Find the Roots/Zeros Using the Rational Roots Test 9x^3+9x^2-16x+4

$9 x^{3} + 9 x^{2} - 16 x + 4$

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.

$9 {(- 2)}^{3} + 9 {(- 2)}^{2} - 16 \cdot - 2 + 4$

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 $3$ .

$9 \cdot - 8 + 9 {(- 2)}^{2} - 16 \cdot - 2 + 4$

Multiply $9$ by $- 8$ .

$- 72 + 9 {(- 2)}^{2} - 16 \cdot - 2 + 4$

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

$- 72 + 9 \cdot 4 - 16 \cdot - 2 + 4$

Multiply $9$ by $4$ .

$- 72 + 36 - 16 \cdot - 2 + 4$

Multiply $- 16$ by $- 2$ .

$- 72 + 36 + 32 + 4$

Simplify by adding numbers.

Add $- 72$ and $36$ .

$- 36 + 32 + 4$

Add $- 36$ and $32$ .

$- 4 + 4$

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.

$\frac{9 x^{3} + 9 x^{2} - 16 x + 4}{x + 2}$

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.

$- 2$	$9$	$9$	$- 16$	$4$
	\xa0			\xa0

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

$- 2$	$9$	$9$	$- 16$	$4$
	\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 $(9)$ .

$- 2$	$9$	$9$	$- 16$	$4$
	\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.

$- 2$	$9$	$9$	$- 16$	$4$
	\xa0	$- 18$		\xa0
	$9$	$- 9$

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 $(- 16)$ .

$- 2$	$9$	$9$	$- 16$	$4$
	\xa0	$- 18$	$18$	\xa0
	$9$	$- 9$

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

$- 2$	$9$	$9$	$- 16$	$4$
	\xa0	$- 18$	$18$	\xa0
	$9$	$- 9$	$2$

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 $(4)$ .

$- 2$	$9$	$9$	$- 16$	$4$
	\xa0	$- 18$	$18$	$- 4$
	$9$	$- 9$	$2$

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

$- 2$	$9$	$9$	$- 16$	$4$
	\xa0	$- 18$	$18$	$- 4$
	$9$	$- 9$	$2$	$0$

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

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

Simplify the quotient polynomial.

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

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 $b = - 9$ .

Factor $- 9$ out of $- 9 x$ .

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

Rewrite $- 9$ 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 $9 x^{3} + 9 x^{2} - 16 x + 4$ 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.

$9 \cdot {0.\overline{3}}^{3} + 9 \cdot {0.\overline{3}}^{2} - 16 \cdot 0.\overline{3} + 4$

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

$9 \cdot 0.\overline{037} + 9 \cdot {0.\overline{3}}^{2} - 16 \cdot 0.\overline{3} + 4$

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

$0.\overline{3} + 9 \cdot {0.\overline{3}}^{2} - 16 \cdot 0.\overline{3} + 4$

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

$0.\overline{3} + 9 \cdot 0.\overline{1} - 16 \cdot 0.\overline{3} + 4$

Multiply $9$ by $0.\overline{1}$ .

$0.\overline{3} + 1 - 16 \cdot 0.\overline{3} + 4$

Add $0.\overline{3}$ and $1$ .

$1.\overline{3} - 16 \cdot 0.\overline{3} + 4$

Multiply $- 16$ by $0.\overline{3}$ .

$1.\overline{3} - 5.33333333 + 4$

Subtract $5.33333333$ from $1.\overline{3}$ .

$- 4 + 4$

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.

$\frac{9 x^{3} + 9 x^{2} - 16 x + 4}{3 x - 1}$

Divide $9 x^{3} + 9 x^{2} - 16 x + 4$ 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 $0$ .

$3 x$

$1$

$9 x^{3}$

$9 x^{2}$

$16 x$

$4$

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

					$3 x^{2}$
	$3 x$	-	$1$		$9 x^{3}$	+	$9 x^{2}$	-	$16 x$	+	$4$

Multiply the new quotient term by the divisor.

				$3 x^{2}$
$3 x$	-	$1$		$9 x^{3}$	+	$9 x^{2}$	-	$16 x$	+	$4$
			+	$9 x^{3}$	-	$3 x^{2}$

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

				$3 x^{2}$
$3 x$	-	$1$		$9 x^{3}$	+	$9 x^{2}$	-	$16 x$	+	$4$
			-	$9 x^{3}$	+	$3 x^{2}$

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

				$3 x^{2}$
$3 x$	-	$1$		$9 x^{3}$	+	$9 x^{2}$	-	$16 x$	+	$4$
			-	$9 x^{3}$	+	$3 x^{2}$
					+	$12 x^{2}$

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

				$3 x^{2}$
$3 x$	-	$1$		$9 x^{3}$	+	$9 x^{2}$	-	$16 x$	+	$4$
			-	$9 x^{3}$	+	$3 x^{2}$
					+	$12 x^{2}$	-	$16 x$

Divide the highest order term in the dividend $12 x^{2}$ by the highest order term in divisor $3 x$ .

				$3 x^{2}$	+	$4 x$
$3 x$	-	$1$		$9 x^{3}$	+	$9 x^{2}$	-	$16 x$	+	$4$
			-	$9 x^{3}$	+	$3 x^{2}$
					+	$12 x^{2}$	-	$16 x$

Multiply the new quotient term by the divisor.

				$3 x^{2}$	+	$4 x$
$3 x$	-	$1$		$9 x^{3}$	+	$9 x^{2}$	-	$16 x$	+	$4$
			-	$9 x^{3}$	+	$3 x^{2}$
					+	$12 x^{2}$	-	$16 x$
					+	$12 x^{2}$	-	$4 x$

The expression needs to be subtracted from the dividend, so change all the signs in $12 x^{2} - 4 x$

				$3 x^{2}$	+	$4 x$
$3 x$	-	$1$		$9 x^{3}$	+	$9 x^{2}$	-	$16 x$	+	$4$
			-	$9 x^{3}$	+	$3 x^{2}$
					+	$12 x^{2}$	-	$16 x$
					-	$12 x^{2}$	+	$4 x$

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

				$3 x^{2}$	+	$4 x$
$3 x$	-	$1$		$9 x^{3}$	+	$9 x^{2}$	-	$16 x$	+	$4$
			-	$9 x^{3}$	+	$3 x^{2}$
					+	$12 x^{2}$	-	$16 x$
					-	$12 x^{2}$	+	$4 x$
							-	$12 x$

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

				$3 x^{2}$	+	$4 x$
$3 x$	-	$1$		$9 x^{3}$	+	$9 x^{2}$	-	$16 x$	+	$4$
			-	$9 x^{3}$	+	$3 x^{2}$
					+	$12 x^{2}$	-	$16 x$
					-	$12 x^{2}$	+	$4 x$
							-	$12 x$	+	$4$

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

				$3 x^{2}$	+	$4 x$	-	$4$
$3 x$	-	$1$		$9 x^{3}$	+	$9 x^{2}$	-	$16 x$	+	$4$
			-	$9 x^{3}$	+	$3 x^{2}$
					+	$12 x^{2}$	-	$16 x$
					-	$12 x^{2}$	+	$4 x$
							-	$12 x$	+	$4$

Multiply the new quotient term by the divisor.

				$3 x^{2}$	+	$4 x$	-	$4$
$3 x$	-	$1$		$9 x^{3}$	+	$9 x^{2}$	-	$16 x$	+	$4$
			-	$9 x^{3}$	+	$3 x^{2}$
					+	$12 x^{2}$	-	$16 x$
					-	$12 x^{2}$	+	$4 x$
							-	$12 x$	+	$4$
							-	$12 x$	+	$4$

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

				$3 x^{2}$	+	$4 x$	-	$4$
$3 x$	-	$1$		$9 x^{3}$	+	$9 x^{2}$	-	$16 x$	+	$4$
			-	$9 x^{3}$	+	$3 x^{2}$
					+	$12 x^{2}$	-	$16 x$
					-	$12 x^{2}$	+	$4 x$
							-	$12 x$	+	$4$
							+	$12 x$	-	$4$

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

				$3 x^{2}$	+	$4 x$	-	$4$
$3 x$	-	$1$		$9 x^{3}$	+	$9 x^{2}$	-	$16 x$	+	$4$
			-	$9 x^{3}$	+	$3 x^{2}$
					+	$12 x^{2}$	-	$16 x$
					-	$12 x^{2}$	+	$4 x$
							-	$12 x$	+	$4$
							+	$12 x$	-	$4$
										$0$

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

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

Write $9 x^{3} + 9 x^{2} - 16 x + 4$ as a set of factors.

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

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 $b = 4$ .

Factor $4$ out of $4 x$ .

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

Rewrite $4$ 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 $x$ .

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 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 $0$ .

$x + 2 = 0$

Subtract $2$ from 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 9x^3+9x^2-16x+4

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