Find the Roots/Zeros Using the Rational Roots Test q(x)=6x^3+19x^2-15x-28

$q (x) = 6 x^{3} + 19 x^{2} - 15 x - 28$

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.

$p = \pm 1, \pm 2, \pm 4, \pm 7, \pm 14, \pm 28$

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

Step 2

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

$\pm 1, \pm \frac{1}{2}, \pm \frac{1}{3}, \pm \frac{1}{6}, \pm 2, \pm \frac{2}{3}, \pm 4, \pm \frac{4}{3}, \pm 7, \pm \frac{7}{2}, \pm \frac{7}{3}, \pm \frac{7}{6}, \pm 14, \pm \frac{14}{3}, \pm 28, \pm \frac{28}{3}$

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.

$6 {(- 1)}^{3} + 19 {(- 1)}^{2} - 15 \cdot - 1 - 28$

Step 4

Simplify the expression. In this case, the expression is equal to $0$ so $x = - 1$ is a root of the polynomial.

Simplify each term.

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

$6 \cdot - 1 + 19 {(- 1)}^{2} - 15 \cdot - 1 - 28$

Multiply $6$ by $- 1$ .

$- 6 + 19 {(- 1)}^{2} - 15 \cdot - 1 - 28$

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

$- 6 + 19 \cdot 1 - 15 \cdot - 1 - 28$

Multiply $19$ by $1$ .

$- 6 + 19 - 15 \cdot - 1 - 28$

Multiply $- 15$ by $- 1$ .

$- 6 + 19 + 15 - 28$

Simplify by adding and subtracting.

Add $- 6$ and $19$ .

$13 + 15 - 28$

Add $13$ and $15$ .

$28 - 28$

Subtract $28$ from $28$ .

$0$

Step 5

Since $- 1$ is a known root, divide the polynomial by $x + 1$ to find the quotient polynomial. This polynomial can then be used to find the remaining roots.

$\frac{6 x^{3} + 19 x^{2} - 15 x - 28}{x + 1}$

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.

$- 1$	$6$	$19$	$- 15$	$- 28$
	\xa0			\xa0

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

$- 1$	$6$	$19$	$- 15$	$- 28$
	\xa0			\xa0
	$6$

Multiply the newest entry in the result $(6)$ by the divisor $(- 1)$ and place the result of $(- 6)$ under the next term in the dividend $(19)$ .

$- 1$	$6$	$19$	$- 15$	$- 28$
	\xa0	$- 6$		\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.

$- 1$	$6$	$19$	$- 15$	$- 28$
	\xa0	$- 6$		\xa0
	$6$	$13$

Multiply the newest entry in the result $(13)$ by the divisor $(- 1)$ and place the result of $(- 13)$ under the next term in the dividend $(- 15)$ .

$- 1$	$6$	$19$	$- 15$	$- 28$
	\xa0	$- 6$	$- 13$	\xa0
	$6$	$13$

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

$- 1$	$6$	$19$	$- 15$	$- 28$
	\xa0	$- 6$	$- 13$	\xa0
	$6$	$13$	$- 28$

Multiply the newest entry in the result $(- 28)$ by the divisor $(- 1)$ and place the result of $(28)$ under the next term in the dividend $(- 28)$ .

$- 1$	$6$	$19$	$- 15$	$- 28$
	\xa0	$- 6$	$- 13$	$28$
	$6$	$13$	$- 28$

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

$- 1$	$6$	$19$	$- 15$	$- 28$
	\xa0	$- 6$	$- 13$	$28$
	$6$	$13$	$- 28$	$0$

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

$6 x^{2} + 13 x - 28$

Step 7

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 $a \cdot c = 6 \cdot - 28 = - 168$ and whose sum is $b = 13$ .

Factor $13$ out of $13 x$ .

$(x + 1) + 6 x^{2} + 13 (x) - 28$

Rewrite $13$ as $- 8$ plus $21$

$(x + 1) + 6 x^{2} + (- 8 + 21) x - 28$

Apply the distributive property.

$(x + 1) + 6 x^{2} - 8 x + 21 x - 28$

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

$(x + 1) (6 x^{2} - 8 x) + 21 x - 28$

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

$(x + 1) + 2 x (3 x - 4) + 7 (3 x - 4)$

Factor the polynomial by factoring out the greatest common factor, $3 x - 4$ .

$(x + 1) + (3 x - 4) (2 x + 7)$

$(x + 1) (3 x - 4) (2 x + 7)$

Step 8

Factor the left side of the equation.

Factor $6 x^{3} + 19 x^{2} - 15 x - 28$ 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.

$p = \pm 1, \pm 28, \pm 2, \pm 14, \pm 4, \pm 7 q = \pm 1, \pm 6, \pm 2, \pm 3$

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

$\pm 1, \pm 0.1\overline{6}, \pm 0.5, \pm 0.\overline{3}, \pm 28, \pm 4.\overline{6}, \pm 14, \pm 9.\overline{3}, \pm 2, \pm 0.\overline{6}, \pm 2.\overline{3}, \pm 7, \pm 4, \pm 1.\overline{3}, \pm 1.1\overline{6}, \pm 3.5$

Substitute $- 1$ and simplify the expression. In this case, the expression is equal to $0$ so $- 1$ is a root of the polynomial.

Substitute $- 1$ into the polynomial.

$6 {(- 1)}^{3} + 19 {(- 1)}^{2} - 15 \cdot - 1 - 28$

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

$6 \cdot - 1 + 19 {(- 1)}^{2} - 15 \cdot - 1 - 28$

Multiply $6$ by $- 1$ .

$- 6 + 19 {(- 1)}^{2} - 15 \cdot - 1 - 28$

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

$- 6 + 19 \cdot 1 - 15 \cdot - 1 - 28$

Multiply $19$ by $1$ .

$- 6 + 19 - 15 \cdot - 1 - 28$

Add $- 6$ and $19$ .

$13 - 15 \cdot - 1 - 28$

Multiply $- 15$ by $- 1$ .

$13 + 15 - 28$

Add $13$ and $15$ .

$28 - 28$

Subtract $28$ from $28$ .

$0$

Since $- 1$ is a known root, divide the polynomial by $x + 1$ to find the quotient polynomial. This polynomial can then be used to find the remaining roots.

$\frac{6 x^{3} + 19 x^{2} - 15 x - 28}{x + 1}$

Divide $6 x^{3} + 19 x^{2} - 15 x - 28$ by $x + 1$ .

Set up the polynomials to be divided. If there is not a term for every exponent, insert one with a value of $0$ .

$x$

$1$

$6 x^{3}$

$19 x^{2}$

$15 x$

$28$

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

					$6 x^{2}$
	$x$	+	$1$		$6 x^{3}$	+	$19 x^{2}$	-	$15 x$	-	$28$

Multiply the new quotient term by the divisor.

				$6 x^{2}$
$x$	+	$1$		$6 x^{3}$	+	$19 x^{2}$	-	$15 x$	-	$28$
			+	$6 x^{3}$	+	$6 x^{2}$

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

				$6 x^{2}$
$x$	+	$1$		$6 x^{3}$	+	$19 x^{2}$	-	$15 x$	-	$28$
			-	$6 x^{3}$	-	$6 x^{2}$

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

				$6 x^{2}$
$x$	+	$1$		$6 x^{3}$	+	$19 x^{2}$	-	$15 x$	-	$28$
			-	$6 x^{3}$	-	$6 x^{2}$
					+	$13 x^{2}$

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

				$6 x^{2}$
$x$	+	$1$		$6 x^{3}$	+	$19 x^{2}$	-	$15 x$	-	$28$
			-	$6 x^{3}$	-	$6 x^{2}$
					+	$13 x^{2}$	-	$15 x$

Divide the highest order term in the dividend $13 x^{2}$ by the highest order term in divisor $x$ .

				$6 x^{2}$	+	$13 x$
$x$	+	$1$		$6 x^{3}$	+	$19 x^{2}$	-	$15 x$	-	$28$
			-	$6 x^{3}$	-	$6 x^{2}$
					+	$13 x^{2}$	-	$15 x$

Multiply the new quotient term by the divisor.

				$6 x^{2}$	+	$13 x$
$x$	+	$1$		$6 x^{3}$	+	$19 x^{2}$	-	$15 x$	-	$28$
			-	$6 x^{3}$	-	$6 x^{2}$
					+	$13 x^{2}$	-	$15 x$
					+	$13 x^{2}$	+	$13 x$

The expression needs to be subtracted from the dividend, so change all the signs in $13 x^{2} + 13 x$

				$6 x^{2}$	+	$13 x$
$x$	+	$1$		$6 x^{3}$	+	$19 x^{2}$	-	$15 x$	-	$28$
			-	$6 x^{3}$	-	$6 x^{2}$
					+	$13 x^{2}$	-	$15 x$
					-	$13 x^{2}$	-	$13 x$

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

				$6 x^{2}$	+	$13 x$
$x$	+	$1$		$6 x^{3}$	+	$19 x^{2}$	-	$15 x$	-	$28$
			-	$6 x^{3}$	-	$6 x^{2}$
					+	$13 x^{2}$	-	$15 x$
					-	$13 x^{2}$	-	$13 x$
							-	$28 x$

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

				$6 x^{2}$	+	$13 x$
$x$	+	$1$		$6 x^{3}$	+	$19 x^{2}$	-	$15 x$	-	$28$
			-	$6 x^{3}$	-	$6 x^{2}$
					+	$13 x^{2}$	-	$15 x$
					-	$13 x^{2}$	-	$13 x$
							-	$28 x$	-	$28$

Divide the highest order term in the dividend $- 28 x$ by the highest order term in divisor $x$ .

				$6 x^{2}$	+	$13 x$	-	$28$
$x$	+	$1$		$6 x^{3}$	+	$19 x^{2}$	-	$15 x$	-	$28$
			-	$6 x^{3}$	-	$6 x^{2}$
					+	$13 x^{2}$	-	$15 x$
					-	$13 x^{2}$	-	$13 x$
							-	$28 x$	-	$28$

Multiply the new quotient term by the divisor.

				$6 x^{2}$	+	$13 x$	-	$28$
$x$	+	$1$		$6 x^{3}$	+	$19 x^{2}$	-	$15 x$	-	$28$
			-	$6 x^{3}$	-	$6 x^{2}$
					+	$13 x^{2}$	-	$15 x$
					-	$13 x^{2}$	-	$13 x$
							-	$28 x$	-	$28$
							-	$28 x$	-	$28$

The expression needs to be subtracted from the dividend, so change all the signs in $- 28 x - 28$

				$6 x^{2}$	+	$13 x$	-	$28$
$x$	+	$1$		$6 x^{3}$	+	$19 x^{2}$	-	$15 x$	-	$28$
			-	$6 x^{3}$	-	$6 x^{2}$
					+	$13 x^{2}$	-	$15 x$
					-	$13 x^{2}$	-	$13 x$
							-	$28 x$	-	$28$
							+	$28 x$	+	$28$

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

				$6 x^{2}$	+	$13 x$	-	$28$
$x$	+	$1$		$6 x^{3}$	+	$19 x^{2}$	-	$15 x$	-	$28$
			-	$6 x^{3}$	-	$6 x^{2}$
					+	$13 x^{2}$	-	$15 x$
					-	$13 x^{2}$	-	$13 x$
							-	$28 x$	-	$28$
							+	$28 x$	+	$28$
										$0$

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

$6 x^{2} + 13 x - 28$

Write $6 x^{3} + 19 x^{2} - 15 x - 28$ as a set of factors.

$(x + 1) (6 x^{2} + 13 x - 28) = 0$

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 $a \cdot c = 6 \cdot - 28 = - 168$ and whose sum is $b = 13$ .

Factor $13$ out of $13 x$ .

$(x + 1) (6 x^{2} + 13 (x) - 28) = 0$

Rewrite $13$ as $- 8$ plus $21$

$(x + 1) (6 x^{2} + (- 8 + 21) x - 28) = 0$

Apply the distributive property.

$(x + 1) (6 x^{2} - 8 x + 21 x - 28) = 0$

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

$(x + 1) ((6 x^{2} - 8 x) + 21 x - 28) = 0$

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

$(x + 1) (2 x (3 x - 4) + 7 (3 x - 4)) = 0$

Factor the polynomial by factoring out the greatest common factor, $3 x - 4$ .

$(x + 1) ((3 x - 4) (2 x + 7)) = 0$

Remove unnecessary parentheses.

$(x + 1) (3 x - 4) (2 x + 7) = 0$

Step 9

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

$x + 1 = 0$

$3 x - 4 = 0$

$2 x + 7 = 0$

Step 10

Set $x + 1$ equal to $0$ and solve for $x$ .

Set $x + 1$ equal to $0$ .

$x + 1 = 0$

Subtract $1$ from both sides of the equation.

$x = - 1$

Step 11

Set $3 x - 4$ equal to $0$ and solve for $x$ .

Set $3 x - 4$ equal to $0$ .

$3 x - 4 = 0$

Solve $3 x - 4 = 0$ for $x$ .

Add $4$ to both sides of the equation.

$3 x = 4$

Divide each term in $3 x = 4$ by $3$ and simplify.

Divide each term in $3 x = 4$ by $3$ .

$\frac{3 x}{3} = \frac{4}{3}$

Simplify the left side.

Cancel the common factor of $3$ .

Cancel the common factor.

$\frac{3 x}{3} = \frac{4}{3}$

Divide $x$ by $1$ .

$x = \frac{4}{3}$

Step 12

Set $2 x + 7$ equal to $0$ and solve for $x$ .

Set $2 x + 7$ equal to $0$ .

$2 x + 7 = 0$

Solve $2 x + 7 = 0$ for $x$ .

Subtract $7$ from both sides of the equation.

$2 x = - 7$

Divide each term in $2 x = - 7$ by $2$ and simplify.

Divide each term in $2 x = - 7$ by $2$ .

$\frac{2 x}{2} = \frac{- 7}{2}$

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

$\frac{2 x}{2} = \frac{- 7}{2}$

Divide $x$ by $1$ .

$x = \frac{- 7}{2}$

Simplify the right side.

Move the negative in front of the fraction.

$x = - \frac{7}{2}$

Step 13

The final solution is all the values that make $(x + 1) (3 x - 4) (2 x + 7) = 0$ true.

$x = - 1, \frac{4}{3}, - \frac{7}{2}$

Step 14

Find the Roots/Zeros Using the Rational Roots Test q(x)=6x^3+19x^2-15x-28

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