Find the Roots/Zeros Using the Rational Roots Test 3x^3-19x^2+33x-9

$3 x^{3} - 19 x^{2} + 33 x - 9$

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.

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

$q = \pm 1, \pm 3$

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

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

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.

$3 {(3)}^{3} - 19 {(3)}^{2} + 33 (3) - 9$

Simplify the expression. In this case, the expression is equal to $0$ so $x = 3$ is a root of the polynomial.

Simplify each term.

Multiply $3$ by ${(3)}^{3}$ by adding the exponents.

Multiply $3$ by ${(3)}^{3}$ .

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

$3^{1} {(3)}^{3} - 19 {(3)}^{2} + 33 (3) - 9$

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$3^{1 + 3} - 19 {(3)}^{2} + 33 (3) - 9$

Add $1$ and $3$ .

$3^{4} - 19 {(3)}^{2} + 33 (3) - 9$

Raise $3$ to the power of $4$ .

$81 - 19 {(3)}^{2} + 33 (3) - 9$

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

$81 - 19 \cdot 9 + 33 (3) - 9$

Multiply $- 19$ by $9$ .

$81 - 171 + 33 (3) - 9$

Multiply $33$ by $3$ .

$81 - 171 + 99 - 9$

Simplify by adding and subtracting.

Subtract $171$ from $81$ .

$- 90 + 99 - 9$

Add $- 90$ and $99$ .

$9 - 9$

Subtract $9$ from $9$ .

$0$

Since $3$ is a known root, divide the polynomial by $x - 3$ to find the quotient polynomial. This polynomial can then be used to find the remaining roots.

$\frac{3 x^{3} - 19 x^{2} + 33 x - 9}{x - 3}$

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.

$3$	$3$	$- 19$	$33$	$- 9$
	\xa0			\xa0

The first number in the dividend $(3)$ is put into the first position of the result area (below the horizontal line).

$3$	$3$	$- 19$	$33$	$- 9$
	\xa0			\xa0
	$3$

Multiply the newest entry in the result $(3)$ by the divisor $(3)$ and place the result of $(9)$ under the next term in the dividend $(- 19)$ .

$3$	$3$	$- 19$	$33$	$- 9$
	\xa0	$9$		\xa0
	$3$

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

$3$	$3$	$- 19$	$33$	$- 9$
	\xa0	$9$		\xa0
	$3$	$- 10$

Multiply the newest entry in the result $(- 10)$ by the divisor $(3)$ and place the result of $(- 30)$ under the next term in the dividend $(33)$ .

$3$	$3$	$- 19$	$33$	$- 9$
	\xa0	$9$	$- 30$	\xa0
	$3$	$- 10$

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

$3$	$3$	$- 19$	$33$	$- 9$
	\xa0	$9$	$- 30$	\xa0
	$3$	$- 10$	$3$

Multiply the newest entry in the result $(3)$ by the divisor $(3)$ and place the result of $(9)$ under the next term in the dividend $(- 9)$ .

$3$	$3$	$- 19$	$33$	$- 9$
	\xa0	$9$	$- 30$	$9$
	$3$	$- 10$	$3$

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

$3$	$3$	$- 19$	$33$	$- 9$
	\xa0	$9$	$- 30$	$9$
	$3$	$- 10$	$3$	$0$

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

$3 x^{2} + - 10 x + 3$

Simplify the quotient polynomial.

$3 x^{2} - 10 x + 3$

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 $a \cdot c = 3 \cdot 3 = 9$ and whose sum is $b = - 10$ .

Factor $- 10$ out of $- 10 x$ .

$(x - 3) + 3 x^{2} - 10 x + 3$

Rewrite $- 10$ as $- 1$ plus $- 9$

$(x - 3) + 3 x^{2} + (- 1 - 9) x + 3$

Apply the distributive property.

$(x - 3) + 3 x^{2} - 1 x - 9 x + 3$

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

$(x - 3) (3 x^{2} - 1 x) - 9 x + 3$

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

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

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

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

$(x - 3) (3 x - 1) (x - 3)$

Factor the left side of the equation.

Factor $3 x^{3} - 19 x^{2} + 33 x - 9$ 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.

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

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

$\pm 1, \pm 0.\overline{3}, \pm 9, \pm 3$

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.

$3 \cdot {0.\overline{3}}^{3} - 19 \cdot {0.\overline{3}}^{2} + 33 \cdot 0.\overline{3} - 9$

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

$3 \cdot 0.\overline{037} - 19 \cdot {0.\overline{3}}^{2} + 33 \cdot 0.\overline{3} - 9$

Multiply $3$ by $0.\overline{037}$ .

$0.\overline{1} - 19 \cdot {0.\overline{3}}^{2} + 33 \cdot 0.\overline{3} - 9$

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

$0.\overline{1} - 19 \cdot 0.\overline{1} + 33 \cdot 0.\overline{3} - 9$

Multiply $- 19$ by $0.\overline{1}$ .

$0.\overline{1} - 2.11111111 + 33 \cdot 0.\overline{3} - 9$

Subtract $2.11111111$ from $0.\overline{1}$ .

$- 2 + 33 \cdot 0.\overline{3} - 9$

Multiply $33$ by $0.\overline{3}$ .

$- 2 + 11 - 9$

Add $- 2$ and $11$ .

$9 - 9$

Subtract $9$ from $9$ .

$0$

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.

$\frac{3 x^{3} - 19 x^{2} + 33 x - 9}{3 x - 1}$

Divide $3 x^{3} - 19 x^{2} + 33 x - 9$ 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 $0$ .

$3 x$

$1$

$3 x^{3}$

$19 x^{2}$

$33 x$

$9$

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

					$x^{2}$
	$3 x$	-	$1$		$3 x^{3}$	-	$19 x^{2}$	+	$33 x$	-	$9$

Multiply the new quotient term by the divisor.

				$x^{2}$
$3 x$	-	$1$		$3 x^{3}$	-	$19 x^{2}$	+	$33 x$	-	$9$
			+	$3 x^{3}$	-	$x^{2}$

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

				$x^{2}$
$3 x$	-	$1$		$3 x^{3}$	-	$19 x^{2}$	+	$33 x$	-	$9$
			-	$3 x^{3}$	+	$x^{2}$

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

				$x^{2}$
$3 x$	-	$1$		$3 x^{3}$	-	$19 x^{2}$	+	$33 x$	-	$9$
			-	$3 x^{3}$	+	$x^{2}$
					-	$18 x^{2}$

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

				$x^{2}$
$3 x$	-	$1$		$3 x^{3}$	-	$19 x^{2}$	+	$33 x$	-	$9$
			-	$3 x^{3}$	+	$x^{2}$
					-	$18 x^{2}$	+	$33 x$

Divide the highest order term in the dividend $- 18 x^{2}$ by the highest order term in divisor $3 x$ .

				$x^{2}$	-	$6 x$
$3 x$	-	$1$		$3 x^{3}$	-	$19 x^{2}$	+	$33 x$	-	$9$
			-	$3 x^{3}$	+	$x^{2}$
					-	$18 x^{2}$	+	$33 x$

Multiply the new quotient term by the divisor.

				$x^{2}$	-	$6 x$
$3 x$	-	$1$		$3 x^{3}$	-	$19 x^{2}$	+	$33 x$	-	$9$
			-	$3 x^{3}$	+	$x^{2}$
					-	$18 x^{2}$	+	$33 x$
					-	$18 x^{2}$	+	$6 x$

The expression needs to be subtracted from the dividend, so change all the signs in $- 18 x^{2} + 6 x$

				$x^{2}$	-	$6 x$
$3 x$	-	$1$		$3 x^{3}$	-	$19 x^{2}$	+	$33 x$	-	$9$
			-	$3 x^{3}$	+	$x^{2}$
					-	$18 x^{2}$	+	$33 x$
					+	$18 x^{2}$	-	$6 x$

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

				$x^{2}$	-	$6 x$
$3 x$	-	$1$		$3 x^{3}$	-	$19 x^{2}$	+	$33 x$	-	$9$
			-	$3 x^{3}$	+	$x^{2}$
					-	$18 x^{2}$	+	$33 x$
					+	$18 x^{2}$	-	$6 x$
							+	$27 x$

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

				$x^{2}$	-	$6 x$
$3 x$	-	$1$		$3 x^{3}$	-	$19 x^{2}$	+	$33 x$	-	$9$
			-	$3 x^{3}$	+	$x^{2}$
					-	$18 x^{2}$	+	$33 x$
					+	$18 x^{2}$	-	$6 x$
							+	$27 x$	-	$9$

Divide the highest order term in the dividend $27 x$ by the highest order term in divisor $3 x$ .

				$x^{2}$	-	$6 x$	+	$9$
$3 x$	-	$1$		$3 x^{3}$	-	$19 x^{2}$	+	$33 x$	-	$9$
			-	$3 x^{3}$	+	$x^{2}$
					-	$18 x^{2}$	+	$33 x$
					+	$18 x^{2}$	-	$6 x$
							+	$27 x$	-	$9$

Multiply the new quotient term by the divisor.

				$x^{2}$	-	$6 x$	+	$9$
$3 x$	-	$1$		$3 x^{3}$	-	$19 x^{2}$	+	$33 x$	-	$9$
			-	$3 x^{3}$	+	$x^{2}$
					-	$18 x^{2}$	+	$33 x$
					+	$18 x^{2}$	-	$6 x$
							+	$27 x$	-	$9$
							+	$27 x$	-	$9$

The expression needs to be subtracted from the dividend, so change all the signs in $27 x - 9$

				$x^{2}$	-	$6 x$	+	$9$
$3 x$	-	$1$		$3 x^{3}$	-	$19 x^{2}$	+	$33 x$	-	$9$
			-	$3 x^{3}$	+	$x^{2}$
					-	$18 x^{2}$	+	$33 x$
					+	$18 x^{2}$	-	$6 x$
							+	$27 x$	-	$9$
							-	$27 x$	+	$9$

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

				$x^{2}$	-	$6 x$	+	$9$
$3 x$	-	$1$		$3 x^{3}$	-	$19 x^{2}$	+	$33 x$	-	$9$
			-	$3 x^{3}$	+	$x^{2}$
					-	$18 x^{2}$	+	$33 x$
					+	$18 x^{2}$	-	$6 x$
							+	$27 x$	-	$9$
							-	$27 x$	+	$9$
										$0$

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

$x^{2} - 6 x + 9$

Write $3 x^{3} - 19 x^{2} + 33 x - 9$ as a set of factors.

$(3 x - 1) (x^{2} - 6 x + 9) = 0$

Factor using the perfect square rule.

Rewrite $9$ as $3^{2}$ .

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

Check that the middle term is two times the product of the numbers being squared in the first term and third term.

$6 x = 2 \cdot x \cdot 3$

Rewrite the polynomial.

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

Factor using the perfect square trinomial rule $a^{2} - 2 a b + b^{2} = {(a - b)}^{2}$ , where $a = x$ and $b = 3$ .

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

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

$3 x - 1 = 0$

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

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 $1$ .

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

Set ${(x - 3)}^{2}$ equal to $0$ and solve for $x$ .

Set ${(x - 3)}^{2}$ equal to $0$ .

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

Solve ${(x - 3)}^{2} = 0$ for $x$ .

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

$x - 3 = 0$

Add $3$ to both sides of the equation.

$x = 3$

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

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

Find the Roots/Zeros Using the Rational Roots Test 3x^3-19x^2+33x-9

Related Questions

Support LCMGCF.com by disabling your adblocker.