{- 17 \cdot 4}{4} + \frac{7}{1} \cdot \frac{4}{4} + \frac{40}{4}$

Multiply $\frac{7}{1}$ by $\frac{4}{4}$ .

$\frac{- 17 \cdot 4}{4} + \frac{7 \cdot 4}{4} + \frac{40}{4}$

Combine the numerators over the common denominator.

$\frac{- 17 \cdot 4 + 7 \cdot 4 + 40}{4}$

Simplify each term.

Multiply $- 17$ by $4$ .

$\frac{- 68 + 7 \cdot 4 + 40}{4}$

Multiply $7$ by $4$ .

$\frac{- 68 + 28 + 40}{4}$

Simplify the expression.

Add $- 68$ and $28$ .

$\frac{- 40 + 40}{4}$

Add $- 40$ and $40$ .

$\frac{0}{4}$

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.

$\frac{6 x^{3} + 37 x^{2} - 34 x + 7}{x - \frac{1}{2}}$

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.

$\frac{1}{2}$	$6$	$37$	$- 34$	$7$
	\xa0			\xa0

The first number in the dividend $(6)$ is put into the first position of the result area (below the horizontal line).

$\frac{1}{2}$	$6$	$37$	$- 34$	$7$
	\xa0			\xa0
	$6$

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 $(37)$ .

$\frac{1}{2}$	$6$	$37$	$- 34$	$7$
	\xa0	$3$		\xa0
	$6$

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

$\frac{1}{2}$	$6$	$37$	$- 34$	$7$
	\xa0	$3$		\xa0
	$6$	$40$

Multiply the newest entry in the result $(40)$ by the divisor $(\frac{1}{2})$ and place the result of $(20)$ under the next term in the dividend $(- 34)$ .

$\frac{1}{2}$	$6$	$37$	$- 34$	$7$
	\xa0	$3$	$20$	\xa0
	$6$	$40$

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

$\frac{1}{2}$	$6$	$37$	$- 34$	$7$
	\xa0	$3$	$20$	\xa0
	$6$	$40$	$- 14$

Multiply the newest entry in the result $(- 14)$ by the divisor $(\frac{1}{2})$ and place the result of $(- 7)$ under the next term in the dividend $(7)$ .

$\frac{1}{2}$	$6$	$37$	$- 34$	$7$
	\xa0	$3$	$20$	$- 7$
	$6$	$40$	$- 14$

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

$\frac{1}{2}$	$6$	$37$	$- 34$	$7$
	\xa0	$3$	$20$	$- 7$
	$6$	$40$	$- 14$	$0$

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

$6 x^{2} + 40 x - 14$

Step 7

Factor $2$ out of $6 x^{2} + 40 x - 14$ .

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

$(x - \frac{1}{2}) (2 (3 x^{2}) + 40 x - 14)$

Factor $2$ out of $40 x$ .

$(x - \frac{1}{2}) (2 (3 x^{2}) + 2 (20 x) - 14)$

Factor $2$ out of $- 14$ .

$(x - \frac{1}{2}) (2 (3 x^{2}) + 2 (20 x) + 2 (- 7))$

Factor $2$ out of $2 (3 x^{2}) + 2 (20 x)$ .

$(x - \frac{1}{2}) (2 (3 x^{2} + 20 x) + 2 (- 7))$

Factor $2$ out of $2 (3 x^{2} + 20 x) + 2 (- 7)$ .

$(x - \frac{1}{2}) (2 (3 x^{2} + 20 x - 7))$

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 $a \cdot c = 3 \cdot - 7 = - 21$ and whose sum is $b = 20$ .

Factor $20$ out of $20 x$ .

$(x - \frac{1}{2}) (2 (3 x^{2} + 20 (x) - 7))$

Rewrite $20$ as $- 1$ plus $21$

$(x - \frac{1}{2}) (2 (3 x^{2} + (- 1 + 21) x - 7))$

Apply the distributive property.

$(x - \frac{1}{2}) (2 (3 x^{2} - 1 x + 21 x - 7))$

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

$(x - \frac{1}{2}) (2 ((3 x^{2} - 1 x) + 21 x - 7))$

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

$(x - \frac{1}{2}) (2 (x (3 x - 1) + 7 (3 x - 1)))$

Factor the polynomial by factoring out the greatest common factor, $3 x - 1$ .

$(x - \frac{1}{2}) (2 ((3 x - 1) (x + 7)))$

Remove unnecessary parentheses.

$(x - \frac{1}{2}) (2 (3 x - 1) (x + 7))$

Step 9

Factor the left side of the equation.

Factor $6 x^{3} + 37 x^{2} - 34 x + 7$ 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.

$p = \pm 1, \pm 7 q = \pm 1, \pm 6, \pm 2, \pm 3$

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

$\pm 1, \pm 0.1\overline{6}, \pm 0.5, \pm 0.\overline{3}, \pm 7, \pm 1.1\overline{6}, \pm 3.5, \pm 2.\overline{3}$

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.

$6 \cdot {0.5}^{3} + 37 \cdot {0.5}^{2} - 34 \cdot 0.5 + 7$

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

$6 \cdot 0.125 + 37 \cdot {0.5}^{2} - 34 \cdot 0.5 + 7$

Multiply $6$ by $0.125$ .

$0.75 + 37 \cdot {0.5}^{2} - 34 \cdot 0.5 + 7$

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

$0.75 + 37 \cdot 0.25 - 34 \cdot 0.5 + 7$

Multiply $37$ by $0.25$ .

$0.75 + 9.25 - 34 \cdot 0.5 + 7$

Add $0.75$ and $9.25$ .

$10 - 34 \cdot 0.5 + 7$

Multiply $- 34$ by $0.5$ .

$10 - 17 + 7$

Subtract $17$ from $10$ .

$- 7 + 7$

Add $- 7$ and $7$ .

$0$

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.

$\frac{6 x^{3} + 37 x^{2} - 34 x + 7}{2 x - 1}$

Divide $6 x^{3} + 37 x^{2} - 34 x + 7$ 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 $0$ .

$2 x$

$1$

$6 x^{3}$

$37 x^{2}$

$34 x$

$7$

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

					$3 x^{2}$
	$2 x$	-	$1$		$6 x^{3}$	+	$37 x^{2}$	-	$34 x$	+	$7$

Multiply the new quotient term by the divisor.

				$3 x^{2}$
$2 x$	-	$1$		$6 x^{3}$	+	$37 x^{2}$	-	$34 x$	+	$7$
			+	$6 x^{3}$	-	$3 x^{2}$

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

				$3 x^{2}$
$2 x$	-	$1$		$6 x^{3}$	+	$37 x^{2}$	-	$34 x$	+	$7$
			-	$6 x^{3}$	+	$3 x^{2}$

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

				$3 x^{2}$
$2 x$	-	$1$		$6 x^{3}$	+	$37 x^{2}$	-	$34 x$	+	$7$
			-	$6 x^{3}$	+	$3 x^{2}$
					+	$40 x^{2}$

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

				$3 x^{2}$
$2 x$	-	$1$		$6 x^{3}$	+	$37 x^{2}$	-	$34 x$	+	$7$
			-	$6 x^{3}$	+	$3 x^{2}$
					+	$40 x^{2}$	-	$34 x$

Divide the highest order term in the dividend $40 x^{2}$ by the highest order term in divisor $2 x$ .

				$3 x^{2}$	+	$20 x$
$2 x$	-	$1$		$6 x^{3}$	+	$37 x^{2}$	-	$34 x$	+	$7$
			-	$6 x^{3}$	+	$3 x^{2}$
					+	$40 x^{2}$	-	$34 x$

Multiply the new quotient term by the divisor.

				$3 x^{2}$	+	$20 x$
$2 x$	-	$1$		$6 x^{3}$	+	$37 x^{2}$	-	$34 x$	+	$7$
			-	$6 x^{3}$	+	$3 x^{2}$
					+	$40 x^{2}$	-	$34 x$
					+	$40 x^{2}$	-	$20 x$

The expression needs to be subtracted from the dividend, so change all the signs in $40 x^{2} - 20 x$

				$3 x^{2}$	+	$20 x$
$2 x$	-	$1$		$6 x^{3}$	+	$37 x^{2}$	-	$34 x$	+	$7$
			-	$6 x^{3}$	+	$3 x^{2}$
					+	$40 x^{2}$	-	$34 x$
					-	$40 x^{2}$	+	$20 x$

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

				$3 x^{2}$	+	$20 x$
$2 x$	-	$1$		$6 x^{3}$	+	$37 x^{2}$	-	$34 x$	+	$7$
			-	$6 x^{3}$	+	$3 x^{2}$
					+	$40 x^{2}$	-	$34 x$
					-	$40 x^{2}$	+	$20 x$
							-	$14 x$

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

				$3 x^{2}$	+	$20 x$
$2 x$	-	$1$		$6 x^{3}$	+	$37 x^{2}$	-	$34 x$	+	$7$
			-	$6 x^{3}$	+	$3 x^{2}$
					+	$40 x^{2}$	-	$34 x$
					-	$40 x^{2}$	+	$20 x$
							-	$14 x$	+	$7$

Divide the highest order term in the dividend $- 14 x$ by the highest order term in divisor $2 x$ .

				$3 x^{2}$	+	$20 x$	-	$7$
$2 x$	-	$1$		$6 x^{3}$	+	$37 x^{2}$	-	$34 x$	+	$7$
			-	$6 x^{3}$	+	$3 x^{2}$
					+	$40 x^{2}$	-	$34 x$
					-	$40 x^{2}$	+	$20 x$
							-	$14 x$	+	$7$

Multiply the new quotient term by the divisor.

				$3 x^{2}$	+	$20 x$	-	$7$
$2 x$	-	$1$		$6 x^{3}$	+	$37 x^{2}$	-	$34 x$	+	$7$
			-	$6 x^{3}$	+	$3 x^{2}$
					+	$40 x^{2}$	-	$34 x$
					-	$40 x^{2}$	+	$20 x$
							-	$14 x$	+	$7$
							-	$14 x$	+	$7$

The expression needs to be subtracted from the dividend, so change all the signs in $- 14 x + 7$

				$3 x^{2}$	+	$20 x$	-	$7$
$2 x$	-	$1$		$6 x^{3}$	+	$37 x^{2}$	-	$34 x$	+	$7$
			-	$6 x^{3}$	+	$3 x^{2}$
					+	$40 x^{2}$	-	$34 x$
					-	$40 x^{2}$	+	$20 x$
							-	$14 x$	+	$7$
							+	$14 x$	-	$7$

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

				$3 x^{2}$	+	$20 x$	-	$7$
$2 x$	-	$1$		$6 x^{3}$	+	$37 x^{2}$	-	$34 x$	+	$7$
			-	$6 x^{3}$	+	$3 x^{2}$
					+	$40 x^{2}$	-	$34 x$
					-	$40 x^{2}$	+	$20 x$
							-	$14 x$	+	$7$
							+	$14 x$	-	$7$
										$0$

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

$3 x^{2} + 20 x - 7$

Write $6 x^{3} + 37 x^{2} - 34 x + 7$ as a set of factors.

$(2 x - 1) (3 x^{2} + 20 x - 7) = 0$

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 $a \cdot c = 3 \cdot - 7 = - 21$ and whose sum is $b = 20$ .

Factor $20$ out of $20 x$ .

$(2 x - 1) (3 x^{2} + 20 (x) - 7) = 0$

Rewrite $20$ as $- 1$ plus $21$

$(2 x - 1) (3 x^{2} + (- 1 + 21) x - 7) = 0$

Apply the distributive property.

$(2 x - 1) (3 x^{2} - 1 x + 21 x - 7) = 0$

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

$(2 x - 1) ((3 x^{2} - 1 x) + 21 x - 7) = 0$

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

$(2 x - 1) (x (3 x - 1) + 7 (3 x - 1)) = 0$

Factor the polynomial by factoring out the greatest common factor, $3 x - 1$ .

$(2 x - 1) ((3 x - 1) (x + 7)) = 0$

Remove unnecessary parentheses.

$(2 x - 1) (3 x - 1) (x + 7) = 0$

Step 10

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

$2 x - 1 = 0$

$3 x - 1 = 0$

$x + 7 = 0$

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 $3 x - 1$ equal to $0$ and solve for $x$ .

Set $3 x - 1$ equal to $0$ .

$3 x - 1 = 0$

Solve $3 x - 1 = 0$ for $x$ .

Add $1$ to both sides of the equation.

$3 x = 1$

Divide each term in $3 x = 1$ by $3$ and simplify.

Divide each term in $3 x = 1$ by $3$ .

$\frac{3 x}{3} = \frac{1}{3}$

Simplify the left side.

Cancel the common factor of $3$ .

Cancel the common factor.

$\frac{3 x}{3} = \frac{1}{3}$

Divide $x$ by $1$ .

$x = \frac{1}{3}$

Step 13

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.

$x = - 7$

Step 14

The final solution is all the values that make $(2 x - 1) (3 x - 1) (x + 7) = 0$ true.

$x = \frac{1}{2}, \frac{1}{3}, - 7$

Step 15

Find the Roots/Zeros Using the Rational Roots Test f(x)=6x^3+37x^2-34x+7

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