Find the Roots/Zeros Using the Rational Roots Test f(x)=x^3-5x^2-61x-55

$f (x) = x^{3} - 5 x^{2} - 61 x - 55$

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.

$p = \pm 1, \pm 5, \pm 11, \pm 55$

$q = \pm 1$

Step 2

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

$\pm 1, \pm 5, \pm 11, \pm 55$

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.

${(- 1)}^{3} - 5 {(- 1)}^{2} - 61 \cdot - 1 - 55$

Step 4

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

Simplify each term.

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

$- 1 - 5 {(- 1)}^{2} - 61 \cdot - 1 - 55$

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

$- 1 - 5 \cdot 1 - 61 \cdot - 1 - 55$

Multiply $- 5$ by $1$ .

$- 1 - 5 - 61 \cdot - 1 - 55$

Multiply $- 61$ by $- 1$ .

$- 1 - 5 + 61 - 55$

Simplify by adding and subtracting.

Subtract $5$ from $- 1$ .

$- 6 + 61 - 55$

Add $- 6$ and $61$ .

$55 - 55$

Subtract $55$ from $55$ .

$0$

Step 5

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.

$\frac{x^{3} - 5 x^{2} - 61 x - 55}{x + 1}$

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.

$- 1$	$1$	$- 5$	$- 61$	$- 55$
	\xa0			\xa0

The first number in the dividend $(1)$ is put into the first position of the result area (below the horizontal line).

$- 1$	$1$	$- 5$	$- 61$	$- 55$
	\xa0			\xa0
	$1$

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 $(- 5)$ .

$- 1$	$1$	$- 5$	$- 61$	$- 55$
	\xa0	$- 1$		\xa0
	$1$

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

$- 1$	$1$	$- 5$	$- 61$	$- 55$
	\xa0	$- 1$		\xa0
	$1$	$- 6$

Multiply the newest entry in the result $(- 6)$ by the divisor $(- 1)$ and place the result of $(6)$ under the next term in the dividend $(- 61)$ .

$- 1$	$1$	$- 5$	$- 61$	$- 55$
	\xa0	$- 1$	$6$	\xa0
	$1$	$- 6$

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

$- 1$	$1$	$- 5$	$- 61$	$- 55$
	\xa0	$- 1$	$6$	\xa0
	$1$	$- 6$	$- 55$

Multiply the newest entry in the result $(- 55)$ by the divisor $(- 1)$ and place the result of $(55)$ under the next term in the dividend $(- 55)$ .

$- 1$	$1$	$- 5$	$- 61$	$- 55$
	\xa0	$- 1$	$6$	$55$
	$1$	$- 6$	$- 55$

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

$- 1$	$1$	$- 5$	$- 61$	$- 55$
	\xa0	$- 1$	$6$	$55$
	$1$	$- 6$	$- 55$	$0$

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

$1 x^{2} + - 6 x - 55$

Simplify the quotient polynomial.

$x^{2} - 6 x - 55$

Step 7

Factor $x^{2} - 6 x - 55$ 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 $- 55$ and whose sum is $- 6$ .

$- 11, 5$

Write the factored form using these integers.

$(x + 1) + (x - 11) (x + 5)$

$(x + 1) (x - 11) (x + 5)$

Step 8

Factor the left side of the equation.

Factor $x^{3} - 5 x^{2} - 61 x - 55$ 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.

$p = \pm 1, \pm 55, \pm 5, \pm 11 q = \pm 1$

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

$\pm 1, \pm 55, \pm 5, \pm 11$

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.

${(- 1)}^{3} - 5 {(- 1)}^{2} - 61 \cdot - 1 - 55$

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

$- 1 - 5 {(- 1)}^{2} - 61 \cdot - 1 - 55$

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

$- 1 - 5 \cdot 1 - 61 \cdot - 1 - 55$

Multiply $- 5$ by $1$ .

$- 1 - 5 - 61 \cdot - 1 - 55$

Subtract $5$ from $- 1$ .

$- 6 - 61 \cdot - 1 - 55$

Multiply $- 61$ by $- 1$ .

$- 6 + 61 - 55$

Add $- 6$ and $61$ .

$55 - 55$

Subtract $55$ from $55$ .

$0$

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.

$\frac{x^{3} - 5 x^{2} - 61 x - 55}{x + 1}$

Divide $x^{3} - 5 x^{2} - 61 x - 55$ 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 $0$ .

$x$

$1$

$x^{3}$

$5 x^{2}$

$61 x$

$55$

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

					$x^{2}$
	$x$	+	$1$		$x^{3}$	-	$5 x^{2}$	-	$61 x$	-	$55$

Multiply the new quotient term by the divisor.

				$x^{2}$
$x$	+	$1$		$x^{3}$	-	$5 x^{2}$	-	$61 x$	-	$55$
			+	$x^{3}$	+	$x^{2}$

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

				$x^{2}$
$x$	+	$1$		$x^{3}$	-	$5 x^{2}$	-	$61 x$	-	$55$
			-	$x^{3}$	-	$x^{2}$

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

				$x^{2}$
$x$	+	$1$		$x^{3}$	-	$5 x^{2}$	-	$61 x$	-	$55$
			-	$x^{3}$	-	$x^{2}$
					-	$6 x^{2}$

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

				$x^{2}$
$x$	+	$1$		$x^{3}$	-	$5 x^{2}$	-	$61 x$	-	$55$
			-	$x^{3}$	-	$x^{2}$
					-	$6 x^{2}$	-	$61 x$

Divide the highest order term in the dividend $- 6 x^{2}$ by the highest order term in divisor $x$ .

				$x^{2}$	-	$6 x$
$x$	+	$1$		$x^{3}$	-	$5 x^{2}$	-	$61 x$	-	$55$
			-	$x^{3}$	-	$x^{2}$
					-	$6 x^{2}$	-	$61 x$

Multiply the new quotient term by the divisor.

				$x^{2}$	-	$6 x$
$x$	+	$1$		$x^{3}$	-	$5 x^{2}$	-	$61 x$	-	$55$
			-	$x^{3}$	-	$x^{2}$
					-	$6 x^{2}$	-	$61 x$
					-	$6 x^{2}$	-	$6 x$

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

				$x^{2}$	-	$6 x$
$x$	+	$1$		$x^{3}$	-	$5 x^{2}$	-	$61 x$	-	$55$
			-	$x^{3}$	-	$x^{2}$
					-	$6 x^{2}$	-	$61 x$
					+	$6 x^{2}$	+	$6 x$

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

				$x^{2}$	-	$6 x$
$x$	+	$1$		$x^{3}$	-	$5 x^{2}$	-	$61 x$	-	$55$
			-	$x^{3}$	-	$x^{2}$
					-	$6 x^{2}$	-	$61 x$
					+	$6 x^{2}$	+	$6 x$
							-	$55 x$

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

				$x^{2}$	-	$6 x$
$x$	+	$1$		$x^{3}$	-	$5 x^{2}$	-	$61 x$	-	$55$
			-	$x^{3}$	-	$x^{2}$
					-	$6 x^{2}$	-	$61 x$
					+	$6 x^{2}$	+	$6 x$
							-	$55 x$	-	$55$

Divide the highest order term in the dividend $- 55 x$ by the highest order term in divisor $x$ .

				$x^{2}$	-	$6 x$	-	$55$
$x$	+	$1$		$x^{3}$	-	$5 x^{2}$	-	$61 x$	-	$55$
			-	$x^{3}$	-	$x^{2}$
					-	$6 x^{2}$	-	$61 x$
					+	$6 x^{2}$	+	$6 x$
							-	$55 x$	-	$55$

Multiply the new quotient term by the divisor.

				$x^{2}$	-	$6 x$	-	$55$
$x$	+	$1$		$x^{3}$	-	$5 x^{2}$	-	$61 x$	-	$55$
			-	$x^{3}$	-	$x^{2}$
					-	$6 x^{2}$	-	$61 x$
					+	$6 x^{2}$	+	$6 x$
							-	$55 x$	-	$55$
							-	$55 x$	-	$55$

The expression needs to be subtracted from the dividend, so change all the signs in $- 55 x - 55$

				$x^{2}$	-	$6 x$	-	$55$
$x$	+	$1$		$x^{3}$	-	$5 x^{2}$	-	$61 x$	-	$55$
			-	$x^{3}$	-	$x^{2}$
					-	$6 x^{2}$	-	$61 x$
					+	$6 x^{2}$	+	$6 x$
							-	$55 x$	-	$55$
							+	$55 x$	+	$55$

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

				$x^{2}$	-	$6 x$	-	$55$
$x$	+	$1$		$x^{3}$	-	$5 x^{2}$	-	$61 x$	-	$55$
			-	$x^{3}$	-	$x^{2}$
					-	$6 x^{2}$	-	$61 x$
					+	$6 x^{2}$	+	$6 x$
							-	$55 x$	-	$55$
							+	$55 x$	+	$55$
										$0$

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

$x^{2} - 6 x - 55$

Write $x^{3} - 5 x^{2} - 61 x - 55$ as a set of factors.

$(x + 1) (x^{2} - 6 x - 55) = 0$

Factor $x^{2} - 6 x - 55$ 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 $- 55$ and whose sum is $- 6$ .

$- 11, 5$

Write the factored form using these integers.

$(x + 1) ((x - 11) (x + 5)) = 0$

Remove unnecessary parentheses.

$(x + 1) (x - 11) (x + 5) = 0$

Step 9

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

$x + 1 = 0$

$x - 11 = 0$

$x + 5 = 0$

Step 10

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

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

$x + 1 = 0$

Subtract $1$ from both sides of the equation.

$x = - 1$

Step 11

Set $x - 11$ equal to $0$ and solve for $x$ .

Set $x - 11$ equal to $0$ .

$x - 11 = 0$

Add $11$ to both sides of the equation.

$x = 11$

Step 12

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 13

The final solution is all the values that make $(x + 1) (x - 11) (x + 5) = 0$ true.

$x = - 1, 11, - 5$

Step 14

Find the Roots/Zeros Using the Rational Roots Test f(x)=x^3-5x^2-61x-55

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