Find the Roots/Zeros Using the Rational Roots Test 12x^3+53x^2-34x+5

$12 x^{3} + 53 x^{2} - 34 x + 5$

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.

$p = \pm 1, \pm 5$

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

Step 2

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

$\pm 1, \pm \frac{1}{2}, \pm \frac{1}{3}, \pm \frac{1}{4}, \pm \frac{1}{6}, \pm \frac{1}{12}, \pm 5, \pm \frac{5}{2}, \pm \frac{5}{3}, \pm \frac{5}{4}, \pm \frac{5}{6}, \pm \frac{5}{12}$

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.

$12 {(\frac{1}{4})}^{3} + 53 {(\frac{1}{4})}^{2} - 34 (\frac{1}{4}) + 5$

Step 4

Simplify the expression. In this case, the expression is equal to $0$ so $x = \frac{1}{4}$ is a root of the polynomial.

Simplify each term.

Apply the product rule to $\frac{1}{4}$ .

$12 \frac{1^{3}}{4^{3}} + 53 {(\frac{1}{4})}^{2} - 34 (\frac{1}{4}) + 5$

One to any power is one.

$12 \frac{1}{4^{3}} + 53 {(\frac{1}{4})}^{2} - 34 (\frac{1}{4}) + 5$

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

$12 (\frac{1}{64}) + 53 {(\frac{1}{4})}^{2} - 34 (\frac{1}{4}) + 5$

Cancel the common factor of $4$ .

Factor $4$ out of $12$ .

$4 (3) \frac{1}{64} + 53 {(\frac{1}{4})}^{2} - 34 (\frac{1}{4}) + 5$

Factor $4$ out of $64$ .

$4 \cdot 3 \frac{1}{4 \cdot 16} + 53 {(\frac{1}{4})}^{2} - 34 (\frac{1}{4}) + 5$

Cancel the common factor.

$4 \cdot 3 \frac{1}{4 \cdot 16} + 53 {(\frac{1}{4})}^{2} - 34 (\frac{1}{4}) + 5$

Rewrite the expression.

$3 (\frac{1}{16}) + 53 {(\frac{1}{4})}^{2} - 34 (\frac{1}{4}) + 5$

Combine $3$ and $\frac{1}{16}$ .

$\frac{3}{16} + 53 {(\frac{1}{4})}^{2} - 34 (\frac{1}{4}) + 5$

Apply the product rule to $\frac{1}{4}$ .

$\frac{3}{16} + 53 \frac{1^{2}}{4^{2}} - 34 (\frac{1}{4}) + 5$

One to any power is one.

$\frac{3}{16} + 53 \frac{1}{4^{2}} - 34 (\frac{1}{4}) + 5$

Raise $4$ to the power of $2$ .

$\frac{3}{16} + 53 (\frac{1}{16}) - 34 (\frac{1}{4}) + 5$

Combine $53$ and $\frac{1}{16}$ .

$\frac{3}{16} + \frac{53}{16} - 34 (\frac{1}{4}) + 5$

Cancel the common factor of $2$ .

Factor $2$ out of $- 34$ .

$\frac{3}{16} + \frac{53}{16} + 2 (- 17) \frac{1}{4} + 5$

Factor $2$ out of $4$ .

$\frac{3}{16} + \frac{53}{16} + 2 \cdot - 17 \frac{1}{2 \cdot 2} + 5$

Cancel the common factor.

$\frac{3}{16} + \frac{53}{16} + 2 \cdot - 17 \frac{1}{2 \cdot 2} + 5$

Rewrite the expression.

$\frac{3}{16} + \frac{53}{16} - 17 (\frac{1}{2}) + 5$

Combine $- 17$ and $\frac{1}{2}$ .

$\frac{3}{16} + \frac{53}{16} + \frac{- 17}{2} + 5$

Move the negative in front of the fraction.

$\frac{3}{16} + \frac{53}{16} - \frac{17}{2} + 5$

Combine fractions.

Combine the numerators over the common denominator.

$5 + \frac{3 + 53}{16} + \frac{- 17}{2}$

Add $3$ and $53$ .

$5 + \frac{56}{16} + \frac{- 17}{2}$

Find the common denominator.

Write $5$ as a fraction with denominator $1$ .

$\frac{5}{1} + \frac{56}{16} + \frac{- 17}{2}$

Multiply $\frac{5}{1}$ by $\frac{16}{16}$ .

$\frac{5}{1} \cdot \frac{16}{16} + \frac{56}{16} + \frac{- 17}{2}$

Multiply $\frac{5}{1}$ by $\frac{16}{16}$ .

$\frac{5 \cdot 16}{16} + \frac{56}{16} + \frac{- 17}{2}$

Multiply $\frac{- 17}{2}$ by $\frac{8}{8}$ .

$\frac{5 \cdot 16}{16} + \frac{56}{16} + \frac{- 17}{2} \cdot \frac{8}{8}$

Multiply $\frac{- 17}{2}$ by $\frac{8}{8}$ .

$\frac{5 \cdot 16}{16} + \frac{56}{16} + \frac{- 17 \cdot 8}{2 \cdot 8}$

Multiply $2$ by $8$ .

$\frac{5 \cdot 16}{16} + \frac{56}{16} + \frac{- 17 \cdot 8}{16}$

Combine the numerators over the common denominator.

$\frac{5 \cdot 16 + 56 - 17 \cdot 8}{16}$

Simplify each term.

Multiply $5$ by $16$ .

$\frac{80 + 56 - 17 \cdot 8}{16}$

Multiply $- 17$ by $8$ .

$\frac{80 + 56 - 136}{16}$

Simplify the expression.

Add $80$ and $56$ .

$\frac{136 - 136}{16}$

Subtract $136$ from $136$ .

$\frac{0}{16}$

Divide $0$ by $16$ .

$0$

Step 5

Since $\frac{1}{4}$ is a known root, divide the polynomial by $x - \frac{1}{4}$ to find the quotient polynomial. This polynomial can then be used to find the remaining roots.

$\frac{12 x^{3} + 53 x^{2} - 34 x + 5}{x - \frac{1}{4}}$

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.

$\frac{1}{4}$	$12$	$53$	$- 34$	$5$
	\xa0			\xa0

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

$\frac{1}{4}$	$12$	$53$	$- 34$	$5$
	\xa0			\xa0
	$12$

Multiply the newest entry in the result $(12)$ by the divisor $(\frac{1}{4})$ and place the result of $(3)$ under the next term in the dividend $(53)$ .

$\frac{1}{4}$	$12$	$53$	$- 34$	$5$
	\xa0	$3$		\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.

$\frac{1}{4}$	$12$	$53$	$- 34$	$5$
	\xa0	$3$		\xa0
	$12$	$56$

Multiply the newest entry in the result $(56)$ by the divisor $(\frac{1}{4})$ and place the result of $(14)$ under the next term in the dividend $(- 34)$ .

$\frac{1}{4}$	$12$	$53$	$- 34$	$5$
	\xa0	$3$	$14$	\xa0
	$12$	$56$

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

$\frac{1}{4}$	$12$	$53$	$- 34$	$5$
	\xa0	$3$	$14$	\xa0
	$12$	$56$	$- 20$

Multiply the newest entry in the result $(- 20)$ by the divisor $(\frac{1}{4})$ and place the result of $(- 5)$ under the next term in the dividend $(5)$ .

$\frac{1}{4}$	$12$	$53$	$- 34$	$5$
	\xa0	$3$	$14$	$- 5$
	$12$	$56$	$- 20$

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

$\frac{1}{4}$	$12$	$53$	$- 34$	$5$
	\xa0	$3$	$14$	$- 5$
	$12$	$56$	$- 20$	$0$

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

$12 x^{2} + 56 x - 20$

Step 7

Factor $4$ out of $12 x^{2} + 56 x - 20$ .

Factor $4$ out of $12 x^{2}$ .

$(x - \frac{1}{4}) (4 (3 x^{2}) + 56 x - 20)$

Factor $4$ out of $56 x$ .

$(x - \frac{1}{4}) (4 (3 x^{2}) + 4 (14 x) - 20)$

Factor $4$ out of $- 20$ .

$(x - \frac{1}{4}) (4 (3 x^{2}) + 4 (14 x) + 4 (- 5))$

Factor $4$ out of $4 (3 x^{2}) + 4 (14 x)$ .

$(x - \frac{1}{4}) (4 (3 x^{2} + 14 x) + 4 (- 5))$

Factor $4$ out of $4 (3 x^{2} + 14 x) + 4 (- 5)$ .

$(x - \frac{1}{4}) (4 (3 x^{2} + 14 x - 5))$

Step 8

Factor.

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 $a \cdot c = 3 \cdot - 5 = - 15$ and whose sum is $b = 14$ .

Factor $14$ out of $14 x$ .

$(x - \frac{1}{4}) (4 (3 x^{2} + 14 (x) - 5))$

Rewrite $14$ as $- 1$ plus $15$

$(x - \frac{1}{4}) (4 (3 x^{2} + (- 1 + 15) x - 5))$

Apply the distributive property.

$(x - \frac{1}{4}) (4 (3 x^{2} - 1 x + 15 x - 5))$

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

$(x - \frac{1}{4}) (4 ((3 x^{2} - 1 x) + 15 x - 5))$

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

$(x - \frac{1}{4}) (4 (x (3 x - 1) + 5 (3 x - 1)))$

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

$(x - \frac{1}{4}) (4 ((3 x - 1) (x + 5)))$

Remove unnecessary parentheses.

$(x - \frac{1}{4}) (4 (3 x - 1) (x + 5))$

Step 9

Factor the left side of the equation.

Factor $12 x^{3} + 53 x^{2} - 34 x + 5$ 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.

$p = \pm 1, \pm 5 q = \pm 1, \pm 12, \pm 2, \pm 6, \pm 3, \pm 4$

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

$\pm 1, \pm 0.08\overline{3}, \pm 0.5, \pm 0.1\overline{6}, \pm 0.\overline{3}, \pm 0.25, \pm 5, \pm 0.41\overline{6}, \pm 2.5, \pm 0.8\overline{3}, \pm 1.\overline{6}, \pm 1.25$

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.

$12 \cdot {0.\overline{3}}^{3} + 53 \cdot {0.\overline{3}}^{2} - 34 \cdot 0.\overline{3} + 5$

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

$12 \cdot 0.\overline{037} + 53 \cdot {0.\overline{3}}^{2} - 34 \cdot 0.\overline{3} + 5$

Multiply $12$ by $0.\overline{037}$ .

$0.\overline{4} + 53 \cdot {0.\overline{3}}^{2} - 34 \cdot 0.\overline{3} + 5$

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

$0.\overline{4} + 53 \cdot 0.\overline{1} - 34 \cdot 0.\overline{3} + 5$

Multiply $53$ by $0.\overline{1}$ .

$0.\overline{4} + 5.\overline{8} - 34 \cdot 0.\overline{3} + 5$

Add $0.\overline{4}$ and $5.\overline{8}$ .

$6.33333333 - 34 \cdot 0.\overline{3} + 5$

Multiply $- 34$ by $0.\overline{3}$ .

$6.33333333 - 11.\overline{3} + 5$

Subtract $11.\overline{3}$ from $6.33333333$ .

$- 5 + 5$

Add $- 5$ and $5$ .

$0$

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.

$\frac{12 x^{3} + 53 x^{2} - 34 x + 5}{3 x - 1}$

Divide $12 x^{3} + 53 x^{2} - 34 x + 5$ 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 $0$ .

$3 x$

$1$

$12 x^{3}$

$53 x^{2}$

$34 x$

$5$

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

					$4 x^{2}$
	$3 x$	-	$1$		$12 x^{3}$	+	$53 x^{2}$	-	$34 x$	+	$5$

Multiply the new quotient term by the divisor.

				$4 x^{2}$
$3 x$	-	$1$		$12 x^{3}$	+	$53 x^{2}$	-	$34 x$	+	$5$
			+	$12 x^{3}$	-	$4 x^{2}$

The expression needs to be subtracted from the dividend, so change all the signs in $12 x^{3} - 4 x^{2}$

				$4 x^{2}$
$3 x$	-	$1$		$12 x^{3}$	+	$53 x^{2}$	-	$34 x$	+	$5$
			-	$12 x^{3}$	+	$4 x^{2}$

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

				$4 x^{2}$
$3 x$	-	$1$		$12 x^{3}$	+	$53 x^{2}$	-	$34 x$	+	$5$
			-	$12 x^{3}$	+	$4 x^{2}$
					+	$57 x^{2}$

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

				$4 x^{2}$
$3 x$	-	$1$		$12 x^{3}$	+	$53 x^{2}$	-	$34 x$	+	$5$
			-	$12 x^{3}$	+	$4 x^{2}$
					+	$57 x^{2}$	-	$34 x$

Divide the highest order term in the dividend $57 x^{2}$ by the highest order term in divisor $3 x$ .

				$4 x^{2}$	+	$19 x$
$3 x$	-	$1$		$12 x^{3}$	+	$53 x^{2}$	-	$34 x$	+	$5$
			-	$12 x^{3}$	+	$4 x^{2}$
					+	$57 x^{2}$	-	$34 x$

Multiply the new quotient term by the divisor.

				$4 x^{2}$	+	$19 x$
$3 x$	-	$1$		$12 x^{3}$	+	$53 x^{2}$	-	$34 x$	+	$5$
			-	$12 x^{3}$	+	$4 x^{2}$
					+	$57 x^{2}$	-	$34 x$
					+	$57 x^{2}$	-	$19 x$

The expression needs to be subtracted from the dividend, so change all the signs in $57 x^{2} - 19 x$

				$4 x^{2}$	+	$19 x$
$3 x$	-	$1$		$12 x^{3}$	+	$53 x^{2}$	-	$34 x$	+	$5$
			-	$12 x^{3}$	+	$4 x^{2}$
					+	$57 x^{2}$	-	$34 x$
					-	$57 x^{2}$	+	$19 x$

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

				$4 x^{2}$	+	$19 x$
$3 x$	-	$1$		$12 x^{3}$	+	$53 x^{2}$	-	$34 x$	+	$5$
			-	$12 x^{3}$	+	$4 x^{2}$
					+	$57 x^{2}$	-	$34 x$
					-	$57 x^{2}$	+	$19 x$
							-	$15 x$

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

				$4 x^{2}$	+	$19 x$
$3 x$	-	$1$		$12 x^{3}$	+	$53 x^{2}$	-	$34 x$	+	$5$
			-	$12 x^{3}$	+	$4 x^{2}$
					+	$57 x^{2}$	-	$34 x$
					-	$57 x^{2}$	+	$19 x$
							-	$15 x$	+	$5$

Divide the highest order term in the dividend $- 15 x$ by the highest order term in divisor $3 x$ .

				$4 x^{2}$	+	$19 x$	-	$5$
$3 x$	-	$1$		$12 x^{3}$	+	$53 x^{2}$	-	$34 x$	+	$5$
			-	$12 x^{3}$	+	$4 x^{2}$
					+	$57 x^{2}$	-	$34 x$
					-	$57 x^{2}$	+	$19 x$
							-	$15 x$	+	$5$

Multiply the new quotient term by the divisor.

				$4 x^{2}$	+	$19 x$	-	$5$
$3 x$	-	$1$		$12 x^{3}$	+	$53 x^{2}$	-	$34 x$	+	$5$
			-	$12 x^{3}$	+	$4 x^{2}$
					+	$57 x^{2}$	-	$34 x$
					-	$57 x^{2}$	+	$19 x$
							-	$15 x$	+	$5$
							-	$15 x$	+	$5$

The expression needs to be subtracted from the dividend, so change all the signs in $- 15 x + 5$

				$4 x^{2}$	+	$19 x$	-	$5$
$3 x$	-	$1$		$12 x^{3}$	+	$53 x^{2}$	-	$34 x$	+	$5$
			-	$12 x^{3}$	+	$4 x^{2}$
					+	$57 x^{2}$	-	$34 x$
					-	$57 x^{2}$	+	$19 x$
							-	$15 x$	+	$5$
							+	$15 x$	-	$5$

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

				$4 x^{2}$	+	$19 x$	-	$5$
$3 x$	-	$1$		$12 x^{3}$	+	$53 x^{2}$	-	$34 x$	+	$5$
			-	$12 x^{3}$	+	$4 x^{2}$
					+	$57 x^{2}$	-	$34 x$
					-	$57 x^{2}$	+	$19 x$
							-	$15 x$	+	$5$
							+	$15 x$	-	$5$
										$0$

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

$4 x^{2} + 19 x - 5$

Write $12 x^{3} + 53 x^{2} - 34 x + 5$ as a set of factors.

$(3 x - 1) (4 x^{2} + 19 x - 5) = 0$

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 $a \cdot c = 4 \cdot - 5 = - 20$ and whose sum is $b = 19$ .

Factor $19$ out of $19 x$ .

$(3 x - 1) (4 x^{2} + 19 (x) - 5) = 0$

Rewrite $19$ as $- 1$ plus $20$

$(3 x - 1) (4 x^{2} + (- 1 + 20) x - 5) = 0$

Apply the distributive property.

$(3 x - 1) (4 x^{2} - 1 x + 20 x - 5) = 0$

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

$(3 x - 1) ((4 x^{2} - 1 x) + 20 x - 5) = 0$

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

$(3 x - 1) (x (4 x - 1) + 5 (4 x - 1)) = 0$

Factor the polynomial by factoring out the greatest common factor, $4 x - 1$ .

$(3 x - 1) ((4 x - 1) (x + 5)) = 0$

Remove unnecessary parentheses.

$(3 x - 1) (4 x - 1) (x + 5) = 0$

Step 10

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

$3 x - 1 = 0$

$4 x - 1 = 0$

$x + 5 = 0$

Step 11

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 12

Set $4 x - 1$ equal to $0$ and solve for $x$ .

Set $4 x - 1$ equal to $0$ .

$4 x - 1 = 0$

Solve $4 x - 1 = 0$ for $x$ .

Add $1$ to both sides of the equation.

$4 x = 1$

Divide each term in $4 x = 1$ by $4$ and simplify.

Divide each term in $4 x = 1$ by $4$ .

$\frac{4 x}{4} = \frac{1}{4}$

Simplify the left side.

Cancel the common factor of $4$ .

Cancel the common factor.

$\frac{4 x}{4} = \frac{1}{4}$

Divide $x$ by $1$ .

$x = \frac{1}{4}$

Step 13

Set $x + 5$ equal to $0$ and solve for $x$ .

Set $x + 5$ equal to $0$ .

$x + 5 = 0$

Subtract $5$ from both sides of the equation.

$x = - 5$

Step 14

The final solution is all the values that make $(3 x - 1) (4 x - 1) (x + 5) = 0$ true.

$x = \frac{1}{3}, \frac{1}{4}, - 5$

Step 15

Find the Roots/Zeros Using the Rational Roots Test 12x^3+53x^2-34x+5

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