Find the Roots/Zeros Using the Rational Roots Test 12x^3-20x^2+9x-1=0

$12 x^{3} - 20 x^{2} + 9 x - 1 = 0$

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.

$p = \pm 1$

$q = \pm 1, \pm 2, \pm 3, \pm 4, \pm 6, \pm 12$

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

$\pm 1, \pm \frac{1}{2}, \pm \frac{1}{3}, \pm \frac{1}{4}, \pm \frac{1}{6}, \pm \frac{1}{12}$

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.

$12 {(1)}^{3} - 20 {(1)}^{2} + 9 (1) - 1$

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

Simplify each term.

One to any power is one.

$12 \cdot 1 - 20 {(1)}^{2} + 9 (1) - 1$

Multiply $12$ by $1$ .

$12 - 20 {(1)}^{2} + 9 (1) - 1$

One to any power is one.

$12 - 20 \cdot 1 + 9 (1) - 1$

Multiply $- 20$ by $1$ .

$12 - 20 + 9 (1) - 1$

Multiply $9$ by $1$ .

$12 - 20 + 9 - 1$

Simplify by adding and subtracting.

Subtract $20$ from $12$ .

$- 8 + 9 - 1$

Add $- 8$ and $9$ .

$1 - 1$

Subtract $1$ from $1$ .

$0$

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.

$\frac{12 x^{3} - 20 x^{2} + 9 x - 1}{x - 1}$

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.

$1$	$12$	$- 20$	$9$	$- 1$
	\xa0			\xa0

The first number in the dividend $(12)$ is put into the first position of the result area (below the horizontal line).

$1$	$12$	$- 20$	$9$	$- 1$
	\xa0			\xa0
	$12$

Multiply the newest entry in the result $(12)$ by the divisor $(1)$ and place the result of $(12)$ under the next term in the dividend $(- 20)$ .

$1$	$12$	$- 20$	$9$	$- 1$
	\xa0	$12$		\xa0
	$12$

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

$1$	$12$	$- 20$	$9$	$- 1$
	\xa0	$12$		\xa0
	$12$	$- 8$

Multiply the newest entry in the result $(- 8)$ by the divisor $(1)$ and place the result of $(- 8)$ under the next term in the dividend $(9)$ .

$1$	$12$	$- 20$	$9$	$- 1$
	\xa0	$12$	$- 8$	\xa0
	$12$	$- 8$

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

$1$	$12$	$- 20$	$9$	$- 1$
	\xa0	$12$	$- 8$	\xa0
	$12$	$- 8$	$1$

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

$1$	$12$	$- 20$	$9$	$- 1$
	\xa0	$12$	$- 8$	$1$
	$12$	$- 8$	$1$

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

$1$	$12$	$- 20$	$9$	$- 1$
	\xa0	$12$	$- 8$	$1$
	$12$	$- 8$	$1$	$0$

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

$12 x^{2} + - 8 x + 1$

Simplify the quotient polynomial.

$12 x^{2} - 8 x + 1$

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 $a \cdot c = 12 \cdot 1 = 12$ and whose sum is $b = - 8$ .

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

$(x - 1) + 12 x^{2} - 8 x + 1$

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

$(x - 1) + 12 x^{2} + (- 2 - 6) x + 1$

Apply the distributive property.

$(x - 1) + 12 x^{2} - 2 x - 6 x + 1$

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

$(x - 1) (12 x^{2} - 2 x) - 6 x + 1$

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

$(x - 1) + 2 x (6 x - 1) - (6 x - 1)$

Factor the polynomial by factoring out the greatest common factor, $6 x - 1$ .

$(x - 1) + (6 x - 1) (2 x - 1)$

$(x - 1) (6 x - 1) (2 x - 1)$

Factor the left side of the equation.

Factor $12 x^{3} - 20 x^{2} + 9 x - 1$ 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.

$p = \pm 1 q = \pm 1, \pm 12, \pm 2, \pm 6, \pm 3, \pm 4$

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

$\pm 1, \pm 0.08\overline{3}, \pm 0.5, \pm 0.1\overline{6}, \pm 0.\overline{3}, \pm 0.25$

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.

$12 \cdot 1^{3} - 20 \cdot 1^{2} + 9 \cdot 1 - 1$

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

$12 \cdot 1 - 20 \cdot 1^{2} + 9 \cdot 1 - 1$

Multiply $12$ by $1$ .

$12 - 20 \cdot 1^{2} + 9 \cdot 1 - 1$

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

$12 - 20 \cdot 1 + 9 \cdot 1 - 1$

Multiply $- 20$ by $1$ .

$12 - 20 + 9 \cdot 1 - 1$

Subtract $20$ from $12$ .

$- 8 + 9 \cdot 1 - 1$

Multiply $9$ by $1$ .

$- 8 + 9 - 1$

Add $- 8$ and $9$ .

$1 - 1$

Subtract $1$ from $1$ .

$0$

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.

$\frac{12 x^{3} - 20 x^{2} + 9 x - 1}{x - 1}$

Divide $12 x^{3} - 20 x^{2} + 9 x - 1$ 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 $0$ .

$x$

$1$

$12 x^{3}$

$20 x^{2}$

$9 x$

$1$

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

					$12 x^{2}$
	$x$	-	$1$		$12 x^{3}$	-	$20 x^{2}$	+	$9 x$	-	$1$

Multiply the new quotient term by the divisor.

				$12 x^{2}$
$x$	-	$1$		$12 x^{3}$	-	$20 x^{2}$	+	$9 x$	-	$1$
			+	$12 x^{3}$	-	$12 x^{2}$

The expression needs to be subtracted from the dividend, so change all the signs in $12 x^{3} - 12 x^{2}$

				$12 x^{2}$
$x$	-	$1$		$12 x^{3}$	-	$20 x^{2}$	+	$9 x$	-	$1$
			-	$12 x^{3}$	+	$12 x^{2}$

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

				$12 x^{2}$
$x$	-	$1$		$12 x^{3}$	-	$20 x^{2}$	+	$9 x$	-	$1$
			-	$12 x^{3}$	+	$12 x^{2}$
					-	$8 x^{2}$

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

				$12 x^{2}$
$x$	-	$1$		$12 x^{3}$	-	$20 x^{2}$	+	$9 x$	-	$1$
			-	$12 x^{3}$	+	$12 x^{2}$
					-	$8 x^{2}$	+	$9 x$

Divide the highest order term in the dividend $- 8 x^{2}$ by the highest order term in divisor $x$ .

				$12 x^{2}$	-	$8 x$
$x$	-	$1$		$12 x^{3}$	-	$20 x^{2}$	+	$9 x$	-	$1$
			-	$12 x^{3}$	+	$12 x^{2}$
					-	$8 x^{2}$	+	$9 x$

Multiply the new quotient term by the divisor.

				$12 x^{2}$	-	$8 x$
$x$	-	$1$		$12 x^{3}$	-	$20 x^{2}$	+	$9 x$	-	$1$
			-	$12 x^{3}$	+	$12 x^{2}$
					-	$8 x^{2}$	+	$9 x$
					-	$8 x^{2}$	+	$8 x$

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

				$12 x^{2}$	-	$8 x$
$x$	-	$1$		$12 x^{3}$	-	$20 x^{2}$	+	$9 x$	-	$1$
			-	$12 x^{3}$	+	$12 x^{2}$
					-	$8 x^{2}$	+	$9 x$
					+	$8 x^{2}$	-	$8 x$

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

				$12 x^{2}$	-	$8 x$
$x$	-	$1$		$12 x^{3}$	-	$20 x^{2}$	+	$9 x$	-	$1$
			-	$12 x^{3}$	+	$12 x^{2}$
					-	$8 x^{2}$	+	$9 x$
					+	$8 x^{2}$	-	$8 x$
							+	$x$

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

				$12 x^{2}$	-	$8 x$
$x$	-	$1$		$12 x^{3}$	-	$20 x^{2}$	+	$9 x$	-	$1$
			-	$12 x^{3}$	+	$12 x^{2}$
					-	$8 x^{2}$	+	$9 x$
					+	$8 x^{2}$	-	$8 x$
							+	$x$	-	$1$

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

				$12 x^{2}$	-	$8 x$	+	$1$
$x$	-	$1$		$12 x^{3}$	-	$20 x^{2}$	+	$9 x$	-	$1$
			-	$12 x^{3}$	+	$12 x^{2}$
					-	$8 x^{2}$	+	$9 x$
					+	$8 x^{2}$	-	$8 x$
							+	$x$	-	$1$

Multiply the new quotient term by the divisor.

				$12 x^{2}$	-	$8 x$	+	$1$
$x$	-	$1$		$12 x^{3}$	-	$20 x^{2}$	+	$9 x$	-	$1$
			-	$12 x^{3}$	+	$12 x^{2}$
					-	$8 x^{2}$	+	$9 x$
					+	$8 x^{2}$	-	$8 x$
							+	$x$	-	$1$
							+	$x$	-	$1$

The expression needs to be subtracted from the dividend, so change all the signs in $x - 1$

				$12 x^{2}$	-	$8 x$	+	$1$
$x$	-	$1$		$12 x^{3}$	-	$20 x^{2}$	+	$9 x$	-	$1$
			-	$12 x^{3}$	+	$12 x^{2}$
					-	$8 x^{2}$	+	$9 x$
					+	$8 x^{2}$	-	$8 x$
							+	$x$	-	$1$
							-	$x$	+	$1$

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

				$12 x^{2}$	-	$8 x$	+	$1$
$x$	-	$1$		$12 x^{3}$	-	$20 x^{2}$	+	$9 x$	-	$1$
			-	$12 x^{3}$	+	$12 x^{2}$
					-	$8 x^{2}$	+	$9 x$
					+	$8 x^{2}$	-	$8 x$
							+	$x$	-	$1$
							-	$x$	+	$1$
										$0$

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

$12 x^{2} - 8 x + 1$

Write $12 x^{3} - 20 x^{2} + 9 x - 1$ as a set of factors.

$(x - 1) (12 x^{2} - 8 x + 1) = 0$

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 $a \cdot c = 12 \cdot 1 = 12$ and whose sum is $b = - 8$ .

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

$(x - 1) (12 x^{2} - 8 x + 1) = 0$

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

$(x - 1) (12 x^{2} + (- 2 - 6) x + 1) = 0$

Apply the distributive property.

$(x - 1) (12 x^{2} - 2 x - 6 x + 1) = 0$

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

$(x - 1) ((12 x^{2} - 2 x) - 6 x + 1) = 0$

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

$(x - 1) (2 x (6 x - 1) - (6 x - 1)) = 0$

Factor the polynomial by factoring out the greatest common factor, $6 x - 1$ .

$(x - 1) ((6 x - 1) (2 x - 1)) = 0$

Remove unnecessary parentheses.

$(x - 1) (6 x - 1) (2 x - 1) = 0$

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

$x - 1 = 0$

$6 x - 1 = 0$

$2 x - 1 = 0$

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

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

$x - 1 = 0$

Add $1$ to both sides of the equation.

$x = 1$

Set $6 x - 1$ equal to $0$ and solve for $x$ .

Set $6 x - 1$ equal to $0$ .

$6 x - 1 = 0$

Solve $6 x - 1 = 0$ for $x$ .

Add $1$ to both sides of the equation.

$6 x = 1$

Divide each term in $6 x = 1$ by $6$ and simplify.

Divide each term in $6 x = 1$ by $6$ .

$\frac{6 x}{6} = \frac{1}{6}$

Simplify the left side.

Cancel the common factor of $6$ .

Cancel the common factor.

$\frac{6 x}{6} = \frac{1}{6}$

Divide $x$ by $1$ .

$x = \frac{1}{6}$

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 $x$ .

Add $1$ to 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 $1$ .

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

The final solution is all the values that make $(x - 1) (6 x - 1) (2 x - 1) = 0$ true.

$x = 1, \frac{1}{6}, \frac{1}{2}$

Find the Roots/Zeros Using the Rational Roots Test 12x^3-20x^2+9x-1=0

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