Find the Roots/Zeros Using the Rational Roots Test f(x)=6x^3+43x^2+6x-7

$f (x) = 6 x^{3} + 43 x^{2} + 6 x - 7$

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.

$6 {(- \frac{1}{2})}^{3} + 43 {(- \frac{1}{2})}^{2} + 6 (- \frac{1}{2}) - 7$

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.

Use the power rule ${(a b)}^{n} = a^{n} b^{n}$ to distribute the exponent.

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

$6 ({(- 1)}^{3} {(\frac{1}{2})}^{3}) + 43 {(- \frac{1}{2})}^{2} + 6 (- \frac{1}{2}) - 7$

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

$6 ({(- 1)}^{3} \frac{1^{3}}{2^{3}}) + 43 {(- \frac{1}{2})}^{2} + 6 (- \frac{1}{2}) - 7$

Raise $- 1$ to the power of $3$ .

$6 (- \frac{1^{3}}{2^{3}}) + 43 {(- \frac{1}{2})}^{2} + 6 (- \frac{1}{2}) - 7$

One to any power is one.

$6 (- \frac{1}{2^{3}}) + 43 {(- \frac{1}{2})}^{2} + 6 (- \frac{1}{2}) - 7$

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

$6 (- \frac{1}{8}) + 43 {(- \frac{1}{2})}^{2} + 6 (- \frac{1}{2}) - 7$

Cancel the common factor of $2$ .

Move the leading negative in $- \frac{1}{8}$ into the numerator.

$6 (\frac{- 1}{8}) + 43 {(- \frac{1}{2})}^{2} + 6 (- \frac{1}{2}) - 7$

Factor $2$ out of $6$ .

$2 (3) \frac{- 1}{8} + 43 {(- \frac{1}{2})}^{2} + 6 (- \frac{1}{2}) - 7$

Factor $2$ out of $8$ .

$2 \cdot 3 \frac{- 1}{2 \cdot 4} + 43 {(- \frac{1}{2})}^{2} + 6 (- \frac{1}{2}) - 7$

Cancel the common factor.

$2 \cdot 3 \frac{- 1}{2 \cdot 4} + 43 {(- \frac{1}{2})}^{2} + 6 (- \frac{1}{2}) - 7$

Rewrite the expression.

$3 (\frac{- 1}{4}) + 43 {(- \frac{1}{2})}^{2} + 6 (- \frac{1}{2}) - 7$

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

$\frac{3 \cdot - 1}{4} + 43 {(- \frac{1}{2})}^{2} + 6 (- \frac{1}{2}) - 7$

Multiply $3$ by $- 1$ .

$\frac{- 3}{4} + 43 {(- \frac{1}{2})}^{2} + 6 (- \frac{1}{2}) - 7$

Move the negative in front of the fraction.

$- \frac{3}{4} + 43 {(- \frac{1}{2})}^{2} + 6 (- \frac{1}{2}) - 7$

Use the power rule ${(a b)}^{n} = a^{n} b^{n}$ to distribute the exponent.

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

$- \frac{3}{4} + 43 ({(- 1)}^{2} {(\frac{1}{2})}^{2}) + 6 (- \frac{1}{2}) - 7$

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

$- \frac{3}{4} + 43 ({(- 1)}^{2} \frac{1^{2}}{2^{2}}) + 6 (- \frac{1}{2}) - 7$

Raise $- 1$ to the power of $2$ .

$- \frac{3}{4} + 43 (1 \frac{1^{2}}{2^{2}}) + 6 (- \frac{1}{2}) - 7$

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

$- \frac{3}{4} + 43 \frac{1^{2}}{2^{2}} + 6 (- \frac{1}{2}) - 7$

One to any power is one.

$- \frac{3}{4} + 43 \frac{1}{2^{2}} + 6 (- \frac{1}{2}) - 7$

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

$- \frac{3}{4} + 43 (\frac{1}{4}) + 6 (- \frac{1}{2}) - 7$

Combine $43$ and $\frac{1}{4}$ .

$- \frac{3}{4} + \frac{43}{4} + 6 (- \frac{1}{2}) - 7$

Cancel the common factor of $2$ .

Move the leading negative in $- \frac{1}{2}$ into the numerator.

$- \frac{3}{4} + \frac{43}{4} + 6 (\frac{- 1}{2}) - 7$

Factor $2$ out of $6$ .

$- \frac{3}{4} + \frac{43}{4} + 2 (3) \frac{- 1}{2} - 7$

Cancel the common factor.

$- \frac{3}{4} + \frac{43}{4} + 2 \cdot 3 \frac{- 1}{2} - 7$

Rewrite the expression.

$- \frac{3}{4} + \frac{43}{4} + 3 \cdot - 1 - 7$

Multiply $3$ by $- 1$ .

$- \frac{3}{4} + \frac{43}{4} - 3 - 7$

Combine fractions.

Combine the numerators over the common denominator.

$- 3 - 7 + \frac{- 3 + 43}{4}$

Add $- 3$ and $43$ .

$- 3 - 7 + \frac{40}{4}$

Find the common denominator.

Write $- 3$ as a fraction with denominator $1$ .

$\frac{- 3}{1} - 7 + \frac{40}{4}$

Multiply $\frac{- 3}{1}$ by $\frac{4}{4}$ .

$\frac{- 3}{1} \cdot \frac{4}{4} - 7 + \frac{40}{4}$

Multiply $\frac{- 3}{1}$ by $\frac{4}{4}$ .

$\frac{- 3 \cdot 4}{4} - 7 + \frac{40}{4}$

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

$\frac{- 3 \cdot 4}{4} + \frac{- 7}{1} + \frac{40}{4}$

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

$\frac{- 3 \cdot 4}{4} + \frac{- 7}{1} \cdot \frac{4}{4} + \frac{40}{4}$

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

$\frac{- 3 \cdot 4}{4} + \frac{- 7 \cdot 4}{4} + \frac{40}{4}$

Combine the numerators over the common denominator.

$\frac{- 3 \cdot 4 - 7 \cdot 4 + 40}{4}$

Simplify each term.

Multiply $- 3$ by $4$ .

$\frac{- 12 - 7 \cdot 4 + 40}{4}$

Multiply $- 7$ by $4$ .

$\frac{- 12 - 28 + 40}{4}$

Simplify the expression.

Subtract $28$ from $- 12$ .

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

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.

$\frac{6 x^{3} + 43 x^{2} + 6 x - 7}{x + \frac{1}{2}}$

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.

$- \frac{1}{2}$	$6$	$43$	$6$	$- 7$
	\xa0			\xa0

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

$- \frac{1}{2}$	$6$	$43$	$6$	$- 7$
	\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 $(43)$ .

$- \frac{1}{2}$	$6$	$43$	$6$	$- 7$
	\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.

$- \frac{1}{2}$	$6$	$43$	$6$	$- 7$
	\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 $(6)$ .

$- \frac{1}{2}$	$6$	$43$	$6$	$- 7$
	\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.

$- \frac{1}{2}$	$6$	$43$	$6$	$- 7$
	\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 $(- 7)$ .

$- \frac{1}{2}$	$6$	$43$	$6$	$- 7$
	\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.

$- \frac{1}{2}$	$6$	$43$	$6$	$- 7$
	\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 $6 x^{3} + 43 x^{2} + 6 x - 7$ 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.

$6 {(- 0.5)}^{3} + 43 {(- 0.5)}^{2} + 6 \cdot - 0.5 - 7$

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

$6 \cdot - 0.125 + 43 {(- 0.5)}^{2} + 6 \cdot - 0.5 - 7$

Multiply $6$ by $- 0.125$ .

$- 0.75 + 43 {(- 0.5)}^{2} + 6 \cdot - 0.5 - 7$

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

$- 0.75 + 43 \cdot 0.25 + 6 \cdot - 0.5 - 7$

Multiply $43$ by $0.25$ .

$- 0.75 + 10.75 + 6 \cdot - 0.5 - 7$

Add $- 0.75$ and $10.75$ .

$10 + 6 \cdot - 0.5 - 7$

Multiply $6$ by $- 0.5$ .

$10 - 3 - 7$

Subtract $3$ from $10$ .

$7 - 7$

Subtract $7$ from $7$ .

$0$

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.

$\frac{6 x^{3} + 43 x^{2} + 6 x - 7}{2 x + 1}$

Divide $6 x^{3} + 43 x^{2} + 6 x - 7$ 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 $0$ .

$2 x$

$1$

$6 x^{3}$

$43 x^{2}$

$6 x$

$7$

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

					$3 x^{2}$
	$2 x$	+	$1$		$6 x^{3}$	+	$43 x^{2}$	+	$6 x$	-	$7$

Multiply the new quotient term by the divisor.

				$3 x^{2}$
$2 x$	+	$1$		$6 x^{3}$	+	$43 x^{2}$	+	$6 x$	-	$7$
			+	$6 x^{3}$	+	$3 x^{2}$

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

				$3 x^{2}$
$2 x$	+	$1$		$6 x^{3}$	+	$43 x^{2}$	+	$6 x$	-	$7$
			-	$6 x^{3}$	-	$3 x^{2}$

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

				$3 x^{2}$
$2 x$	+	$1$		$6 x^{3}$	+	$43 x^{2}$	+	$6 x$	-	$7$
			-	$6 x^{3}$	-	$3 x^{2}$
					+	$40 x^{2}$

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

				$3 x^{2}$
$2 x$	+	$1$		$6 x^{3}$	+	$43 x^{2}$	+	$6 x$	-	$7$
			-	$6 x^{3}$	-	$3 x^{2}$
					+	$40 x^{2}$	+	$6 x$

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

				$3 x^{2}$	+	$20 x$
$2 x$	+	$1$		$6 x^{3}$	+	$43 x^{2}$	+	$6 x$	-	$7$
			-	$6 x^{3}$	-	$3 x^{2}$
					+	$40 x^{2}$	+	$6 x$

Multiply the new quotient term by the divisor.

				$3 x^{2}$	+	$20 x$
$2 x$	+	$1$		$6 x^{3}$	+	$43 x^{2}$	+	$6 x$	-	$7$
			-	$6 x^{3}$	-	$3 x^{2}$
					+	$40 x^{2}$	+	$6 x$
					+	$40 x^{2}$	+	$20 x$

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

				$3 x^{2}$	+	$20 x$
$2 x$	+	$1$		$6 x^{3}$	+	$43 x^{2}$	+	$6 x$	-	$7$
			-	$6 x^{3}$	-	$3 x^{2}$
					+	$40 x^{2}$	+	$6 x$
					-	$40 x^{2}$	-	$20 x$

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

				$3 x^{2}$	+	$20 x$
$2 x$	+	$1$		$6 x^{3}$	+	$43 x^{2}$	+	$6 x$	-	$7$
			-	$6 x^{3}$	-	$3 x^{2}$
					+	$40 x^{2}$	+	$6 x$
					-	$40 x^{2}$	-	$20 x$
							-	$14 x$

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

				$3 x^{2}$	+	$20 x$
$2 x$	+	$1$		$6 x^{3}$	+	$43 x^{2}$	+	$6 x$	-	$7$
			-	$6 x^{3}$	-	$3 x^{2}$
					+	$40 x^{2}$	+	$6 x$
					-	$40 x^{2}$	-	$20 x$
							-	$14 x$	-	$7$

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

				$3 x^{2}$	+	$20 x$	-	$7$
$2 x$	+	$1$		$6 x^{3}$	+	$43 x^{2}$	+	$6 x$	-	$7$
			-	$6 x^{3}$	-	$3 x^{2}$
					+	$40 x^{2}$	+	$6 x$
					-	$40 x^{2}$	-	$20 x$
							-	$14 x$	-	$7$

Multiply the new quotient term by the divisor.

				$3 x^{2}$	+	$20 x$	-	$7$
$2 x$	+	$1$		$6 x^{3}$	+	$43 x^{2}$	+	$6 x$	-	$7$
			-	$6 x^{3}$	-	$3 x^{2}$
					+	$40 x^{2}$	+	$6 x$
					-	$40 x^{2}$	-	$20 x$
							-	$14 x$	-	$7$
							-	$14 x$	-	$7$

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

				$3 x^{2}$	+	$20 x$	-	$7$
$2 x$	+	$1$		$6 x^{3}$	+	$43 x^{2}$	+	$6 x$	-	$7$
			-	$6 x^{3}$	-	$3 x^{2}$
					+	$40 x^{2}$	+	$6 x$
					-	$40 x^{2}$	-	$20 x$
							-	$14 x$	-	$7$
							+	$14 x$	+	$7$

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

				$3 x^{2}$	+	$20 x$	-	$7$
$2 x$	+	$1$		$6 x^{3}$	+	$43 x^{2}$	+	$6 x$	-	$7$
			-	$6 x^{3}$	-	$3 x^{2}$
					+	$40 x^{2}$	+	$6 x$
					-	$40 x^{2}$	-	$20 x$
							-	$14 x$	-	$7$
							+	$14 x$	+	$7$
										$0$

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

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

Write $6 x^{3} + 43 x^{2} + 6 x - 7$ 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 $x$ .

Subtract $1$ from 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 $1$ .

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

Simplify the right side.

Move the negative in front of the fraction.

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

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+43x^2+6x-7

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