Find the Roots/Zeros Using the Rational Roots Test f(x)=x^3+11x^2+40x+48

$f (x) = x^{3} + 11 x^{2} + 40 x + 48$

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.

$p = \pm 1, \pm 2, \pm 3, \pm 4, \pm 6, \pm 8, \pm 12, \pm 16, \pm 24, \pm 48$

$q = \pm 1$

Step 2

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

$\pm 1, \pm 2, \pm 3, \pm 4, \pm 6, \pm 8, \pm 12, \pm 16, \pm 24, \pm 48$

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.

${(- 3)}^{3} + 11 {(- 3)}^{2} + 40 (- 3) + 48$

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 $3$ .

$- 27 + 11 {(- 3)}^{2} + 40 (- 3) + 48$

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

$- 27 + 11 \cdot 9 + 40 (- 3) + 48$

Multiply $11$ by $9$ .

$- 27 + 99 + 40 (- 3) + 48$

Multiply $40$ by $- 3$ .

$- 27 + 99 - 120 + 48$

Simplify by adding and subtracting.

Add $- 27$ and $99$ .

$72 - 120 + 48$

Subtract $120$ from $72$ .

$- 48 + 48$

Add $- 48$ and $48$ .

$0$

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.

$\frac{x^{3} + 11 x^{2} + 40 x + 48}{x + 3}$

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.

$- 3$	$1$	$11$	$40$	$48$
	\xa0			\xa0

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

$- 3$	$1$	$11$	$40$	$48$
	\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 $(11)$ .

$- 3$	$1$	$11$	$40$	$48$
	\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.

$- 3$	$1$	$11$	$40$	$48$
	\xa0	$- 3$		\xa0
	$1$	$8$

Multiply the newest entry in the result $(8)$ by the divisor $(- 3)$ and place the result of $(- 24)$ under the next term in the dividend $(40)$ .

$- 3$	$1$	$11$	$40$	$48$
	\xa0	$- 3$	$- 24$	\xa0
	$1$	$8$

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

$- 3$	$1$	$11$	$40$	$48$
	\xa0	$- 3$	$- 24$	\xa0
	$1$	$8$	$16$

Multiply the newest entry in the result $(16)$ by the divisor $(- 3)$ and place the result of $(- 48)$ under the next term in the dividend $(48)$ .

$- 3$	$1$	$11$	$40$	$48$
	\xa0	$- 3$	$- 24$	$- 48$
	$1$	$8$	$16$

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

$- 3$	$1$	$11$	$40$	$48$
	\xa0	$- 3$	$- 24$	$- 48$
	$1$	$8$	$16$	$0$

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

$1 x^{2} + 8 x + 16$

Simplify the quotient polynomial.

$x^{2} + 8 x + 16$

Step 7

Factor using the perfect square rule.

Rewrite $16$ as $4^{2}$ .

$(x + 3) + x^{2} + 8 x + 4^{2}$

Check that the middle term is two times the product of the numbers being squared in the first term and third term.

$8 x = 2 \cdot x \cdot 4$

Rewrite the polynomial.

$(x + 3) + x^{2} + 2 \cdot x \cdot 4 + 4^{2}$

Factor using the perfect square trinomial rule $a^{2} + 2 a b + b^{2} = {(a + b)}^{2}$ , where $a = x$ and $b = 4$ .

$(x + 3) + {(x + 4)}^{2}$

$(x + 3) {(x + 4)}^{2}$

Step 8

Factor the left side of the equation.

Factor $x^{3} + 11 x^{2} + 40 x + 48$ 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.

$p = \pm 1, \pm 48, \pm 2, \pm 24, \pm 3, \pm 16, \pm 4, \pm 12, \pm 6, \pm 8 q = \pm 1$

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

$\pm 1, \pm 48, \pm 2, \pm 24, \pm 3, \pm 16, \pm 4, \pm 12, \pm 6, \pm 8$

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.

${(- 3)}^{3} + 11 {(- 3)}^{2} + 40 \cdot - 3 + 48$

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

$- 27 + 11 {(- 3)}^{2} + 40 \cdot - 3 + 48$

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

$- 27 + 11 \cdot 9 + 40 \cdot - 3 + 48$

Multiply $11$ by $9$ .

$- 27 + 99 + 40 \cdot - 3 + 48$

Add $- 27$ and $99$ .

$72 + 40 \cdot - 3 + 48$

Multiply $40$ by $- 3$ .

$72 - 120 + 48$

Subtract $120$ from $72$ .

$- 48 + 48$

Add $- 48$ and $48$ .

$0$

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.

$\frac{x^{3} + 11 x^{2} + 40 x + 48}{x + 3}$

Divide $x^{3} + 11 x^{2} + 40 x + 48$ 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 $0$ .

$x$

$3$

$x^{3}$

$11 x^{2}$

$40 x$

$48$

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

					$x^{2}$
	$x$	+	$3$		$x^{3}$	+	$11 x^{2}$	+	$40 x$	+	$48$

Multiply the new quotient term by the divisor.

				$x^{2}$
$x$	+	$3$		$x^{3}$	+	$11 x^{2}$	+	$40 x$	+	$48$
			+	$x^{3}$	+	$3 x^{2}$

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

				$x^{2}$
$x$	+	$3$		$x^{3}$	+	$11 x^{2}$	+	$40 x$	+	$48$
			-	$x^{3}$	-	$3 x^{2}$

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

				$x^{2}$
$x$	+	$3$		$x^{3}$	+	$11 x^{2}$	+	$40 x$	+	$48$
			-	$x^{3}$	-	$3 x^{2}$
					+	$8 x^{2}$

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

				$x^{2}$
$x$	+	$3$		$x^{3}$	+	$11 x^{2}$	+	$40 x$	+	$48$
			-	$x^{3}$	-	$3 x^{2}$
					+	$8 x^{2}$	+	$40 x$

Divide the highest order term in the dividend $8 x^{2}$ by the highest order term in divisor $x$ .

				$x^{2}$	+	$8 x$
$x$	+	$3$		$x^{3}$	+	$11 x^{2}$	+	$40 x$	+	$48$
			-	$x^{3}$	-	$3 x^{2}$
					+	$8 x^{2}$	+	$40 x$

Multiply the new quotient term by the divisor.

				$x^{2}$	+	$8 x$
$x$	+	$3$		$x^{3}$	+	$11 x^{2}$	+	$40 x$	+	$48$
			-	$x^{3}$	-	$3 x^{2}$
					+	$8 x^{2}$	+	$40 x$
					+	$8 x^{2}$	+	$24 x$

The expression needs to be subtracted from the dividend, so change all the signs in $8 x^{2} + 24 x$

				$x^{2}$	+	$8 x$
$x$	+	$3$		$x^{3}$	+	$11 x^{2}$	+	$40 x$	+	$48$
			-	$x^{3}$	-	$3 x^{2}$
					+	$8 x^{2}$	+	$40 x$
					-	$8 x^{2}$	-	$24 x$

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

				$x^{2}$	+	$8 x$
$x$	+	$3$		$x^{3}$	+	$11 x^{2}$	+	$40 x$	+	$48$
			-	$x^{3}$	-	$3 x^{2}$
					+	$8 x^{2}$	+	$40 x$
					-	$8 x^{2}$	-	$24 x$
							+	$16 x$

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

				$x^{2}$	+	$8 x$
$x$	+	$3$		$x^{3}$	+	$11 x^{2}$	+	$40 x$	+	$48$
			-	$x^{3}$	-	$3 x^{2}$
					+	$8 x^{2}$	+	$40 x$
					-	$8 x^{2}$	-	$24 x$
							+	$16 x$	+	$48$

Divide the highest order term in the dividend $16 x$ by the highest order term in divisor $x$ .

				$x^{2}$	+	$8 x$	+	$16$
$x$	+	$3$		$x^{3}$	+	$11 x^{2}$	+	$40 x$	+	$48$
			-	$x^{3}$	-	$3 x^{2}$
					+	$8 x^{2}$	+	$40 x$
					-	$8 x^{2}$	-	$24 x$
							+	$16 x$	+	$48$

Multiply the new quotient term by the divisor.

				$x^{2}$	+	$8 x$	+	$16$
$x$	+	$3$		$x^{3}$	+	$11 x^{2}$	+	$40 x$	+	$48$
			-	$x^{3}$	-	$3 x^{2}$
					+	$8 x^{2}$	+	$40 x$
					-	$8 x^{2}$	-	$24 x$
							+	$16 x$	+	$48$
							+	$16 x$	+	$48$

The expression needs to be subtracted from the dividend, so change all the signs in $16 x + 48$

				$x^{2}$	+	$8 x$	+	$16$
$x$	+	$3$		$x^{3}$	+	$11 x^{2}$	+	$40 x$	+	$48$
			-	$x^{3}$	-	$3 x^{2}$
					+	$8 x^{2}$	+	$40 x$
					-	$8 x^{2}$	-	$24 x$
							+	$16 x$	+	$48$
							-	$16 x$	-	$48$

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

				$x^{2}$	+	$8 x$	+	$16$
$x$	+	$3$		$x^{3}$	+	$11 x^{2}$	+	$40 x$	+	$48$
			-	$x^{3}$	-	$3 x^{2}$
					+	$8 x^{2}$	+	$40 x$
					-	$8 x^{2}$	-	$24 x$
							+	$16 x$	+	$48$
							-	$16 x$	-	$48$
										$0$

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

$x^{2} + 8 x + 16$

Write $x^{3} + 11 x^{2} + 40 x + 48$ as a set of factors.

$(x + 3) (x^{2} + 8 x + 16) = 0$

Factor using the perfect square rule.

Rewrite $16$ as $4^{2}$ .

$(x + 3) (x^{2} + 8 x + 4^{2}) = 0$

Check that the middle term is two times the product of the numbers being squared in the first term and third term.

$8 x = 2 \cdot x \cdot 4$

Rewrite the polynomial.

$(x + 3) (x^{2} + 2 \cdot x \cdot 4 + 4^{2}) = 0$

Factor using the perfect square trinomial rule $a^{2} + 2 a b + b^{2} = {(a + b)}^{2}$ , where $a = x$ and $b = 4$ .

$(x + 3) {(x + 4)}^{2} = 0$

Step 9

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

$x + 3 = 0$

${(x + 4)}^{2} = 0$

Step 10

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 11

Set ${(x + 4)}^{2}$ equal to $0$ and solve for $x$ .

Set ${(x + 4)}^{2}$ equal to $0$ .

${(x + 4)}^{2} = 0$

Solve ${(x + 4)}^{2} = 0$ for $x$ .

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

$x + 4 = 0$

Subtract $4$ from both sides of the equation.

$x = - 4$

Step 12

The final solution is all the values that make $(x + 3) {(x + 4)}^{2} = 0$ true.

$x = - 3, - 4$

Step 13

Find the Roots/Zeros Using the Rational Roots Test f(x)=x^3+11x^2+40x+48

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