Find the Roots/Zeros Using the Rational Roots Test f(x)=x^4-7x^3+8x^2+28x-48

$f (x) = x^{4} - 7 x^{3} + 8 x^{2} + 28 x - 48$

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.

${(2)}^{4} - 7 {(2)}^{3} + 8 {(2)}^{2} + 28 (2) - 48$

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 $4$ .

$16 - 7 {(2)}^{3} + 8 {(2)}^{2} + 28 (2) - 48$

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

$16 - 7 \cdot 8 + 8 {(2)}^{2} + 28 (2) - 48$

Multiply $- 7$ by $8$ .

$16 - 56 + 8 {(2)}^{2} + 28 (2) - 48$

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

$16 - 56 + 8 \cdot 4 + 28 (2) - 48$

Multiply $8$ by $4$ .

$16 - 56 + 32 + 28 (2) - 48$

Multiply $28$ by $2$ .

$16 - 56 + 32 + 56 - 48$

Simplify by adding and subtracting.

Subtract $56$ from $16$ .

$- 40 + 32 + 56 - 48$

Add $- 40$ and $32$ .

$- 8 + 56 - 48$

Add $- 8$ and $56$ .

$48 - 48$

Subtract $48$ from $48$ .

$0$

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.

$\frac{x^{4} - 7 x^{3} + 8 x^{2} + 28 x - 48}{x - 2}$

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.

$2$	$1$	$- 7$	$8$	$28$	$- 48$
	\xa0				\xa0

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

$2$	$1$	$- 7$	$8$	$28$	$- 48$
	\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 $(- 7)$ .

$2$	$1$	$- 7$	$8$	$28$	$- 48$
	\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.

$2$	$1$	$- 7$	$8$	$28$	$- 48$
	\xa0	$2$			\xa0
	$1$	$- 5$

Multiply the newest entry in the result $(- 5)$ by the divisor $(2)$ and place the result of $(- 10)$ under the next term in the dividend $(8)$ .

$2$	$1$	$- 7$	$8$	$28$	$- 48$
	\xa0	$2$	$- 10$		\xa0
	$1$	$- 5$

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

$2$	$1$	$- 7$	$8$	$28$	$- 48$
	\xa0	$2$	$- 10$		\xa0
	$1$	$- 5$	$- 2$

Multiply the newest entry in the result $(- 2)$ by the divisor $(2)$ and place the result of $(- 4)$ under the next term in the dividend $(28)$ .

$2$	$1$	$- 7$	$8$	$28$	$- 48$
	\xa0	$2$	$- 10$	$- 4$	\xa0
	$1$	$- 5$	$- 2$

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

$2$	$1$	$- 7$	$8$	$28$	$- 48$
	\xa0	$2$	$- 10$	$- 4$	\xa0
	$1$	$- 5$	$- 2$	$24$

Multiply the newest entry in the result $(24)$ by the divisor $(2)$ and place the result of $(48)$ under the next term in the dividend $(- 48)$ .

$2$	$1$	$- 7$	$8$	$28$	$- 48$
	\xa0	$2$	$- 10$	$- 4$	$48$
	$1$	$- 5$	$- 2$	$24$

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

$2$	$1$	$- 7$	$8$	$28$	$- 48$
	\xa0	$2$	$- 10$	$- 4$	$48$
	$1$	$- 5$	$- 2$	$24$	$0$

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

$1 x^{3} + - 5 x^{2} + (- 2) x + 24$

Simplify the quotient polynomial.

$x^{3} - 5 x^{2} - 2 x + 24$

Step 7

Solve the equation to find any remaining roots.

Factor the left side of the equation.

Factor $x^{3} - 5 x^{2} - 2 x + 24$ 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.

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

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

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

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.

${(- 2)}^{3} - 5 {(- 2)}^{2} - 2 \cdot - 2 + 24$

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

$- 8 - 5 {(- 2)}^{2} - 2 \cdot - 2 + 24$

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

$- 8 - 5 \cdot 4 - 2 \cdot - 2 + 24$

Multiply $- 5$ by $4$ .

$- 8 - 20 - 2 \cdot - 2 + 24$

Subtract $20$ from $- 8$ .

$- 28 - 2 \cdot - 2 + 24$

Multiply $- 2$ by $- 2$ .

$- 28 + 4 + 24$

Add $- 28$ and $4$ .

$- 24 + 24$

Add $- 24$ and $24$ .

$0$

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.

$\frac{x^{3} - 5 x^{2} - 2 x + 24}{x + 2}$

Divide $x^{3} - 5 x^{2} - 2 x + 24$ 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 $0$ .

$x$

$2$

$x^{3}$

$5 x^{2}$

$2 x$

$24$

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

					$x^{2}$
	$x$	+	$2$		$x^{3}$	-	$5 x^{2}$	-	$2 x$	+	$24$

Multiply the new quotient term by the divisor.

				$x^{2}$
$x$	+	$2$		$x^{3}$	-	$5 x^{2}$	-	$2 x$	+	$24$
			+	$x^{3}$	+	$2 x^{2}$

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

				$x^{2}$
$x$	+	$2$		$x^{3}$	-	$5 x^{2}$	-	$2 x$	+	$24$
			-	$x^{3}$	-	$2 x^{2}$

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

				$x^{2}$
$x$	+	$2$		$x^{3}$	-	$5 x^{2}$	-	$2 x$	+	$24$
			-	$x^{3}$	-	$2 x^{2}$
					-	$7 x^{2}$

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

				$x^{2}$
$x$	+	$2$		$x^{3}$	-	$5 x^{2}$	-	$2 x$	+	$24$
			-	$x^{3}$	-	$2 x^{2}$
					-	$7 x^{2}$	-	$2 x$

Divide the highest order term in the dividend $- 7 x^{2}$ by the highest order term in divisor $x$ .

				$x^{2}$	-	$7 x$
$x$	+	$2$		$x^{3}$	-	$5 x^{2}$	-	$2 x$	+	$24$
			-	$x^{3}$	-	$2 x^{2}$
					-	$7 x^{2}$	-	$2 x$

Multiply the new quotient term by the divisor.

				$x^{2}$	-	$7 x$
$x$	+	$2$		$x^{3}$	-	$5 x^{2}$	-	$2 x$	+	$24$
			-	$x^{3}$	-	$2 x^{2}$
					-	$7 x^{2}$	-	$2 x$
					-	$7 x^{2}$	-	$14 x$

The expression needs to be subtracted from the dividend, so change all the signs in $- 7 x^{2} - 14 x$

				$x^{2}$	-	$7 x$
$x$	+	$2$		$x^{3}$	-	$5 x^{2}$	-	$2 x$	+	$24$
			-	$x^{3}$	-	$2 x^{2}$
					-	$7 x^{2}$	-	$2 x$
					+	$7 x^{2}$	+	$14 x$

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

				$x^{2}$	-	$7 x$
$x$	+	$2$		$x^{3}$	-	$5 x^{2}$	-	$2 x$	+	$24$
			-	$x^{3}$	-	$2 x^{2}$
					-	$7 x^{2}$	-	$2 x$
					+	$7 x^{2}$	+	$14 x$
							+	$12 x$

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

				$x^{2}$	-	$7 x$
$x$	+	$2$		$x^{3}$	-	$5 x^{2}$	-	$2 x$	+	$24$
			-	$x^{3}$	-	$2 x^{2}$
					-	$7 x^{2}$	-	$2 x$
					+	$7 x^{2}$	+	$14 x$
							+	$12 x$	+	$24$

Divide the highest order term in the dividend $12 x$ by the highest order term in divisor $x$ .

				$x^{2}$	-	$7 x$	+	$12$
$x$	+	$2$		$x^{3}$	-	$5 x^{2}$	-	$2 x$	+	$24$
			-	$x^{3}$	-	$2 x^{2}$
					-	$7 x^{2}$	-	$2 x$
					+	$7 x^{2}$	+	$14 x$
							+	$12 x$	+	$24$

Multiply the new quotient term by the divisor.

				$x^{2}$	-	$7 x$	+	$12$
$x$	+	$2$		$x^{3}$	-	$5 x^{2}$	-	$2 x$	+	$24$
			-	$x^{3}$	-	$2 x^{2}$
					-	$7 x^{2}$	-	$2 x$
					+	$7 x^{2}$	+	$14 x$
							+	$12 x$	+	$24$
							+	$12 x$	+	$24$

The expression needs to be subtracted from the dividend, so change all the signs in $12 x + 24$

				$x^{2}$	-	$7 x$	+	$12$
$x$	+	$2$		$x^{3}$	-	$5 x^{2}$	-	$2 x$	+	$24$
			-	$x^{3}$	-	$2 x^{2}$
					-	$7 x^{2}$	-	$2 x$
					+	$7 x^{2}$	+	$14 x$
							+	$12 x$	+	$24$
							-	$12 x$	-	$24$

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

				$x^{2}$	-	$7 x$	+	$12$
$x$	+	$2$		$x^{3}$	-	$5 x^{2}$	-	$2 x$	+	$24$
			-	$x^{3}$	-	$2 x^{2}$
					-	$7 x^{2}$	-	$2 x$
					+	$7 x^{2}$	+	$14 x$
							+	$12 x$	+	$24$
							-	$12 x$	-	$24$
										$0$

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

$x^{2} - 7 x + 12$

Write $x^{3} - 5 x^{2} - 2 x + 24$ as a set of factors.

$(x + 2) (x^{2} - 7 x + 12) = 0$

Factor $x^{2} - 7 x + 12$ 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 $- 7$ .

$- 4, - 3$

Write the factored form using these integers.

$(x + 2) ((x - 4) (x - 3)) = 0$

Remove unnecessary parentheses.

$(x + 2) (x - 4) (x - 3) = 0$

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

$x + 2 = 0$

$x - 4 = 0$

$x - 3 = 0$

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

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

$x + 2 = 0$

Subtract $2$ from both sides of the equation.

$x = - 2$

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.

$x = 4$

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.

$x = 3$

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

$x = - 2, 4, 3$

Step 8

The polynomial can be written as a set of linear factors.

$(x - 2) (x + 2) (x - 4) (x - 3)$

Step 9

These are the roots (zeros) of the polynomial $x^{4} - 7 x^{3} + 8 x^{2} + 28 x - 48$ .

$x = 2, - 2, 4, 3$

Step 10

Find the Roots/Zeros Using the Rational Roots Test f(x)=x^4-7x^3+8x^2+28x-48

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