Find the Roots/Zeros Using the Rational Roots Test h(x)=9x^3-21x^2+4x+4

$h (x) = 9 x^{3} - 21 x^{2} + 4 x + 4$

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.

$p = \pm 1, \pm 2, \pm 4$

$q = \pm 1, \pm 3, \pm 9$

Step 2

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

$\pm 1, \pm \frac{1}{3}, \pm \frac{1}{9}, \pm 2, \pm \frac{2}{3}, \pm \frac{2}{9}, \pm 4, \pm \frac{4}{3}, \pm \frac{4}{9}$

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.

$9 {(2)}^{3} - 21 {(2)}^{2} + 4 (2) + 4$

Step 4

Simplify the expression. In this case, the expression is equal to $0$ so $x = 2$ is a root of the polynomial.

Simplify each term.

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

$9 \cdot 8 - 21 {(2)}^{2} + 4 (2) + 4$

Multiply $9$ by $8$ .

$72 - 21 {(2)}^{2} + 4 (2) + 4$

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

$72 - 21 \cdot 4 + 4 (2) + 4$

Multiply $- 21$ by $4$ .

$72 - 84 + 4 (2) + 4$

Multiply $4$ by $2$ .

$72 - 84 + 8 + 4$

Simplify by adding and subtracting.

Subtract $84$ from $72$ .

$- 12 + 8 + 4$

Add $- 12$ and $8$ .

$- 4 + 4$

Add $- 4$ and $4$ .

$0$

Step 5

Since $2$ is a known root, divide the polynomial by $x - 2$ to find the quotient polynomial. This polynomial can then be used to find the remaining roots.

$\frac{9 x^{3} - 21 x^{2} + 4 x + 4}{x - 2}$

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.

$2$	$9$	$- 21$	$4$	$4$
	\xa0			\xa0

The first number in the dividend $(9)$ is put into the first position of the result area (below the horizontal line).

$2$	$9$	$- 21$	$4$	$4$
	\xa0			\xa0
	$9$

Multiply the newest entry in the result $(9)$ by the divisor $(2)$ and place the result of $(18)$ under the next term in the dividend $(- 21)$ .

$2$	$9$	$- 21$	$4$	$4$
	\xa0	$18$		\xa0
	$9$

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

$2$	$9$	$- 21$	$4$	$4$
	\xa0	$18$		\xa0
	$9$	$- 3$

Multiply the newest entry in the result $(- 3)$ by the divisor $(2)$ and place the result of $(- 6)$ under the next term in the dividend $(4)$ .

$2$	$9$	$- 21$	$4$	$4$
	\xa0	$18$	$- 6$	\xa0
	$9$	$- 3$

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

$2$	$9$	$- 21$	$4$	$4$
	\xa0	$18$	$- 6$	\xa0
	$9$	$- 3$	$- 2$

Multiply the newest entry in the result $(- 2)$ by the divisor $(2)$ and place the result of $(- 4)$ under the next term in the dividend $(4)$ .

$2$	$9$	$- 21$	$4$	$4$
	\xa0	$18$	$- 6$	$- 4$
	$9$	$- 3$	$- 2$

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

$2$	$9$	$- 21$	$4$	$4$
	\xa0	$18$	$- 6$	$- 4$
	$9$	$- 3$	$- 2$	$0$

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

$9 x^{2} + - 3 x - 2$

Simplify the quotient polynomial.

$9 x^{2} - 3 x - 2$

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 $a \cdot c = 9 \cdot - 2 = - 18$ and whose sum is $b = - 3$ .

Factor $- 3$ out of $- 3 x$ .

$(x - 2) + 9 x^{2} - 3 x - 2$

Rewrite $- 3$ as $3$ plus $- 6$

$(x - 2) + 9 x^{2} + (3 - 6) x - 2$

Apply the distributive property.

$(x - 2) + 9 x^{2} + 3 x - 6 x - 2$

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

$(x - 2) (9 x^{2} + 3 x) - 6 x - 2$

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

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

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

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

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

Step 8

Factor the left side of the equation.

Factor $9 x^{3} - 21 x^{2} + 4 x + 4$ 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.

$p = \pm 1, \pm 4, \pm 2 q = \pm 1, \pm 9, \pm 3$

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

$\pm 1, \pm 0.\overline{1}, \pm 0.\overline{3}, \pm 4, \pm 0.\overline{4}, \pm 1.\overline{3}, \pm 2, \pm 0.\overline{2}, \pm 0.\overline{6}$

Substitute $- 0.\overline{3}$ and simplify the expression. In this case, the expression is equal to $0$ so $- 0.\overline{3}$ is a root of the polynomial.

Substitute $- 0.\overline{3}$ into the polynomial.

$9 {(- 0.\overline{3})}^{3} - 21 {(- 0.\overline{3})}^{2} + 4 \cdot - 0.\overline{3} + 4$

Raise $- 0.\overline{3}$ to the power of $3$ .

$9 \cdot - 0.\overline{037} - 21 {(- 0.\overline{3})}^{2} + 4 \cdot - 0.\overline{3} + 4$

Multiply $9$ by $- 0.\overline{037}$ .

$- 0.\overline{3} - 21 {(- 0.\overline{3})}^{2} + 4 \cdot - 0.\overline{3} + 4$

Raise $- 0.\overline{3}$ to the power of $2$ .

$- 0.\overline{3} - 21 \cdot 0.\overline{1} + 4 \cdot - 0.\overline{3} + 4$

Multiply $- 21$ by $0.\overline{1}$ .

$- 0.\overline{3} - 2.\overline{3} + 4 \cdot - 0.\overline{3} + 4$

Subtract $2.\overline{3}$ from $- 0.\overline{3}$ .

$- 2.\overline{6} + 4 \cdot - 0.\overline{3} + 4$

Multiply $4$ by $- 0.\overline{3}$ .

$- 2.\overline{6} - 1.\overline{3} + 4$

Subtract $1.\overline{3}$ from $- 2.\overline{6}$ .

$- 4 + 4$

Add $- 4$ and $4$ .

$0$

Since $- 0.\overline{3}$ is a known root, divide the polynomial by $3 x + 1$ to find the quotient polynomial. This polynomial can then be used to find the remaining roots.

$\frac{9 x^{3} - 21 x^{2} + 4 x + 4}{3 x + 1}$

Divide $9 x^{3} - 21 x^{2} + 4 x + 4$ by $3 x + 1$ .

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

$3 x$

$1$

$9 x^{3}$

$21 x^{2}$

$4 x$

$4$

Divide the highest order term in the dividend $9 x^{3}$ by the highest order term in divisor $3 x$ .

					$3 x^{2}$
	$3 x$	+	$1$		$9 x^{3}$	-	$21 x^{2}$	+	$4 x$	+	$4$

Multiply the new quotient term by the divisor.

				$3 x^{2}$
$3 x$	+	$1$		$9 x^{3}$	-	$21 x^{2}$	+	$4 x$	+	$4$
			+	$9 x^{3}$	+	$3 x^{2}$

The expression needs to be subtracted from the dividend, so change all the signs in $9 x^{3} + 3 x^{2}$

				$3 x^{2}$
$3 x$	+	$1$		$9 x^{3}$	-	$21 x^{2}$	+	$4 x$	+	$4$
			-	$9 x^{3}$	-	$3 x^{2}$

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

				$3 x^{2}$
$3 x$	+	$1$		$9 x^{3}$	-	$21 x^{2}$	+	$4 x$	+	$4$
			-	$9 x^{3}$	-	$3 x^{2}$
					-	$24 x^{2}$

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

				$3 x^{2}$
$3 x$	+	$1$		$9 x^{3}$	-	$21 x^{2}$	+	$4 x$	+	$4$
			-	$9 x^{3}$	-	$3 x^{2}$
					-	$24 x^{2}$	+	$4 x$

Divide the highest order term in the dividend $- 24 x^{2}$ by the highest order term in divisor $3 x$ .

				$3 x^{2}$	-	$8 x$
$3 x$	+	$1$		$9 x^{3}$	-	$21 x^{2}$	+	$4 x$	+	$4$
			-	$9 x^{3}$	-	$3 x^{2}$
					-	$24 x^{2}$	+	$4 x$

Multiply the new quotient term by the divisor.

				$3 x^{2}$	-	$8 x$
$3 x$	+	$1$		$9 x^{3}$	-	$21 x^{2}$	+	$4 x$	+	$4$
			-	$9 x^{3}$	-	$3 x^{2}$
					-	$24 x^{2}$	+	$4 x$
					-	$24 x^{2}$	-	$8 x$

The expression needs to be subtracted from the dividend, so change all the signs in $- 24 x^{2} - 8 x$

				$3 x^{2}$	-	$8 x$
$3 x$	+	$1$		$9 x^{3}$	-	$21 x^{2}$	+	$4 x$	+	$4$
			-	$9 x^{3}$	-	$3 x^{2}$
					-	$24 x^{2}$	+	$4 x$
					+	$24 x^{2}$	+	$8 x$

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

				$3 x^{2}$	-	$8 x$
$3 x$	+	$1$		$9 x^{3}$	-	$21 x^{2}$	+	$4 x$	+	$4$
			-	$9 x^{3}$	-	$3 x^{2}$
					-	$24 x^{2}$	+	$4 x$
					+	$24 x^{2}$	+	$8 x$
							+	$12 x$

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

				$3 x^{2}$	-	$8 x$
$3 x$	+	$1$		$9 x^{3}$	-	$21 x^{2}$	+	$4 x$	+	$4$
			-	$9 x^{3}$	-	$3 x^{2}$
					-	$24 x^{2}$	+	$4 x$
					+	$24 x^{2}$	+	$8 x$
							+	$12 x$	+	$4$

Divide the highest order term in the dividend $12 x$ by the highest order term in divisor $3 x$ .

				$3 x^{2}$	-	$8 x$	+	$4$
$3 x$	+	$1$		$9 x^{3}$	-	$21 x^{2}$	+	$4 x$	+	$4$
			-	$9 x^{3}$	-	$3 x^{2}$
					-	$24 x^{2}$	+	$4 x$
					+	$24 x^{2}$	+	$8 x$
							+	$12 x$	+	$4$

Multiply the new quotient term by the divisor.

				$3 x^{2}$	-	$8 x$	+	$4$
$3 x$	+	$1$		$9 x^{3}$	-	$21 x^{2}$	+	$4 x$	+	$4$
			-	$9 x^{3}$	-	$3 x^{2}$
					-	$24 x^{2}$	+	$4 x$
					+	$24 x^{2}$	+	$8 x$
							+	$12 x$	+	$4$
							+	$12 x$	+	$4$

The expression needs to be subtracted from the dividend, so change all the signs in $12 x + 4$

				$3 x^{2}$	-	$8 x$	+	$4$
$3 x$	+	$1$		$9 x^{3}$	-	$21 x^{2}$	+	$4 x$	+	$4$
			-	$9 x^{3}$	-	$3 x^{2}$
					-	$24 x^{2}$	+	$4 x$
					+	$24 x^{2}$	+	$8 x$
							+	$12 x$	+	$4$
							-	$12 x$	-	$4$

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

				$3 x^{2}$	-	$8 x$	+	$4$
$3 x$	+	$1$		$9 x^{3}$	-	$21 x^{2}$	+	$4 x$	+	$4$
			-	$9 x^{3}$	-	$3 x^{2}$
					-	$24 x^{2}$	+	$4 x$
					+	$24 x^{2}$	+	$8 x$
							+	$12 x$	+	$4$
							-	$12 x$	-	$4$
										$0$

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

$3 x^{2} - 8 x + 4$

Write $9 x^{3} - 21 x^{2} + 4 x + 4$ as a set of factors.

$(3 x + 1) (3 x^{2} - 8 x + 4) = 0$

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 $a \cdot c = 3 \cdot 4 = 12$ and whose sum is $b = - 8$ .

Factor $- 8$ out of $- 8 x$ .

$(3 x + 1) (3 x^{2} - 8 x + 4) = 0$

Rewrite $- 8$ as $- 2$ plus $- 6$

$(3 x + 1) (3 x^{2} + (- 2 - 6) x + 4) = 0$

Apply the distributive property.

$(3 x + 1) (3 x^{2} - 2 x - 6 x + 4) = 0$

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

$(3 x + 1) ((3 x^{2} - 2 x) - 6 x + 4) = 0$

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

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

Factor the polynomial by factoring out the greatest common factor, $3 x - 2$ .

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

Remove unnecessary parentheses.

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

Step 9

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

$3 x + 1 = 0$

$3 x - 2 = 0$

$x - 2 = 0$

Step 10

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 $x$ .

Subtract $1$ from 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 $1$ .

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

Simplify the right side.

Move the negative in front of the fraction.

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

Step 11

Set $3 x - 2$ equal to $0$ and solve for $x$ .

Set $3 x - 2$ equal to $0$ .

$3 x - 2 = 0$

Solve $3 x - 2 = 0$ for $x$ .

Add $2$ to both sides of the equation.

$3 x = 2$

Divide each term in $3 x = 2$ by $3$ and simplify.

Divide each term in $3 x = 2$ by $3$ .

$\frac{3 x}{3} = \frac{2}{3}$

Simplify the left side.

Cancel the common factor of $3$ .

Cancel the common factor.

$\frac{3 x}{3} = \frac{2}{3}$

Divide $x$ by $1$ .

$x = \frac{2}{3}$

Step 12

Set $x - 2$ equal to $0$ and solve for $x$ .

Set $x - 2$ equal to $0$ .

$x - 2 = 0$

Add $2$ to both sides of the equation.

$x = 2$

Step 13

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

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

Step 14

Find the Roots/Zeros Using the Rational Roots Test h(x)=9x^3-21x^2+4x+4

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