Find the Roots/Zeros Using the Rational Roots Test P(x)=x^4-5x^3+6x^2+4x-8

$P (x) = x^{4} - 5 x^{3} + 6 x^{2} + 4 x - 8$

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.

$p = \pm 1, \pm 2, \pm 4, \pm 8$

$q = \pm 1$

Step 2

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

$\pm 1, \pm 2, \pm 4, \pm 8$

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.

${(- 1)}^{4} - 5 {(- 1)}^{3} + 6 {(- 1)}^{2} + 4 (- 1) - 8$

Step 4

Simplify the expression. In this case, the expression is equal to $0$ so $x = - 1$ is a root of the polynomial.

Simplify each term.

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

$1 - 5 {(- 1)}^{3} + 6 {(- 1)}^{2} + 4 (- 1) - 8$

Raise $- 1$ to the power of $3$ .

$1 - 5 \cdot - 1 + 6 {(- 1)}^{2} + 4 (- 1) - 8$

Multiply $- 5$ by $- 1$ .

$1 + 5 + 6 {(- 1)}^{2} + 4 (- 1) - 8$

Raise $- 1$ to the power of $2$ .

$1 + 5 + 6 \cdot 1 + 4 (- 1) - 8$

Multiply $6$ by $1$ .

$1 + 5 + 6 + 4 (- 1) - 8$

Multiply $4$ by $- 1$ .

$1 + 5 + 6 - 4 - 8$

Simplify by adding and subtracting.

Add $1$ and $5$ .

$6 + 6 - 4 - 8$

Add $6$ and $6$ .

$12 - 4 - 8$

Subtract $4$ from $12$ .

$8 - 8$

Subtract $8$ from $8$ .

$0$

Step 5

Since $- 1$ is a known root, divide the polynomial by $x + 1$ to find the quotient polynomial. This polynomial can then be used to find the remaining roots.

$\frac{x^{4} - 5 x^{3} + 6 x^{2} + 4 x - 8}{x + 1}$

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.

$- 1$	$1$	$- 5$	$6$	$4$	$- 8$
	\xa0				\xa0

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

$- 1$	$1$	$- 5$	$6$	$4$	$- 8$
	\xa0				\xa0
	$1$

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

$- 1$	$1$	$- 5$	$6$	$4$	$- 8$
	\xa0	$- 1$			\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.

$- 1$	$1$	$- 5$	$6$	$4$	$- 8$
	\xa0	$- 1$			\xa0
	$1$	$- 6$

Multiply the newest entry in the result $(- 6)$ by the divisor $(- 1)$ and place the result of $(6)$ under the next term in the dividend $(6)$ .

$- 1$	$1$	$- 5$	$6$	$4$	$- 8$
	\xa0	$- 1$	$6$		\xa0
	$1$	$- 6$

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

$- 1$	$1$	$- 5$	$6$	$4$	$- 8$
	\xa0	$- 1$	$6$		\xa0
	$1$	$- 6$	$12$

Multiply the newest entry in the result $(12)$ by the divisor $(- 1)$ and place the result of $(- 12)$ under the next term in the dividend $(4)$ .

$- 1$	$1$	$- 5$	$6$	$4$	$- 8$
	\xa0	$- 1$	$6$	$- 12$	\xa0
	$1$	$- 6$	$12$

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

$- 1$	$1$	$- 5$	$6$	$4$	$- 8$
	\xa0	$- 1$	$6$	$- 12$	\xa0
	$1$	$- 6$	$12$	$- 8$

Multiply the newest entry in the result $(- 8)$ by the divisor $(- 1)$ and place the result of $(8)$ under the next term in the dividend $(- 8)$ .

$- 1$	$1$	$- 5$	$6$	$4$	$- 8$
	\xa0	$- 1$	$6$	$- 12$	$8$
	$1$	$- 6$	$12$	$- 8$

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

$- 1$	$1$	$- 5$	$6$	$4$	$- 8$
	\xa0	$- 1$	$6$	$- 12$	$8$
	$1$	$- 6$	$12$	$- 8$	$0$

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

$1 x^{3} + - 6 x^{2} + (12) x - 8$

Simplify the quotient polynomial.

$x^{3} - 6 x^{2} + 12 x - 8$

Step 7

Solve the equation to find any remaining roots.

Factor the left side of the equation.

Factor $x^{3} - 6 x^{2} + 12 x - 8$ 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.

$p = \pm 1, \pm 8, \pm 2, \pm 4 q = \pm 1$

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

$\pm 1, \pm 8, \pm 2, \pm 4$

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.

$2^{3} - 6 \cdot 2^{2} + 12 \cdot 2 - 8$

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

$8 - 6 \cdot 2^{2} + 12 \cdot 2 - 8$

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

$8 - 6 \cdot 4 + 12 \cdot 2 - 8$

Multiply $- 6$ by $4$ .

$8 - 24 + 12 \cdot 2 - 8$

Subtract $24$ from $8$ .

$- 16 + 12 \cdot 2 - 8$

Multiply $12$ by $2$ .

$- 16 + 24 - 8$

Add $- 16$ and $24$ .

$8 - 8$

Subtract $8$ from $8$ .

$0$

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.

$\frac{x^{3} - 6 x^{2} + 12 x - 8}{x - 2}$

Divide $x^{3} - 6 x^{2} + 12 x - 8$ 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 $0$ .

$x$

$2$

$x^{3}$

$6 x^{2}$

$12 x$

$8$

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

					$x^{2}$
	$x$	-	$2$		$x^{3}$	-	$6 x^{2}$	+	$12 x$	-	$8$

Multiply the new quotient term by the divisor.

				$x^{2}$
$x$	-	$2$		$x^{3}$	-	$6 x^{2}$	+	$12 x$	-	$8$
			+	$x^{3}$	-	$2 x^{2}$

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

				$x^{2}$
$x$	-	$2$		$x^{3}$	-	$6 x^{2}$	+	$12 x$	-	$8$
			-	$x^{3}$	+	$2 x^{2}$

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

				$x^{2}$
$x$	-	$2$		$x^{3}$	-	$6 x^{2}$	+	$12 x$	-	$8$
			-	$x^{3}$	+	$2 x^{2}$
					-	$4 x^{2}$

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

				$x^{2}$
$x$	-	$2$		$x^{3}$	-	$6 x^{2}$	+	$12 x$	-	$8$
			-	$x^{3}$	+	$2 x^{2}$
					-	$4 x^{2}$	+	$12 x$

Divide the highest order term in the dividend $- 4 x^{2}$ by the highest order term in divisor $x$ .

				$x^{2}$	-	$4 x$
$x$	-	$2$		$x^{3}$	-	$6 x^{2}$	+	$12 x$	-	$8$
			-	$x^{3}$	+	$2 x^{2}$
					-	$4 x^{2}$	+	$12 x$

Multiply the new quotient term by the divisor.

				$x^{2}$	-	$4 x$
$x$	-	$2$		$x^{3}$	-	$6 x^{2}$	+	$12 x$	-	$8$
			-	$x^{3}$	+	$2 x^{2}$
					-	$4 x^{2}$	+	$12 x$
					-	$4 x^{2}$	+	$8 x$

The expression needs to be subtracted from the dividend, so change all the signs in $- 4 x^{2} + 8 x$

				$x^{2}$	-	$4 x$
$x$	-	$2$		$x^{3}$	-	$6 x^{2}$	+	$12 x$	-	$8$
			-	$x^{3}$	+	$2 x^{2}$
					-	$4 x^{2}$	+	$12 x$
					+	$4 x^{2}$	-	$8 x$

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

				$x^{2}$	-	$4 x$
$x$	-	$2$		$x^{3}$	-	$6 x^{2}$	+	$12 x$	-	$8$
			-	$x^{3}$	+	$2 x^{2}$
					-	$4 x^{2}$	+	$12 x$
					+	$4 x^{2}$	-	$8 x$
							+	$4 x$

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

				$x^{2}$	-	$4 x$
$x$	-	$2$		$x^{3}$	-	$6 x^{2}$	+	$12 x$	-	$8$
			-	$x^{3}$	+	$2 x^{2}$
					-	$4 x^{2}$	+	$12 x$
					+	$4 x^{2}$	-	$8 x$
							+	$4 x$	-	$8$

Divide the highest order term in the dividend $4 x$ by the highest order term in divisor $x$ .

				$x^{2}$	-	$4 x$	+	$4$
$x$	-	$2$		$x^{3}$	-	$6 x^{2}$	+	$12 x$	-	$8$
			-	$x^{3}$	+	$2 x^{2}$
					-	$4 x^{2}$	+	$12 x$
					+	$4 x^{2}$	-	$8 x$
							+	$4 x$	-	$8$

Multiply the new quotient term by the divisor.

				$x^{2}$	-	$4 x$	+	$4$
$x$	-	$2$		$x^{3}$	-	$6 x^{2}$	+	$12 x$	-	$8$
			-	$x^{3}$	+	$2 x^{2}$
					-	$4 x^{2}$	+	$12 x$
					+	$4 x^{2}$	-	$8 x$
							+	$4 x$	-	$8$
							+	$4 x$	-	$8$

The expression needs to be subtracted from the dividend, so change all the signs in $4 x - 8$

				$x^{2}$	-	$4 x$	+	$4$
$x$	-	$2$		$x^{3}$	-	$6 x^{2}$	+	$12 x$	-	$8$
			-	$x^{3}$	+	$2 x^{2}$
					-	$4 x^{2}$	+	$12 x$
					+	$4 x^{2}$	-	$8 x$
							+	$4 x$	-	$8$
							-	$4 x$	+	$8$

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

				$x^{2}$	-	$4 x$	+	$4$
$x$	-	$2$		$x^{3}$	-	$6 x^{2}$	+	$12 x$	-	$8$
			-	$x^{3}$	+	$2 x^{2}$
					-	$4 x^{2}$	+	$12 x$
					+	$4 x^{2}$	-	$8 x$
							+	$4 x$	-	$8$
							-	$4 x$	+	$8$
										$0$

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

$x^{2} - 4 x + 4$

Write $x^{3} - 6 x^{2} + 12 x - 8$ as a set of factors.

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

Factor using the perfect square rule.

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

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

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

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

Rewrite the polynomial.

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

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

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

Combine like factors.

Raise $x - 2$ to the power of $1$ .

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

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

${(x - 2)}^{1 + 2} = 0$

Add $1$ and $2$ .

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

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

$x - 2 = 0$

Add $2$ to both sides of the equation.

$x = 2$

Step 8

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

$(x + 1) (x - 2)$

Step 9

These are the roots (zeros) of the polynomial $x^{4} - 5 x^{3} + 6 x^{2} + 4 x - 8$ .

$x = - 1, 2$

Step 10

Find the Roots/Zeros Using the Rational Roots Test P(x)=x^4-5x^3+6x^2+4x-8

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