Find the Roots/Zeros Using the Rational Roots Test f(x)=x^4+2x^3-10x^2-18x+9

$f (x) = x^{4} + 2 x^{3} - 10 x^{2} - 18 x + 9$

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.

$p = \pm 1, \pm 3, \pm 9$

$q = \pm 1$

Step 2

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

$\pm 1, \pm 3, \pm 9$

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.

${(3)}^{4} + 2 {(3)}^{3} - 10 {(3)}^{2} - 18 \cdot 3 + 9$

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 $4$ .

$81 + 2 {(3)}^{3} - 10 {(3)}^{2} - 18 \cdot 3 + 9$

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

$81 + 2 \cdot 27 - 10 {(3)}^{2} - 18 \cdot 3 + 9$

Multiply $2$ by $27$ .

$81 + 54 - 10 {(3)}^{2} - 18 \cdot 3 + 9$

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

$81 + 54 - 10 \cdot 9 - 18 \cdot 3 + 9$

Multiply $- 10$ by $9$ .

$81 + 54 - 90 - 18 \cdot 3 + 9$

Multiply $- 18$ by $3$ .

$81 + 54 - 90 - 54 + 9$

Simplify by adding and subtracting.

Add $81$ and $54$ .

$135 - 90 - 54 + 9$

Subtract $90$ from $135$ .

$45 - 54 + 9$

Subtract $54$ from $45$ .

$- 9 + 9$

Add $- 9$ and $9$ .

$0$

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.

$\frac{x^{4} + 2 x^{3} - 10 x^{2} - 18 x + 9}{x - 3}$

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.

$3$	$1$	$2$	$- 10$	$- 18$	$9$
	\xa0				\xa0

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

$3$	$1$	$2$	$- 10$	$- 18$	$9$
	\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 $(2)$ .

$3$	$1$	$2$	$- 10$	$- 18$	$9$
	\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.

$3$	$1$	$2$	$- 10$	$- 18$	$9$
	\xa0	$3$			\xa0
	$1$	$5$

Multiply the newest entry in the result $(5)$ by the divisor $(3)$ and place the result of $(15)$ under the next term in the dividend $(- 10)$ .

$3$	$1$	$2$	$- 10$	$- 18$	$9$
	\xa0	$3$	$15$		\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.

$3$	$1$	$2$	$- 10$	$- 18$	$9$
	\xa0	$3$	$15$		\xa0
	$1$	$5$	$5$

Multiply the newest entry in the result $(5)$ by the divisor $(3)$ and place the result of $(15)$ under the next term in the dividend $(- 18)$ .

$3$	$1$	$2$	$- 10$	$- 18$	$9$
	\xa0	$3$	$15$	$15$	\xa0
	$1$	$5$	$5$

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

$3$	$1$	$2$	$- 10$	$- 18$	$9$
	\xa0	$3$	$15$	$15$	\xa0
	$1$	$5$	$5$	$- 3$

Multiply the newest entry in the result $(- 3)$ by the divisor $(3)$ and place the result of $(- 9)$ under the next term in the dividend $(9)$ .

$3$	$1$	$2$	$- 10$	$- 18$	$9$
	\xa0	$3$	$15$	$15$	$- 9$
	$1$	$5$	$5$	$- 3$

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

$3$	$1$	$2$	$- 10$	$- 18$	$9$
	\xa0	$3$	$15$	$15$	$- 9$
	$1$	$5$	$5$	$- 3$	$0$

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

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

Simplify the quotient polynomial.

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

Step 7

Solve the equation to find any remaining roots.

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

$p = \pm 1, \pm 3 q = \pm 1$

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

$\pm 1, \pm 3$

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.

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

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

$- 27 + 5 {(- 3)}^{2} + 5 \cdot - 3 - 3$

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

$- 27 + 5 \cdot 9 + 5 \cdot - 3 - 3$

Multiply $5$ by $9$ .

$- 27 + 45 + 5 \cdot - 3 - 3$

Add $- 27$ and $45$ .

$18 + 5 \cdot - 3 - 3$

Multiply $5$ by $- 3$ .

$18 - 15 - 3$

Subtract $15$ from $18$ .

$3 - 3$

Subtract $3$ from $3$ .

$0$

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.

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

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

$x$

$3$

$x^{3}$

$5 x^{2}$

$5 x$

$3$

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

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

Multiply the new quotient term by the divisor.

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

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

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

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

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

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

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

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

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

Multiply the new quotient term by the divisor.

				$x^{2}$	+	$2 x$
$x$	+	$3$		$x^{3}$	+	$5 x^{2}$	+	$5 x$	-	$3$
			-	$x^{3}$	-	$3 x^{2}$
					+	$2 x^{2}$	+	$5 x$
					+	$2 x^{2}$	+	$6 x$

The expression needs to be subtracted from the dividend, so change all the signs in $2 x^{2} + 6 x$

				$x^{2}$	+	$2 x$
$x$	+	$3$		$x^{3}$	+	$5 x^{2}$	+	$5 x$	-	$3$
			-	$x^{3}$	-	$3 x^{2}$
					+	$2 x^{2}$	+	$5 x$
					-	$2 x^{2}$	-	$6 x$

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

				$x^{2}$	+	$2 x$
$x$	+	$3$		$x^{3}$	+	$5 x^{2}$	+	$5 x$	-	$3$
			-	$x^{3}$	-	$3 x^{2}$
					+	$2 x^{2}$	+	$5 x$
					-	$2 x^{2}$	-	$6 x$
							-	$x$

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

				$x^{2}$	+	$2 x$
$x$	+	$3$		$x^{3}$	+	$5 x^{2}$	+	$5 x$	-	$3$
			-	$x^{3}$	-	$3 x^{2}$
					+	$2 x^{2}$	+	$5 x$
					-	$2 x^{2}$	-	$6 x$
							-	$x$	-	$3$

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

				$x^{2}$	+	$2 x$	-	$1$
$x$	+	$3$		$x^{3}$	+	$5 x^{2}$	+	$5 x$	-	$3$
			-	$x^{3}$	-	$3 x^{2}$
					+	$2 x^{2}$	+	$5 x$
					-	$2 x^{2}$	-	$6 x$
							-	$x$	-	$3$

Multiply the new quotient term by the divisor.

				$x^{2}$	+	$2 x$	-	$1$
$x$	+	$3$		$x^{3}$	+	$5 x^{2}$	+	$5 x$	-	$3$
			-	$x^{3}$	-	$3 x^{2}$
					+	$2 x^{2}$	+	$5 x$
					-	$2 x^{2}$	-	$6 x$
							-	$x$	-	$3$
							-	$x$	-	$3$

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

				$x^{2}$	+	$2 x$	-	$1$
$x$	+	$3$		$x^{3}$	+	$5 x^{2}$	+	$5 x$	-	$3$
			-	$x^{3}$	-	$3 x^{2}$
					+	$2 x^{2}$	+	$5 x$
					-	$2 x^{2}$	-	$6 x$
							-	$x$	-	$3$
							+	$x$	+	$3$

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

				$x^{2}$	+	$2 x$	-	$1$
$x$	+	$3$		$x^{3}$	+	$5 x^{2}$	+	$5 x$	-	$3$
			-	$x^{3}$	-	$3 x^{2}$
					+	$2 x^{2}$	+	$5 x$
					-	$2 x^{2}$	-	$6 x$
							-	$x$	-	$3$
							+	$x$	+	$3$
										$0$

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

$x^{2} + 2 x - 1$

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

$(x + 3) (x^{2} + 2 x - 1) = 0$

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

$x + 3 = 0$

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

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.

$x = - 3$

Set $x^{2} + 2 x - 1$ equal to $0$ and solve for $x$ .

Set $x^{2} + 2 x - 1$ equal to $0$ .

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

Solve $x^{2} + 2 x - 1 = 0$ for $x$ .

Use the quadratic formula to find the solutions.

$\frac{- b \pm \sqrt{b^{2} - 4 (a c)}}{2 a}$

Substitute the values $a = 1$ , $b = 2$ , and $c = - 1$ into the quadratic formula and solve for $x$ .

$\frac{- 2 \pm \sqrt{2^{2} - 4 \cdot (1 \cdot - 1)}}{2 \cdot 1}$

Simplify.

Simplify the numerator.

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

$x = \frac{- 2 \pm \sqrt{4 - 4 \cdot 1 \cdot - 1}}{2 \cdot 1}$

Multiply $- 4 \cdot 1 \cdot - 1$ .

Multiply $- 4$ by $1$ .

$x = \frac{- 2 \pm \sqrt{4 - 4 \cdot - 1}}{2 \cdot 1}$

Multiply $- 4$ by $- 1$ .

$x = \frac{- 2 \pm \sqrt{4 + 4}}{2 \cdot 1}$

Add $4$ and $4$ .

$x = \frac{- 2 \pm \sqrt{8}}{2 \cdot 1}$

Rewrite $8$ as $2^{2} \cdot 2$ .

Factor $4$ out of $8$ .

$x = \frac{- 2 \pm \sqrt{4 (2)}}{2 \cdot 1}$

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

$x = \frac{- 2 \pm \sqrt{2^{2} \cdot 2}}{2 \cdot 1}$

Pull terms out from under the radical.

$x = \frac{- 2 \pm 2 \sqrt{2}}{2 \cdot 1}$

Multiply $2$ by $1$ .

$x = \frac{- 2 \pm 2 \sqrt{2}}{2}$

Simplify $\frac{- 2 \pm 2 \sqrt{2}}{2}$ .

$x = - 1 \pm \sqrt{2}$

Simplify the expression to solve for the $+$ portion of the $\pm$ .

Simplify the numerator.

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

$x = \frac{- 2 \pm \sqrt{4 - 4 \cdot 1 \cdot - 1}}{2 \cdot 1}$

Multiply $- 4 \cdot 1 \cdot - 1$ .

Multiply $- 4$ by $1$ .

$x = \frac{- 2 \pm \sqrt{4 - 4 \cdot - 1}}{2 \cdot 1}$

Multiply $- 4$ by $- 1$ .

$x = \frac{- 2 \pm \sqrt{4 + 4}}{2 \cdot 1}$

Add $4$ and $4$ .

$x = \frac{- 2 \pm \sqrt{8}}{2 \cdot 1}$

Rewrite $8$ as $2^{2} \cdot 2$ .

Factor $4$ out of $8$ .

$x = \frac{- 2 \pm \sqrt{4 (2)}}{2 \cdot 1}$

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

$x = \frac{- 2 \pm \sqrt{2^{2} \cdot 2}}{2 \cdot 1}$

Pull terms out from under the radical.

$x = \frac{- 2 \pm 2 \sqrt{2}}{2 \cdot 1}$

Multiply $2$ by $1$ .

$x = \frac{- 2 \pm 2 \sqrt{2}}{2}$

Simplify $\frac{- 2 \pm 2 \sqrt{2}}{2}$ .

$x = - 1 \pm \sqrt{2}$

Change the $\pm$ to $+$ .

$x = - 1 + \sqrt{2}$

Simplify the expression to solve for the $-$ portion of the $\pm$ .

Simplify the numerator.

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

$x = \frac{- 2 \pm \sqrt{4 - 4 \cdot 1 \cdot - 1}}{2 \cdot 1}$

Multiply $- 4 \cdot 1 \cdot - 1$ .

Multiply $- 4$ by $1$ .

$x = \frac{- 2 \pm \sqrt{4 - 4 \cdot - 1}}{2 \cdot 1}$

Multiply $- 4$ by $- 1$ .

$x = \frac{- 2 \pm \sqrt{4 + 4}}{2 \cdot 1}$

Add $4$ and $4$ .

$x = \frac{- 2 \pm \sqrt{8}}{2 \cdot 1}$

Rewrite $8$ as $2^{2} \cdot 2$ .

Factor $4$ out of $8$ .

$x = \frac{- 2 \pm \sqrt{4 (2)}}{2 \cdot 1}$

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

$x = \frac{- 2 \pm \sqrt{2^{2} \cdot 2}}{2 \cdot 1}$

Pull terms out from under the radical.

$x = \frac{- 2 \pm 2 \sqrt{2}}{2 \cdot 1}$

Multiply $2$ by $1$ .

$x = \frac{- 2 \pm 2 \sqrt{2}}{2}$

Simplify $\frac{- 2 \pm 2 \sqrt{2}}{2}$ .

$x = - 1 \pm \sqrt{2}$

Change the $\pm$ to $-$ .

$x = - 1 - \sqrt{2}$

The final answer is the combination of both solutions.

$x = - 1 + \sqrt{2}, - 1 - \sqrt{2}$

The final solution is all the values that make $(x + 3) (x^{2} + 2 x - 1) = 0$ true.

$x = - 3, - 1 + \sqrt{2}, - 1 - \sqrt{2}$

Step 8

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

$(x - 3) (x + 3) (x + 1 - \sqrt{2}) (x + 1 + \sqrt{2})$

Step 9

These are the roots (zeros) of the polynomial $x^{4} + 2 x^{3} - 10 x^{2} - 18 x + 9$ .

$x = 3, - 3, - 1 + \sqrt{2}, - 1 - \sqrt{2}$

Step 10

The result can be shown in multiple forms.

Exact Form:

$x = 3, - 3, - 1 + \sqrt{2}, - 1 - \sqrt{2}$

Decimal Form:

$x = 3, - 3, 0.41421356 \dots, - 2.41421356 \dots$

Step 11

Find the Roots/Zeros Using the Rational Roots Test f(x)=x^4+2x^3-10x^2-18x+9

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