Find the Roots/Zeros Using the Rational Roots Test x^4-5x^3-36x^2+272x-448

$x^{4} - 5 x^{3} - 36 x^{2} + 272 x - 448$

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.

$p = \pm 1, \pm 2, \pm 4, \pm 7, \pm 8, \pm 14, \pm 16, \pm 28, \pm 32, \pm 56, \pm 64, \pm 112, \pm 224, \pm 448$

$q = \pm 1$

Step 2

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

$\pm 1, \pm 2, \pm 4, \pm 7, \pm 8, \pm 14, \pm 16, \pm 28, \pm 32, \pm 56, \pm 64, \pm 112, \pm 224, \pm 448$

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.

${(4)}^{4} - 5 {(4)}^{3} - 36 {(4)}^{2} + 272 (4) - 448$

Step 4

Simplify the expression. In this case, the expression is equal to $0$ so $x = 4$ is a root of the polynomial.

Simplify each term.

Raise $4$ to the power of $4$ .

$256 - 5 {(4)}^{3} - 36 {(4)}^{2} + 272 (4) - 448$

Raise $4$ to the power of $3$ .

$256 - 5 \cdot 64 - 36 {(4)}^{2} + 272 (4) - 448$

Multiply $- 5$ by $64$ .

$256 - 320 - 36 {(4)}^{2} + 272 (4) - 448$

Raise $4$ to the power of $2$ .

$256 - 320 - 36 \cdot 16 + 272 (4) - 448$

Multiply $- 36$ by $16$ .

$256 - 320 - 576 + 272 (4) - 448$

Multiply $272$ by $4$ .

$256 - 320 - 576 + 1088 - 448$

Simplify by adding and subtracting.

Subtract $320$ from $256$ .

$- 64 - 576 + 1088 - 448$

Subtract $576$ from $- 64$ .

$- 640 + 1088 - 448$

Add $- 640$ and $1088$ .

$448 - 448$

Subtract $448$ from $448$ .

$0$

Step 5

Since $4$ is a known root, divide the polynomial by $x - 4$ to find the quotient polynomial. This polynomial can then be used to find the remaining roots.

$\frac{x^{4} - 5 x^{3} - 36 x^{2} + 272 x - 448}{x - 4}$

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.

$4$	$1$	$- 5$	$- 36$	$272$	$- 448$
	\xa0				\xa0

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

$4$	$1$	$- 5$	$- 36$	$272$	$- 448$
	\xa0				\xa0
	$1$

Multiply the newest entry in the result $(1)$ by the divisor $(4)$ and place the result of $(4)$ under the next term in the dividend $(- 5)$ .

$4$	$1$	$- 5$	$- 36$	$272$	$- 448$
	\xa0	$4$			\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.

$4$	$1$	$- 5$	$- 36$	$272$	$- 448$
	\xa0	$4$			\xa0
	$1$	$- 1$

Multiply the newest entry in the result $(- 1)$ by the divisor $(4)$ and place the result of $(- 4)$ under the next term in the dividend $(- 36)$ .

$4$	$1$	$- 5$	$- 36$	$272$	$- 448$
	\xa0	$4$	$- 4$		\xa0
	$1$	$- 1$

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

$4$	$1$	$- 5$	$- 36$	$272$	$- 448$
	\xa0	$4$	$- 4$		\xa0
	$1$	$- 1$	$- 40$

Multiply the newest entry in the result $(- 40)$ by the divisor $(4)$ and place the result of $(- 160)$ under the next term in the dividend $(272)$ .

$4$	$1$	$- 5$	$- 36$	$272$	$- 448$
	\xa0	$4$	$- 4$	$- 160$	\xa0
	$1$	$- 1$	$- 40$

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

$4$	$1$	$- 5$	$- 36$	$272$	$- 448$
	\xa0	$4$	$- 4$	$- 160$	\xa0
	$1$	$- 1$	$- 40$	$112$

Multiply the newest entry in the result $(112)$ by the divisor $(4)$ and place the result of $(448)$ under the next term in the dividend $(- 448)$ .

$4$	$1$	$- 5$	$- 36$	$272$	$- 448$
	\xa0	$4$	$- 4$	$- 160$	$448$
	$1$	$- 1$	$- 40$	$112$

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

$4$	$1$	$- 5$	$- 36$	$272$	$- 448$
	\xa0	$4$	$- 4$	$- 160$	$448$
	$1$	$- 1$	$- 40$	$112$	$0$

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

$1 x^{3} + - 1 x^{2} + (- 40) x + 112$

Simplify the quotient polynomial.

$x^{3} - x^{2} - 40 x + 112$

Step 7

Solve the equation to find any remaining roots.

Factor the left side of the equation.

Factor $x^{3} - x^{2} - 40 x + 112$ 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.

$p = \pm 1, \pm 112, \pm 2, \pm 56, \pm 4, \pm 28, \pm 7, \pm 16, \pm 8, \pm 14 q = \pm 1$

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

$\pm 1, \pm 112, \pm 2, \pm 56, \pm 4, \pm 28, \pm 7, \pm 16, \pm 8, \pm 14$

Substitute $4$ and simplify the expression. In this case, the expression is equal to $0$ so $4$ is a root of the polynomial.

Substitute $4$ into the polynomial.

$4^{3} - 4^{2} - 40 \cdot 4 + 112$

Raise $4$ to the power of $3$ .

$64 - 4^{2} - 40 \cdot 4 + 112$

Raise $4$ to the power of $2$ .

$64 - 1 \cdot 16 - 40 \cdot 4 + 112$

Multiply $- 1$ by $16$ .

$64 - 16 - 40 \cdot 4 + 112$

Subtract $16$ from $64$ .

$48 - 40 \cdot 4 + 112$

Multiply $- 40$ by $4$ .

$48 - 160 + 112$

Subtract $160$ from $48$ .

$- 112 + 112$

Add $- 112$ and $112$ .

$0$

Since $4$ is a known root, divide the polynomial by $x - 4$ to find the quotient polynomial. This polynomial can then be used to find the remaining roots.

$\frac{x^{3} - x^{2} - 40 x + 112}{x - 4}$

Divide $x^{3} - x^{2} - 40 x + 112$ by $x - 4$ .

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

$x$

$4$

$x^{3}$

$x^{2}$

$40 x$

$112$

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

					$x^{2}$
	$x$	-	$4$		$x^{3}$	-	$x^{2}$	-	$40 x$	+	$112$

Multiply the new quotient term by the divisor.

				$x^{2}$
$x$	-	$4$		$x^{3}$	-	$x^{2}$	-	$40 x$	+	$112$
			+	$x^{3}$	-	$4 x^{2}$

The expression needs to be subtracted from the dividend, so change all the signs in $x^{3} - 4 x^{2}$

				$x^{2}$
$x$	-	$4$		$x^{3}$	-	$x^{2}$	-	$40 x$	+	$112$
			-	$x^{3}$	+	$4 x^{2}$

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

				$x^{2}$
$x$	-	$4$		$x^{3}$	-	$x^{2}$	-	$40 x$	+	$112$
			-	$x^{3}$	+	$4 x^{2}$
					+	$3 x^{2}$

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

				$x^{2}$
$x$	-	$4$		$x^{3}$	-	$x^{2}$	-	$40 x$	+	$112$
			-	$x^{3}$	+	$4 x^{2}$
					+	$3 x^{2}$	-	$40 x$

Divide the highest order term in the dividend $3 x^{2}$ by the highest order term in divisor $x$ .

				$x^{2}$	+	$3 x$
$x$	-	$4$		$x^{3}$	-	$x^{2}$	-	$40 x$	+	$112$
			-	$x^{3}$	+	$4 x^{2}$
					+	$3 x^{2}$	-	$40 x$

Multiply the new quotient term by the divisor.

				$x^{2}$	+	$3 x$
$x$	-	$4$		$x^{3}$	-	$x^{2}$	-	$40 x$	+	$112$
			-	$x^{3}$	+	$4 x^{2}$
					+	$3 x^{2}$	-	$40 x$
					+	$3 x^{2}$	-	$12 x$

The expression needs to be subtracted from the dividend, so change all the signs in $3 x^{2} - 12 x$

				$x^{2}$	+	$3 x$
$x$	-	$4$		$x^{3}$	-	$x^{2}$	-	$40 x$	+	$112$
			-	$x^{3}$	+	$4 x^{2}$
					+	$3 x^{2}$	-	$40 x$
					-	$3 x^{2}$	+	$12 x$

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

				$x^{2}$	+	$3 x$
$x$	-	$4$		$x^{3}$	-	$x^{2}$	-	$40 x$	+	$112$
			-	$x^{3}$	+	$4 x^{2}$
					+	$3 x^{2}$	-	$40 x$
					-	$3 x^{2}$	+	$12 x$
							-	$28 x$

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

				$x^{2}$	+	$3 x$
$x$	-	$4$		$x^{3}$	-	$x^{2}$	-	$40 x$	+	$112$
			-	$x^{3}$	+	$4 x^{2}$
					+	$3 x^{2}$	-	$40 x$
					-	$3 x^{2}$	+	$12 x$
							-	$28 x$	+	$112$

Divide the highest order term in the dividend $- 28 x$ by the highest order term in divisor $x$ .

				$x^{2}$	+	$3 x$	-	$28$
$x$	-	$4$		$x^{3}$	-	$x^{2}$	-	$40 x$	+	$112$
			-	$x^{3}$	+	$4 x^{2}$
					+	$3 x^{2}$	-	$40 x$
					-	$3 x^{2}$	+	$12 x$
							-	$28 x$	+	$112$

Multiply the new quotient term by the divisor.

				$x^{2}$	+	$3 x$	-	$28$
$x$	-	$4$		$x^{3}$	-	$x^{2}$	-	$40 x$	+	$112$
			-	$x^{3}$	+	$4 x^{2}$
					+	$3 x^{2}$	-	$40 x$
					-	$3 x^{2}$	+	$12 x$
							-	$28 x$	+	$112$
							-	$28 x$	+	$112$

The expression needs to be subtracted from the dividend, so change all the signs in $- 28 x + 112$

				$x^{2}$	+	$3 x$	-	$28$
$x$	-	$4$		$x^{3}$	-	$x^{2}$	-	$40 x$	+	$112$
			-	$x^{3}$	+	$4 x^{2}$
					+	$3 x^{2}$	-	$40 x$
					-	$3 x^{2}$	+	$12 x$
							-	$28 x$	+	$112$
							+	$28 x$	-	$112$

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

				$x^{2}$	+	$3 x$	-	$28$
$x$	-	$4$		$x^{3}$	-	$x^{2}$	-	$40 x$	+	$112$
			-	$x^{3}$	+	$4 x^{2}$
					+	$3 x^{2}$	-	$40 x$
					-	$3 x^{2}$	+	$12 x$
							-	$28 x$	+	$112$
							+	$28 x$	-	$112$
										$0$

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

$x^{2} + 3 x - 28$

Write $x^{3} - x^{2} - 40 x + 112$ as a set of factors.

$(x - 4) (x^{2} + 3 x - 28) = 0$

Factor $x^{2} + 3 x - 28$ 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 $- 28$ and whose sum is $3$ .

$- 4, 7$

Write the factored form using these integers.

$(x - 4) ((x - 4) (x + 7)) = 0$

Remove unnecessary parentheses.

$(x - 4) (x - 4) (x + 7) = 0$

Combine like factors.

Raise $x - 4$ to the power of $1$ .

$(x - 4) (x - 4) (x + 7) = 0$

Raise $x - 4$ to the power of $1$ .

$(x - 4) (x - 4) (x + 7) = 0$

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

${(x - 4)}^{1 + 1} (x + 7) = 0$

Add $1$ and $1$ .

${(x - 4)}^{2} (x + 7) = 0$

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

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

$x + 7 = 0$

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 $0$ .

$x - 4 = 0$

Add $4$ to both sides of the equation.

$x = 4$

Set $x + 7$ equal to $0$ and solve for $x$ .

Set $x + 7$ equal to $0$ .

$x + 7 = 0$

Subtract $7$ from both sides of the equation.

$x = - 7$

The final solution is all the values that make ${(x - 4)}^{2} (x + 7) = 0$ true.

$x = 4, - 7$

Step 8

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

$(x - 4) (x + 7)$

Step 9

These are the roots (zeros) of the polynomial $x^{4} - 5 x^{3} - 36 x^{2} + 272 x - 448$ .

$x = 4, - 7$

Step 10

Find the Roots/Zeros Using the Rational Roots Test x^4-5x^3-36x^2+272x-448

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