Find the Roots/Zeros Using the Rational Roots Test x^3-11x^2+38x-40

$x^{3} - 11 x^{2} + 38 x - 40$

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.

$p = \pm 1, \pm 2, \pm 4, \pm 5, \pm 8, \pm 10, \pm 20, \pm 40$

$q = \pm 1$

Step 2

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

$\pm 1, \pm 2, \pm 4, \pm 5, \pm 8, \pm 10, \pm 20, \pm 40$

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.

${(2)}^{3} - 11 {(2)}^{2} + 38 (2) - 40$

Step 4

Simplify the expression. In this case, the expression is equal to $0$ so $x = 2$ is a root of the polynomial.

Simplify each term.

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

$8 - 11 {(2)}^{2} + 38 (2) - 40$

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

$8 - 11 \cdot 4 + 38 (2) - 40$

Multiply $- 11$ by $4$ .

$8 - 44 + 38 (2) - 40$

Multiply $38$ by $2$ .

$8 - 44 + 76 - 40$

Simplify by adding and subtracting.

Subtract $44$ from $8$ .

$- 36 + 76 - 40$

Add $- 36$ and $76$ .

$40 - 40$

Subtract $40$ from $40$ .

$0$

Step 5

Since $2$ is a known root, divide the polynomial by $x - 2$ to find the quotient polynomial. This polynomial can then be used to find the remaining roots.

$\frac{x^{3} - 11 x^{2} + 38 x - 40}{x - 2}$

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.

$2$	$1$	$- 11$	$38$	$- 40$
	\xa0			\xa0

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

$2$	$1$	$- 11$	$38$	$- 40$
	\xa0			\xa0
	$1$

Multiply the newest entry in the result $(1)$ by the divisor $(2)$ and place the result of $(2)$ under the next term in the dividend $(- 11)$ .

$2$	$1$	$- 11$	$38$	$- 40$
	\xa0	$2$		\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.

$2$	$1$	$- 11$	$38$	$- 40$
	\xa0	$2$		\xa0
	$1$	$- 9$

Multiply the newest entry in the result $(- 9)$ by the divisor $(2)$ and place the result of $(- 18)$ under the next term in the dividend $(38)$ .

$2$	$1$	$- 11$	$38$	$- 40$
	\xa0	$2$	$- 18$	\xa0
	$1$	$- 9$

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

$2$	$1$	$- 11$	$38$	$- 40$
	\xa0	$2$	$- 18$	\xa0
	$1$	$- 9$	$20$

Multiply the newest entry in the result $(20)$ by the divisor $(2)$ and place the result of $(40)$ under the next term in the dividend $(- 40)$ .

$2$	$1$	$- 11$	$38$	$- 40$
	\xa0	$2$	$- 18$	$40$
	$1$	$- 9$	$20$

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

$2$	$1$	$- 11$	$38$	$- 40$
	\xa0	$2$	$- 18$	$40$
	$1$	$- 9$	$20$	$0$

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

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

Simplify the quotient polynomial.

$x^{2} - 9 x + 20$

Step 7

Factor $x^{2} - 9 x + 20$ 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 $20$ and whose sum is $- 9$ .

$- 5, - 4$

Write the factored form using these integers.

$(x - 2) + (x - 5) (x - 4)$

$(x - 2) (x - 5) (x - 4)$

Step 8

Factor the left side of the equation.

Factor $x^{3} - 11 x^{2} + 38 x - 40$ 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.

$p = \pm 1, \pm 40, \pm 2, \pm 20, \pm 4, \pm 10, \pm 5, \pm 8 q = \pm 1$

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

$\pm 1, \pm 40, \pm 2, \pm 20, \pm 4, \pm 10, \pm 5, \pm 8$

Substitute $2$ and simplify the expression. In this case, the expression is equal to $0$ so $2$ is a root of the polynomial.

Substitute $2$ into the polynomial.

$2^{3} - 11 \cdot 2^{2} + 38 \cdot 2 - 40$

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

$8 - 11 \cdot 2^{2} + 38 \cdot 2 - 40$

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

$8 - 11 \cdot 4 + 38 \cdot 2 - 40$

Multiply $- 11$ by $4$ .

$8 - 44 + 38 \cdot 2 - 40$

Subtract $44$ from $8$ .

$- 36 + 38 \cdot 2 - 40$

Multiply $38$ by $2$ .

$- 36 + 76 - 40$

Add $- 36$ and $76$ .

$40 - 40$

Subtract $40$ from $40$ .

$0$

Since $2$ is a known root, divide the polynomial by $x - 2$ to find the quotient polynomial. This polynomial can then be used to find the remaining roots.

$\frac{x^{3} - 11 x^{2} + 38 x - 40}{x - 2}$

Divide $x^{3} - 11 x^{2} + 38 x - 40$ by $x - 2$ .

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

$x$

$2$

$x^{3}$

$11 x^{2}$

$38 x$

$40$

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

					$x^{2}$
	$x$	-	$2$		$x^{3}$	-	$11 x^{2}$	+	$38 x$	-	$40$

Multiply the new quotient term by the divisor.

				$x^{2}$
$x$	-	$2$		$x^{3}$	-	$11 x^{2}$	+	$38 x$	-	$40$
			+	$x^{3}$	-	$2 x^{2}$

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

				$x^{2}$
$x$	-	$2$		$x^{3}$	-	$11 x^{2}$	+	$38 x$	-	$40$
			-	$x^{3}$	+	$2 x^{2}$

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

				$x^{2}$
$x$	-	$2$		$x^{3}$	-	$11 x^{2}$	+	$38 x$	-	$40$
			-	$x^{3}$	+	$2 x^{2}$
					-	$9 x^{2}$

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

				$x^{2}$
$x$	-	$2$		$x^{3}$	-	$11 x^{2}$	+	$38 x$	-	$40$
			-	$x^{3}$	+	$2 x^{2}$
					-	$9 x^{2}$	+	$38 x$

Divide the highest order term in the dividend $- 9 x^{2}$ by the highest order term in divisor $x$ .

				$x^{2}$	-	$9 x$
$x$	-	$2$		$x^{3}$	-	$11 x^{2}$	+	$38 x$	-	$40$
			-	$x^{3}$	+	$2 x^{2}$
					-	$9 x^{2}$	+	$38 x$

Multiply the new quotient term by the divisor.

				$x^{2}$	-	$9 x$
$x$	-	$2$		$x^{3}$	-	$11 x^{2}$	+	$38 x$	-	$40$
			-	$x^{3}$	+	$2 x^{2}$
					-	$9 x^{2}$	+	$38 x$
					-	$9 x^{2}$	+	$18 x$

The expression needs to be subtracted from the dividend, so change all the signs in $- 9 x^{2} + 18 x$

				$x^{2}$	-	$9 x$
$x$	-	$2$		$x^{3}$	-	$11 x^{2}$	+	$38 x$	-	$40$
			-	$x^{3}$	+	$2 x^{2}$
					-	$9 x^{2}$	+	$38 x$
					+	$9 x^{2}$	-	$18 x$

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

				$x^{2}$	-	$9 x$
$x$	-	$2$		$x^{3}$	-	$11 x^{2}$	+	$38 x$	-	$40$
			-	$x^{3}$	+	$2 x^{2}$
					-	$9 x^{2}$	+	$38 x$
					+	$9 x^{2}$	-	$18 x$
							+	$20 x$

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

				$x^{2}$	-	$9 x$
$x$	-	$2$		$x^{3}$	-	$11 x^{2}$	+	$38 x$	-	$40$
			-	$x^{3}$	+	$2 x^{2}$
					-	$9 x^{2}$	+	$38 x$
					+	$9 x^{2}$	-	$18 x$
							+	$20 x$	-	$40$

Divide the highest order term in the dividend $20 x$ by the highest order term in divisor $x$ .

				$x^{2}$	-	$9 x$	+	$20$
$x$	-	$2$		$x^{3}$	-	$11 x^{2}$	+	$38 x$	-	$40$
			-	$x^{3}$	+	$2 x^{2}$
					-	$9 x^{2}$	+	$38 x$
					+	$9 x^{2}$	-	$18 x$
							+	$20 x$	-	$40$

Multiply the new quotient term by the divisor.

				$x^{2}$	-	$9 x$	+	$20$
$x$	-	$2$		$x^{3}$	-	$11 x^{2}$	+	$38 x$	-	$40$
			-	$x^{3}$	+	$2 x^{2}$
					-	$9 x^{2}$	+	$38 x$
					+	$9 x^{2}$	-	$18 x$
							+	$20 x$	-	$40$
							+	$20 x$	-	$40$

The expression needs to be subtracted from the dividend, so change all the signs in $20 x - 40$

				$x^{2}$	-	$9 x$	+	$20$
$x$	-	$2$		$x^{3}$	-	$11 x^{2}$	+	$38 x$	-	$40$
			-	$x^{3}$	+	$2 x^{2}$
					-	$9 x^{2}$	+	$38 x$
					+	$9 x^{2}$	-	$18 x$
							+	$20 x$	-	$40$
							-	$20 x$	+	$40$

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

				$x^{2}$	-	$9 x$	+	$20$
$x$	-	$2$		$x^{3}$	-	$11 x^{2}$	+	$38 x$	-	$40$
			-	$x^{3}$	+	$2 x^{2}$
					-	$9 x^{2}$	+	$38 x$
					+	$9 x^{2}$	-	$18 x$
							+	$20 x$	-	$40$
							-	$20 x$	+	$40$
										$0$

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

$x^{2} - 9 x + 20$

Write $x^{3} - 11 x^{2} + 38 x - 40$ as a set of factors.

$(x - 2) (x^{2} - 9 x + 20) = 0$

Factor $x^{2} - 9 x + 20$ 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 $20$ and whose sum is $- 9$ .

$- 5, - 4$

Write the factored form using these integers.

$(x - 2) ((x - 5) (x - 4)) = 0$

Remove unnecessary parentheses.

$(x - 2) (x - 5) (x - 4) = 0$

Step 9

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

$x - 2 = 0$

$x - 5 = 0$

$x - 4 = 0$

Step 10

Set $x - 2$ equal to $0$ and solve for $x$ .

Set $x - 2$ equal to $0$ .

$x - 2 = 0$

Add $2$ to both sides of the equation.

$x = 2$

Step 11

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

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

$x - 5 = 0$

Add $5$ to both sides of the equation.

$x = 5$

Step 12

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 13

The final solution is all the values that make $(x - 2) (x - 5) (x - 4) = 0$ true.

$x = 2, 5, 4$

Step 14

Find the Roots/Zeros Using the Rational Roots Test x^3-11x^2+38x-40

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