Find the Roots/Zeros Using the Rational Roots Test f(x)=x^3-5x^2-18x+72

$f (x) = x^{3} - 5 x^{2} - 18 x + 72$

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.

$p = \pm 1, \pm 2, \pm 3, \pm 4, \pm 6, \pm 8, \pm 9, \pm 12, \pm 18, \pm 24, \pm 36, \pm 72$

$q = \pm 1$

Step 2

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

$\pm 1, \pm 2, \pm 3, \pm 4, \pm 6, \pm 8, \pm 9, \pm 12, \pm 18, \pm 24, \pm 36, \pm 72$

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.

${(3)}^{3} - 5 {(3)}^{2} - 18 \cdot 3 + 72$

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 $3$ .

$27 - 5 {(3)}^{2} - 18 \cdot 3 + 72$

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

$27 - 5 \cdot 9 - 18 \cdot 3 + 72$

Multiply $- 5$ by $9$ .

$27 - 45 - 18 \cdot 3 + 72$

Multiply $- 18$ by $3$ .

$27 - 45 - 54 + 72$

Simplify by adding and subtracting.

Subtract $45$ from $27$ .

$- 18 - 54 + 72$

Subtract $54$ from $- 18$ .

$- 72 + 72$

Add $- 72$ and $72$ .

$0$

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.

$\frac{x^{3} - 5 x^{2} - 18 x + 72}{x - 3}$

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.

$3$	$1$	$- 5$	$- 18$	$72$
	\xa0			\xa0

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

$3$	$1$	$- 5$	$- 18$	$72$
	\xa0			\xa0
	$1$

Multiply the newest entry in the result $(1)$ by the divisor $(3)$ and place the result of $(3)$ under the next term in the dividend $(- 5)$ .

$3$	$1$	$- 5$	$- 18$	$72$
	\xa0	$3$		\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.

$3$	$1$	$- 5$	$- 18$	$72$
	\xa0	$3$		\xa0
	$1$	$- 2$

Multiply the newest entry in the result $(- 2)$ by the divisor $(3)$ and place the result of $(- 6)$ under the next term in the dividend $(- 18)$ .

$3$	$1$	$- 5$	$- 18$	$72$
	\xa0	$3$	$- 6$	\xa0
	$1$	$- 2$

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

$3$	$1$	$- 5$	$- 18$	$72$
	\xa0	$3$	$- 6$	\xa0
	$1$	$- 2$	$- 24$

Multiply the newest entry in the result $(- 24)$ by the divisor $(3)$ and place the result of $(- 72)$ under the next term in the dividend $(72)$ .

$3$	$1$	$- 5$	$- 18$	$72$
	\xa0	$3$	$- 6$	$- 72$
	$1$	$- 2$	$- 24$

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

$3$	$1$	$- 5$	$- 18$	$72$
	\xa0	$3$	$- 6$	$- 72$
	$1$	$- 2$	$- 24$	$0$

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

$1 x^{2} + - 2 x - 24$

Simplify the quotient polynomial.

$x^{2} - 2 x - 24$

Step 7

Factor $x^{2} - 2 x - 24$ 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 $- 24$ and whose sum is $- 2$ .

$- 6, 4$

Write the factored form using these integers.

$(x - 3) + (x - 6) (x + 4)$

$(x - 3) (x - 6) (x + 4)$

Step 8

Factor the left side of the equation.

Factor $x^{3} - 5 x^{2} - 18 x + 72$ 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.

$p = \pm 1, \pm 72, \pm 2, \pm 36, \pm 3, \pm 24, \pm 4, \pm 18, \pm 6, \pm 12, \pm 8, \pm 9 q = \pm 1$

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

$\pm 1, \pm 72, \pm 2, \pm 36, \pm 3, \pm 24, \pm 4, \pm 18, \pm 6, \pm 12, \pm 8, \pm 9$

Substitute $3$ and simplify the expression. In this case, the expression is equal to $0$ so $3$ is a root of the polynomial.

Substitute $3$ into the polynomial.

$3^{3} - 5 \cdot 3^{2} - 18 \cdot 3 + 72$

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

$27 - 5 \cdot 3^{2} - 18 \cdot 3 + 72$

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

$27 - 5 \cdot 9 - 18 \cdot 3 + 72$

Multiply $- 5$ by $9$ .

$27 - 45 - 18 \cdot 3 + 72$

Subtract $45$ from $27$ .

$- 18 - 18 \cdot 3 + 72$

Multiply $- 18$ by $3$ .

$- 18 - 54 + 72$

Subtract $54$ from $- 18$ .

$- 72 + 72$

Add $- 72$ and $72$ .

$0$

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.

$\frac{x^{3} - 5 x^{2} - 18 x + 72}{x - 3}$

Divide $x^{3} - 5 x^{2} - 18 x + 72$ by $x - 3$ .

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

$x$

$3$

$x^{3}$

$5 x^{2}$

$18 x$

$72$

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

					$x^{2}$
	$x$	-	$3$		$x^{3}$	-	$5 x^{2}$	-	$18 x$	+	$72$

Multiply the new quotient term by the divisor.

				$x^{2}$
$x$	-	$3$		$x^{3}$	-	$5 x^{2}$	-	$18 x$	+	$72$
			+	$x^{3}$	-	$3 x^{2}$

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

				$x^{2}$
$x$	-	$3$		$x^{3}$	-	$5 x^{2}$	-	$18 x$	+	$72$
			-	$x^{3}$	+	$3 x^{2}$

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

				$x^{2}$
$x$	-	$3$		$x^{3}$	-	$5 x^{2}$	-	$18 x$	+	$72$
			-	$x^{3}$	+	$3 x^{2}$
					-	$2 x^{2}$

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

				$x^{2}$
$x$	-	$3$		$x^{3}$	-	$5 x^{2}$	-	$18 x$	+	$72$
			-	$x^{3}$	+	$3 x^{2}$
					-	$2 x^{2}$	-	$18 x$

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

				$x^{2}$	-	$2 x$
$x$	-	$3$		$x^{3}$	-	$5 x^{2}$	-	$18 x$	+	$72$
			-	$x^{3}$	+	$3 x^{2}$
					-	$2 x^{2}$	-	$18 x$

Multiply the new quotient term by the divisor.

				$x^{2}$	-	$2 x$
$x$	-	$3$		$x^{3}$	-	$5 x^{2}$	-	$18 x$	+	$72$
			-	$x^{3}$	+	$3 x^{2}$
					-	$2 x^{2}$	-	$18 x$
					-	$2 x^{2}$	+	$6 x$

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

				$x^{2}$	-	$2 x$
$x$	-	$3$		$x^{3}$	-	$5 x^{2}$	-	$18 x$	+	$72$
			-	$x^{3}$	+	$3 x^{2}$
					-	$2 x^{2}$	-	$18 x$
					+	$2 x^{2}$	-	$6 x$

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

				$x^{2}$	-	$2 x$
$x$	-	$3$		$x^{3}$	-	$5 x^{2}$	-	$18 x$	+	$72$
			-	$x^{3}$	+	$3 x^{2}$
					-	$2 x^{2}$	-	$18 x$
					+	$2 x^{2}$	-	$6 x$
							-	$24 x$

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

				$x^{2}$	-	$2 x$
$x$	-	$3$		$x^{3}$	-	$5 x^{2}$	-	$18 x$	+	$72$
			-	$x^{3}$	+	$3 x^{2}$
					-	$2 x^{2}$	-	$18 x$
					+	$2 x^{2}$	-	$6 x$
							-	$24 x$	+	$72$

Divide the highest order term in the dividend $- 24 x$ by the highest order term in divisor $x$ .

				$x^{2}$	-	$2 x$	-	$24$
$x$	-	$3$		$x^{3}$	-	$5 x^{2}$	-	$18 x$	+	$72$
			-	$x^{3}$	+	$3 x^{2}$
					-	$2 x^{2}$	-	$18 x$
					+	$2 x^{2}$	-	$6 x$
							-	$24 x$	+	$72$

Multiply the new quotient term by the divisor.

				$x^{2}$	-	$2 x$	-	$24$
$x$	-	$3$		$x^{3}$	-	$5 x^{2}$	-	$18 x$	+	$72$
			-	$x^{3}$	+	$3 x^{2}$
					-	$2 x^{2}$	-	$18 x$
					+	$2 x^{2}$	-	$6 x$
							-	$24 x$	+	$72$
							-	$24 x$	+	$72$

The expression needs to be subtracted from the dividend, so change all the signs in $- 24 x + 72$

				$x^{2}$	-	$2 x$	-	$24$
$x$	-	$3$		$x^{3}$	-	$5 x^{2}$	-	$18 x$	+	$72$
			-	$x^{3}$	+	$3 x^{2}$
					-	$2 x^{2}$	-	$18 x$
					+	$2 x^{2}$	-	$6 x$
							-	$24 x$	+	$72$
							+	$24 x$	-	$72$

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

				$x^{2}$	-	$2 x$	-	$24$
$x$	-	$3$		$x^{3}$	-	$5 x^{2}$	-	$18 x$	+	$72$
			-	$x^{3}$	+	$3 x^{2}$
					-	$2 x^{2}$	-	$18 x$
					+	$2 x^{2}$	-	$6 x$
							-	$24 x$	+	$72$
							+	$24 x$	-	$72$
										$0$

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

$x^{2} - 2 x - 24$

Write $x^{3} - 5 x^{2} - 18 x + 72$ as a set of factors.

$(x - 3) (x^{2} - 2 x - 24) = 0$

Factor $x^{2} - 2 x - 24$ 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 $- 24$ and whose sum is $- 2$ .

$- 6, 4$

Write the factored form using these integers.

$(x - 3) ((x - 6) (x + 4)) = 0$

Remove unnecessary parentheses.

$(x - 3) (x - 6) (x + 4) = 0$

Step 9

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

$x - 3 = 0$

$x - 6 = 0$

$x + 4 = 0$

Step 10

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

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

$x - 3 = 0$

Add $3$ to both sides of the equation.

$x = 3$

Step 11

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

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

$x - 6 = 0$

Add $6$ to both sides of the equation.

$x = 6$

Step 12

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

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

$x + 4 = 0$

Subtract $4$ from both sides of the equation.

$x = - 4$

Step 13

The final solution is all the values that make $(x - 3) (x - 6) (x + 4) = 0$ true.

$x = 3, 6, - 4$

Step 14

Find the Roots/Zeros Using the Rational Roots Test f(x)=x^3-5x^2-18x+72

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