Find the Roots/Zeros Using the Rational Roots Test 5x^3-6x^2-59x+12

$5 x^{3} - 6 x^{2} - 59 x + 12$

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.

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

$q = \pm 1, \pm 5$

Step 2

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

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

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.

$5 {(- 3)}^{3} - 6 {(- 3)}^{2} - 59 \cdot - 3 + 12$

Step 4

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

Simplify each term.

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

$5 \cdot - 27 - 6 {(- 3)}^{2} - 59 \cdot - 3 + 12$

Multiply $5$ by $- 27$ .

$- 135 - 6 {(- 3)}^{2} - 59 \cdot - 3 + 12$

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

$- 135 - 6 \cdot 9 - 59 \cdot - 3 + 12$

Multiply $- 6$ by $9$ .

$- 135 - 54 - 59 \cdot - 3 + 12$

Multiply $- 59$ by $- 3$ .

$- 135 - 54 + 177 + 12$

Simplify by adding and subtracting.

Subtract $54$ from $- 135$ .

$- 189 + 177 + 12$

Add $- 189$ and $177$ .

$- 12 + 12$

Add $- 12$ and $12$ .

$0$

Step 5

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.

$\frac{5 x^{3} - 6 x^{2} - 59 x + 12}{x + 3}$

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.

$- 3$	$5$	$- 6$	$- 59$	$12$
	\xa0			\xa0

The first number in the dividend $(5)$ is put into the first position of the result area (below the horizontal line).

$- 3$	$5$	$- 6$	$- 59$	$12$
	\xa0			\xa0
	$5$

Multiply the newest entry in the result $(5)$ by the divisor $(- 3)$ and place the result of $(- 15)$ under the next term in the dividend $(- 6)$ .

$- 3$	$5$	$- 6$	$- 59$	$12$
	\xa0	$- 15$		\xa0
	$5$

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

$- 3$	$5$	$- 6$	$- 59$	$12$
	\xa0	$- 15$		\xa0
	$5$	$- 21$

Multiply the newest entry in the result $(- 21)$ by the divisor $(- 3)$ and place the result of $(63)$ under the next term in the dividend $(- 59)$ .

$- 3$	$5$	$- 6$	$- 59$	$12$
	\xa0	$- 15$	$63$	\xa0
	$5$	$- 21$

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

$- 3$	$5$	$- 6$	$- 59$	$12$
	\xa0	$- 15$	$63$	\xa0
	$5$	$- 21$	$4$

Multiply the newest entry in the result $(4)$ by the divisor $(- 3)$ and place the result of $(- 12)$ under the next term in the dividend $(12)$ .

$- 3$	$5$	$- 6$	$- 59$	$12$
	\xa0	$- 15$	$63$	$- 12$
	$5$	$- 21$	$4$

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

$- 3$	$5$	$- 6$	$- 59$	$12$
	\xa0	$- 15$	$63$	$- 12$
	$5$	$- 21$	$4$	$0$

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

$5 x^{2} + - 21 x + 4$

Simplify the quotient polynomial.

$5 x^{2} - 21 x + 4$

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 $a \cdot c = 5 \cdot 4 = 20$ and whose sum is $b = - 21$ .

Factor $- 21$ out of $- 21 x$ .

$(x + 3) + 5 x^{2} - 21 x + 4$

Rewrite $- 21$ as $- 1$ plus $- 20$

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

Apply the distributive property.

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

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

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

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

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

Factor the polynomial by factoring out the greatest common factor, $5 x - 1$ .

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

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

Step 8

Factor the left side of the equation.

Factor $5 x^{3} - 6 x^{2} - 59 x + 12$ 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.

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

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

$\pm 1, \pm 0.2, \pm 12, \pm 2.4, \pm 2, \pm 0.4, \pm 6, \pm 1.2, \pm 3, \pm 0.6, \pm 4, \pm 0.8$

Substitute $0.2$ and simplify the expression. In this case, the expression is equal to $0$ so $0.2$ is a root of the polynomial.

Substitute $0.2$ into the polynomial.

$5 \cdot {0.2}^{3} - 6 \cdot {0.2}^{2} - 59 \cdot 0.2 + 12$

Raise $0.2$ to the power of $3$ .

$5 \cdot 0.008 - 6 \cdot {0.2}^{2} - 59 \cdot 0.2 + 12$

Multiply $5$ by $0.008$ .

$0.04 - 6 \cdot {0.2}^{2} - 59 \cdot 0.2 + 12$

Raise $0.2$ to the power of $2$ .

$0.04 - 6 \cdot 0.04 - 59 \cdot 0.2 + 12$

Multiply $- 6$ by $0.04$ .

$0.04 - 0.24 - 59 \cdot 0.2 + 12$

Subtract $0.24$ from $0.04$ .

$- 0.2 - 59 \cdot 0.2 + 12$

Multiply $- 59$ by $0.2$ .

$- 0.2 - 11.8 + 12$

Subtract $11.8$ from $- 0.2$ .

$- 12 + 12$

Add $- 12$ and $12$ .

$0$

Since $0.2$ is a known root, divide the polynomial by $5 x - 1$ to find the quotient polynomial. This polynomial can then be used to find the remaining roots.

$\frac{5 x^{3} - 6 x^{2} - 59 x + 12}{5 x - 1}$

Divide $5 x^{3} - 6 x^{2} - 59 x + 12$ by $5 x - 1$ .

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

$5 x$

$1$

$5 x^{3}$

$6 x^{2}$

$59 x$

$12$

Divide the highest order term in the dividend $5 x^{3}$ by the highest order term in divisor $5 x$ .

					$x^{2}$
	$5 x$	-	$1$		$5 x^{3}$	-	$6 x^{2}$	-	$59 x$	+	$12$

Multiply the new quotient term by the divisor.

				$x^{2}$
$5 x$	-	$1$		$5 x^{3}$	-	$6 x^{2}$	-	$59 x$	+	$12$
			+	$5 x^{3}$	-	$x^{2}$

The expression needs to be subtracted from the dividend, so change all the signs in $5 x^{3} - x^{2}$

				$x^{2}$
$5 x$	-	$1$		$5 x^{3}$	-	$6 x^{2}$	-	$59 x$	+	$12$
			-	$5 x^{3}$	+	$x^{2}$

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

				$x^{2}$
$5 x$	-	$1$		$5 x^{3}$	-	$6 x^{2}$	-	$59 x$	+	$12$
			-	$5 x^{3}$	+	$x^{2}$
					-	$5 x^{2}$

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

				$x^{2}$
$5 x$	-	$1$		$5 x^{3}$	-	$6 x^{2}$	-	$59 x$	+	$12$
			-	$5 x^{3}$	+	$x^{2}$
					-	$5 x^{2}$	-	$59 x$

Divide the highest order term in the dividend $- 5 x^{2}$ by the highest order term in divisor $5 x$ .

				$x^{2}$	-	$x$
$5 x$	-	$1$		$5 x^{3}$	-	$6 x^{2}$	-	$59 x$	+	$12$
			-	$5 x^{3}$	+	$x^{2}$
					-	$5 x^{2}$	-	$59 x$

Multiply the new quotient term by the divisor.

				$x^{2}$	-	$x$
$5 x$	-	$1$		$5 x^{3}$	-	$6 x^{2}$	-	$59 x$	+	$12$
			-	$5 x^{3}$	+	$x^{2}$
					-	$5 x^{2}$	-	$59 x$
					-	$5 x^{2}$	+	$x$

The expression needs to be subtracted from the dividend, so change all the signs in $- 5 x^{2} + x$

				$x^{2}$	-	$x$
$5 x$	-	$1$		$5 x^{3}$	-	$6 x^{2}$	-	$59 x$	+	$12$
			-	$5 x^{3}$	+	$x^{2}$
					-	$5 x^{2}$	-	$59 x$
					+	$5 x^{2}$	-	$x$

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

				$x^{2}$	-	$x$
$5 x$	-	$1$		$5 x^{3}$	-	$6 x^{2}$	-	$59 x$	+	$12$
			-	$5 x^{3}$	+	$x^{2}$
					-	$5 x^{2}$	-	$59 x$
					+	$5 x^{2}$	-	$x$
							-	$60 x$

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

				$x^{2}$	-	$x$
$5 x$	-	$1$		$5 x^{3}$	-	$6 x^{2}$	-	$59 x$	+	$12$
			-	$5 x^{3}$	+	$x^{2}$
					-	$5 x^{2}$	-	$59 x$
					+	$5 x^{2}$	-	$x$
							-	$60 x$	+	$12$

Divide the highest order term in the dividend $- 60 x$ by the highest order term in divisor $5 x$ .

				$x^{2}$	-	$x$	-	$12$
$5 x$	-	$1$		$5 x^{3}$	-	$6 x^{2}$	-	$59 x$	+	$12$
			-	$5 x^{3}$	+	$x^{2}$
					-	$5 x^{2}$	-	$59 x$
					+	$5 x^{2}$	-	$x$
							-	$60 x$	+	$12$

Multiply the new quotient term by the divisor.

				$x^{2}$	-	$x$	-	$12$
$5 x$	-	$1$		$5 x^{3}$	-	$6 x^{2}$	-	$59 x$	+	$12$
			-	$5 x^{3}$	+	$x^{2}$
					-	$5 x^{2}$	-	$59 x$
					+	$5 x^{2}$	-	$x$
							-	$60 x$	+	$12$
							-	$60 x$	+	$12$

The expression needs to be subtracted from the dividend, so change all the signs in $- 60 x + 12$

				$x^{2}$	-	$x$	-	$12$
$5 x$	-	$1$		$5 x^{3}$	-	$6 x^{2}$	-	$59 x$	+	$12$
			-	$5 x^{3}$	+	$x^{2}$
					-	$5 x^{2}$	-	$59 x$
					+	$5 x^{2}$	-	$x$
							-	$60 x$	+	$12$
							+	$60 x$	-	$12$

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

				$x^{2}$	-	$x$	-	$12$
$5 x$	-	$1$		$5 x^{3}$	-	$6 x^{2}$	-	$59 x$	+	$12$
			-	$5 x^{3}$	+	$x^{2}$
					-	$5 x^{2}$	-	$59 x$
					+	$5 x^{2}$	-	$x$
							-	$60 x$	+	$12$
							+	$60 x$	-	$12$
										$0$

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

$x^{2} - x - 12$

Write $5 x^{3} - 6 x^{2} - 59 x + 12$ as a set of factors.

$(5 x - 1) (x^{2} - x - 12) = 0$

Factor $x^{2} - x - 12$ using the AC method.

Consider the form $x^{2} + b x + c$ . Find a pair of integers whose product is $c$ and whose sum is $b$ . In this case, whose product is $- 12$ and whose sum is $- 1$ .

$- 4, 3$

Write the factored form using these integers.

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

Remove unnecessary parentheses.

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

Step 9

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

$5 x - 1 = 0$

$x - 4 = 0$

$x + 3 = 0$

Step 10

Set $5 x - 1$ equal to $0$ and solve for $x$ .

Set $5 x - 1$ equal to $0$ .

$5 x - 1 = 0$

Solve $5 x - 1 = 0$ for $x$ .

Add $1$ to both sides of the equation.

$5 x = 1$

Divide each term in $5 x = 1$ by $5$ and simplify.

Divide each term in $5 x = 1$ by $5$ .

$\frac{5 x}{5} = \frac{1}{5}$

Simplify the left side.

Cancel the common factor of $5$ .

Cancel the common factor.

$\frac{5 x}{5} = \frac{1}{5}$

Divide $x$ by $1$ .

$x = \frac{1}{5}$

Step 11

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

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

$x - 4 = 0$

Add $4$ to both sides of the equation.

$x = 4$

Step 12

Set $x + 3$ equal to $0$ and solve for $x$ .

Set $x + 3$ equal to $0$ .

$x + 3 = 0$

Subtract $3$ from both sides of the equation.

$x = - 3$

Step 13

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

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

Step 14

Find the Roots/Zeros Using the Rational Roots Test 5x^3-6x^2-59x+12

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