Find the Roots/Zeros Using the Rational Roots Test x^4-7x^3-12x^2+176x-320

$x^{4} - 7 x^{3} - 12 x^{2} + 176 x - 320$

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.

$p = \pm 1, \pm 2, \pm 4, \pm 5, \pm 8, \pm 10, \pm 16, \pm 20, \pm 32, \pm 40, \pm 64, \pm 80, \pm 160, \pm 320$

$q = \pm 1$

Step 2

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

$\pm 1, \pm 2, \pm 4, \pm 5, \pm 8, \pm 10, \pm 16, \pm 20, \pm 32, \pm 40, \pm 64, \pm 80, \pm 160, \pm 320$

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.

${(4)}^{4} - 7 {(4)}^{3} - 12 {(4)}^{2} + 176 (4) - 320$

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 $4$ .

$256 - 7 {(4)}^{3} - 12 {(4)}^{2} + 176 (4) - 320$

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

$256 - 7 \cdot 64 - 12 {(4)}^{2} + 176 (4) - 320$

Multiply $- 7$ by $64$ .

$256 - 448 - 12 {(4)}^{2} + 176 (4) - 320$

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

$256 - 448 - 12 \cdot 16 + 176 (4) - 320$

Multiply $- 12$ by $16$ .

$256 - 448 - 192 + 176 (4) - 320$

Multiply $176$ by $4$ .

$256 - 448 - 192 + 704 - 320$

Simplify by adding and subtracting.

Subtract $448$ from $256$ .

$- 192 - 192 + 704 - 320$

Subtract $192$ from $- 192$ .

$- 384 + 704 - 320$

Add $- 384$ and $704$ .

$320 - 320$

Subtract $320$ from $320$ .

$0$

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.

$\frac{x^{4} - 7 x^{3} - 12 x^{2} + 176 x - 320}{x - 4}$

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.

$4$	$1$	$- 7$	$- 12$	$176$	$- 320$
	\xa0				\xa0

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

$4$	$1$	$- 7$	$- 12$	$176$	$- 320$
	\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 $(- 7)$ .

$4$	$1$	$- 7$	$- 12$	$176$	$- 320$
	\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.

$4$	$1$	$- 7$	$- 12$	$176$	$- 320$
	\xa0	$4$			\xa0
	$1$	$- 3$

Multiply the newest entry in the result $(- 3)$ by the divisor $(4)$ and place the result of $(- 12)$ under the next term in the dividend $(- 12)$ .

$4$	$1$	$- 7$	$- 12$	$176$	$- 320$
	\xa0	$4$	$- 12$		\xa0
	$1$	$- 3$

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

$4$	$1$	$- 7$	$- 12$	$176$	$- 320$
	\xa0	$4$	$- 12$		\xa0
	$1$	$- 3$	$- 24$

Multiply the newest entry in the result $(- 24)$ by the divisor $(4)$ and place the result of $(- 96)$ under the next term in the dividend $(176)$ .

$4$	$1$	$- 7$	$- 12$	$176$	$- 320$
	\xa0	$4$	$- 12$	$- 96$	\xa0
	$1$	$- 3$	$- 24$

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

$4$	$1$	$- 7$	$- 12$	$176$	$- 320$
	\xa0	$4$	$- 12$	$- 96$	\xa0
	$1$	$- 3$	$- 24$	$80$

Multiply the newest entry in the result $(80)$ by the divisor $(4)$ and place the result of $(320)$ under the next term in the dividend $(- 320)$ .

$4$	$1$	$- 7$	$- 12$	$176$	$- 320$
	\xa0	$4$	$- 12$	$- 96$	$320$
	$1$	$- 3$	$- 24$	$80$

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

$4$	$1$	$- 7$	$- 12$	$176$	$- 320$
	\xa0	$4$	$- 12$	$- 96$	$320$
	$1$	$- 3$	$- 24$	$80$	$0$

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

$1 x^{3} + - 3 x^{2} + (- 24) x + 80$

Simplify the quotient polynomial.

$x^{3} - 3 x^{2} - 24 x + 80$

Step 7

Solve the equation to find any remaining roots.

Factor the left side of the equation.

Factor $x^{3} - 3 x^{2} - 24 x + 80$ 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.

$p = \pm 1, \pm 80, \pm 2, \pm 40, \pm 4, \pm 20, \pm 5, \pm 16, \pm 8, \pm 10 q = \pm 1$

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

$\pm 1, \pm 80, \pm 2, \pm 40, \pm 4, \pm 20, \pm 5, \pm 16, \pm 8, \pm 10$

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.

$4^{3} - 3 \cdot 4^{2} - 24 \cdot 4 + 80$

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

$64 - 3 \cdot 4^{2} - 24 \cdot 4 + 80$

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

$64 - 3 \cdot 16 - 24 \cdot 4 + 80$

Multiply $- 3$ by $16$ .

$64 - 48 - 24 \cdot 4 + 80$

Subtract $48$ from $64$ .

$16 - 24 \cdot 4 + 80$

Multiply $- 24$ by $4$ .

$16 - 96 + 80$

Subtract $96$ from $16$ .

$- 80 + 80$

Add $- 80$ and $80$ .

$0$

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.

$\frac{x^{3} - 3 x^{2} - 24 x + 80}{x - 4}$

Divide $x^{3} - 3 x^{2} - 24 x + 80$ 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 $0$ .

$x$

$4$

$x^{3}$

$3 x^{2}$

$24 x$

$80$

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

					$x^{2}$
	$x$	-	$4$		$x^{3}$	-	$3 x^{2}$	-	$24 x$	+	$80$

Multiply the new quotient term by the divisor.

				$x^{2}$
$x$	-	$4$		$x^{3}$	-	$3 x^{2}$	-	$24 x$	+	$80$
			+	$x^{3}$	-	$4 x^{2}$

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

				$x^{2}$
$x$	-	$4$		$x^{3}$	-	$3 x^{2}$	-	$24 x$	+	$80$
			-	$x^{3}$	+	$4 x^{2}$

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

				$x^{2}$
$x$	-	$4$		$x^{3}$	-	$3 x^{2}$	-	$24 x$	+	$80$
			-	$x^{3}$	+	$4 x^{2}$
					+	$x^{2}$

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

				$x^{2}$
$x$	-	$4$		$x^{3}$	-	$3 x^{2}$	-	$24 x$	+	$80$
			-	$x^{3}$	+	$4 x^{2}$
					+	$x^{2}$	-	$24 x$

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

				$x^{2}$	+	$x$
$x$	-	$4$		$x^{3}$	-	$3 x^{2}$	-	$24 x$	+	$80$
			-	$x^{3}$	+	$4 x^{2}$
					+	$x^{2}$	-	$24 x$

Multiply the new quotient term by the divisor.

				$x^{2}$	+	$x$
$x$	-	$4$		$x^{3}$	-	$3 x^{2}$	-	$24 x$	+	$80$
			-	$x^{3}$	+	$4 x^{2}$
					+	$x^{2}$	-	$24 x$
					+	$x^{2}$	-	$4 x$

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

				$x^{2}$	+	$x$
$x$	-	$4$		$x^{3}$	-	$3 x^{2}$	-	$24 x$	+	$80$
			-	$x^{3}$	+	$4 x^{2}$
					+	$x^{2}$	-	$24 x$
					-	$x^{2}$	+	$4 x$

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

				$x^{2}$	+	$x$
$x$	-	$4$		$x^{3}$	-	$3 x^{2}$	-	$24 x$	+	$80$
			-	$x^{3}$	+	$4 x^{2}$
					+	$x^{2}$	-	$24 x$
					-	$x^{2}$	+	$4 x$
							-	$20 x$

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

				$x^{2}$	+	$x$
$x$	-	$4$		$x^{3}$	-	$3 x^{2}$	-	$24 x$	+	$80$
			-	$x^{3}$	+	$4 x^{2}$
					+	$x^{2}$	-	$24 x$
					-	$x^{2}$	+	$4 x$
							-	$20 x$	+	$80$

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

				$x^{2}$	+	$x$	-	$20$
$x$	-	$4$		$x^{3}$	-	$3 x^{2}$	-	$24 x$	+	$80$
			-	$x^{3}$	+	$4 x^{2}$
					+	$x^{2}$	-	$24 x$
					-	$x^{2}$	+	$4 x$
							-	$20 x$	+	$80$

Multiply the new quotient term by the divisor.

				$x^{2}$	+	$x$	-	$20$
$x$	-	$4$		$x^{3}$	-	$3 x^{2}$	-	$24 x$	+	$80$
			-	$x^{3}$	+	$4 x^{2}$
					+	$x^{2}$	-	$24 x$
					-	$x^{2}$	+	$4 x$
							-	$20 x$	+	$80$
							-	$20 x$	+	$80$

The expression needs to be subtracted from the dividend, so change all the signs in $- 20 x + 80$

				$x^{2}$	+	$x$	-	$20$
$x$	-	$4$		$x^{3}$	-	$3 x^{2}$	-	$24 x$	+	$80$
			-	$x^{3}$	+	$4 x^{2}$
					+	$x^{2}$	-	$24 x$
					-	$x^{2}$	+	$4 x$
							-	$20 x$	+	$80$
							+	$20 x$	-	$80$

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

				$x^{2}$	+	$x$	-	$20$
$x$	-	$4$		$x^{3}$	-	$3 x^{2}$	-	$24 x$	+	$80$
			-	$x^{3}$	+	$4 x^{2}$
					+	$x^{2}$	-	$24 x$
					-	$x^{2}$	+	$4 x$
							-	$20 x$	+	$80$
							+	$20 x$	-	$80$
										$0$

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

$x^{2} + x - 20$

Write $x^{3} - 3 x^{2} - 24 x + 80$ as a set of factors.

$(x - 4) (x^{2} + x - 20) = 0$

Factor $x^{2} + x - 20$ 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 $1$ .

$- 4, 5$

Write the factored form using these integers.

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

Remove unnecessary parentheses.

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

Combine like factors.

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

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

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

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

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

${(x - 4)}^{1 + 1} (x + 5) = 0$

Add $1$ and $1$ .

${(x - 4)}^{2} (x + 5) = 0$

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

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

$x + 5 = 0$

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 $x + 5$ equal to $0$ and solve for $x$ .

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

$x + 5 = 0$

Subtract $5$ from both sides of the equation.

$x = - 5$

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

$x = 4, - 5$

Step 8

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

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

Step 9

These are the roots (zeros) of the polynomial $x^{4} - 7 x^{3} - 12 x^{2} + 176 x - 320$ .

$x = 4, - 5$

Step 10

Find the Roots/Zeros Using the Rational Roots Test x^4-7x^3-12x^2+176x-320

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