Find the Roots/Zeros Using the Rational Roots Test f(x)=10x^3-17x^2-97x+20

$f (x) = 10 x^{3} - 17 x^{2} - 97 x + 20$

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.

$p = \pm 1, \pm 2, \pm 4, \pm 5, \pm 10, \pm 20$

$q = \pm 1, \pm 2, \pm 5, \pm 10$

Step 2

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

$\pm 1, \pm \frac{1}{2}, \pm \frac{1}{5}, \pm \frac{1}{10}, \pm 2, \pm \frac{2}{5}, \pm 4, \pm \frac{4}{5}, \pm 5, \pm \frac{5}{2}, \pm 10, \pm 20$

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.

$10 {(4)}^{3} - 17 {(4)}^{2} - 97 \cdot 4 + 20$

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 $3$ .

$10 \cdot 64 - 17 {(4)}^{2} - 97 \cdot 4 + 20$

Multiply $10$ by $64$ .

$640 - 17 {(4)}^{2} - 97 \cdot 4 + 20$

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

$640 - 17 \cdot 16 - 97 \cdot 4 + 20$

Multiply $- 17$ by $16$ .

$640 - 272 - 97 \cdot 4 + 20$

Multiply $- 97$ by $4$ .

$640 - 272 - 388 + 20$

Simplify by adding and subtracting.

Subtract $272$ from $640$ .

$368 - 388 + 20$

Subtract $388$ from $368$ .

$- 20 + 20$

Add $- 20$ and $20$ .

$0$

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.

$\frac{10 x^{3} - 17 x^{2} - 97 x + 20}{x - 4}$

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.

$4$	$10$	$- 17$	$- 97$	$20$
	\xa0			\xa0

The first number in the dividend $(10)$ is put into the first position of the result area (below the horizontal line).

$4$	$10$	$- 17$	$- 97$	$20$
	\xa0			\xa0
	$10$

Multiply the newest entry in the result $(10)$ by the divisor $(4)$ and place the result of $(40)$ under the next term in the dividend $(- 17)$ .

$4$	$10$	$- 17$	$- 97$	$20$
	\xa0	$40$		\xa0
	$10$

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

$4$	$10$	$- 17$	$- 97$	$20$
	\xa0	$40$		\xa0
	$10$	$23$

Multiply the newest entry in the result $(23)$ by the divisor $(4)$ and place the result of $(92)$ under the next term in the dividend $(- 97)$ .

$4$	$10$	$- 17$	$- 97$	$20$
	\xa0	$40$	$92$	\xa0
	$10$	$23$

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

$4$	$10$	$- 17$	$- 97$	$20$
	\xa0	$40$	$92$	\xa0
	$10$	$23$	$- 5$

Multiply the newest entry in the result $(- 5)$ by the divisor $(4)$ and place the result of $(- 20)$ under the next term in the dividend $(20)$ .

$4$	$10$	$- 17$	$- 97$	$20$
	\xa0	$40$	$92$	$- 20$
	$10$	$23$	$- 5$

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

$4$	$10$	$- 17$	$- 97$	$20$
	\xa0	$40$	$92$	$- 20$
	$10$	$23$	$- 5$	$0$

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

$10 x^{2} + 23 x - 5$

Step 7

Factor by grouping.

For a polynomial of the form $a x^{2} + b x + c$ , rewrite the middle term as a sum of two terms whose product is $a \cdot c = 10 \cdot - 5 = - 50$ and whose sum is $b = 23$ .

Factor $23$ out of $23 x$ .

$(x - 4) + 10 x^{2} + 23 (x) - 5$

Rewrite $23$ as $- 2$ plus $25$

$(x - 4) + 10 x^{2} + (- 2 + 25) x - 5$

Apply the distributive property.

$(x - 4) + 10 x^{2} - 2 x + 25 x - 5$

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

$(x - 4) (10 x^{2} - 2 x) + 25 x - 5$

Factor out the greatest common factor (GCF) from each group.

$(x - 4) + 2 x (5 x - 1) + 5 (5 x - 1)$

Factor the polynomial by factoring out the greatest common factor, $5 x - 1$ .

$(x - 4) + (5 x - 1) (2 x + 5)$

$(x - 4) (5 x - 1) (2 x + 5)$

Step 8

Factor the left side of the equation.

Factor $10 x^{3} - 17 x^{2} - 97 x + 20$ 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.

$p = \pm 1, \pm 20, \pm 2, \pm 10, \pm 4, \pm 5 q = \pm 1, \pm 10, \pm 2, \pm 5$

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

$\pm 1, \pm 0.1, \pm 0.5, \pm 0.2, \pm 20, \pm 2, \pm 10, \pm 4, \pm 0.4, \pm 5, \pm 0.8, \pm 2.5$

Substitute $0.2$ and simplify the expression. In this case, the expression is equal to $0$ so $0.2$ is a root of the polynomial.

Substitute $0.2$ into the polynomial.

$10 \cdot {0.2}^{3} - 17 \cdot {0.2}^{2} - 97 \cdot 0.2 + 20$

Raise $0.2$ to the power of $3$ .

$10 \cdot 0.008 - 17 \cdot {0.2}^{2} - 97 \cdot 0.2 + 20$

Multiply $10$ by $0.008$ .

$0.08 - 17 \cdot {0.2}^{2} - 97 \cdot 0.2 + 20$

Raise $0.2$ to the power of $2$ .

$0.08 - 17 \cdot 0.04 - 97 \cdot 0.2 + 20$

Multiply $- 17$ by $0.04$ .

$0.08 - 0.68 - 97 \cdot 0.2 + 20$

Subtract $0.68$ from $0.08$ .

$- 0.6 - 97 \cdot 0.2 + 20$

Multiply $- 97$ by $0.2$ .

$- 0.6 - 19.4 + 20$

Subtract $19.4$ from $- 0.6$ .

$- 20 + 20$

Add $- 20$ and $20$ .

$0$

Since $0.2$ is a known root, divide the polynomial by $5 x - 1$ to find the quotient polynomial. This polynomial can then be used to find the remaining roots.

$\frac{10 x^{3} - 17 x^{2} - 97 x + 20}{5 x - 1}$

Divide $10 x^{3} - 17 x^{2} - 97 x + 20$ by $5 x - 1$ .

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

$5 x$

$1$

$10 x^{3}$

$17 x^{2}$

$97 x$

$20$

Divide the highest order term in the dividend $10 x^{3}$ by the highest order term in divisor $5 x$ .

					$2 x^{2}$
	$5 x$	-	$1$		$10 x^{3}$	-	$17 x^{2}$	-	$97 x$	+	$20$

Multiply the new quotient term by the divisor.

				$2 x^{2}$
$5 x$	-	$1$		$10 x^{3}$	-	$17 x^{2}$	-	$97 x$	+	$20$
			+	$10 x^{3}$	-	$2 x^{2}$

The expression needs to be subtracted from the dividend, so change all the signs in $10 x^{3} - 2 x^{2}$

				$2 x^{2}$
$5 x$	-	$1$		$10 x^{3}$	-	$17 x^{2}$	-	$97 x$	+	$20$
			-	$10 x^{3}$	+	$2 x^{2}$

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

				$2 x^{2}$
$5 x$	-	$1$		$10 x^{3}$	-	$17 x^{2}$	-	$97 x$	+	$20$
			-	$10 x^{3}$	+	$2 x^{2}$
					-	$15 x^{2}$

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

				$2 x^{2}$
$5 x$	-	$1$		$10 x^{3}$	-	$17 x^{2}$	-	$97 x$	+	$20$
			-	$10 x^{3}$	+	$2 x^{2}$
					-	$15 x^{2}$	-	$97 x$

Divide the highest order term in the dividend $- 15 x^{2}$ by the highest order term in divisor $5 x$ .

				$2 x^{2}$	-	$3 x$
$5 x$	-	$1$		$10 x^{3}$	-	$17 x^{2}$	-	$97 x$	+	$20$
			-	$10 x^{3}$	+	$2 x^{2}$
					-	$15 x^{2}$	-	$97 x$

Multiply the new quotient term by the divisor.

				$2 x^{2}$	-	$3 x$
$5 x$	-	$1$		$10 x^{3}$	-	$17 x^{2}$	-	$97 x$	+	$20$
			-	$10 x^{3}$	+	$2 x^{2}$
					-	$15 x^{2}$	-	$97 x$
					-	$15 x^{2}$	+	$3 x$

The expression needs to be subtracted from the dividend, so change all the signs in $- 15 x^{2} + 3 x$

				$2 x^{2}$	-	$3 x$
$5 x$	-	$1$		$10 x^{3}$	-	$17 x^{2}$	-	$97 x$	+	$20$
			-	$10 x^{3}$	+	$2 x^{2}$
					-	$15 x^{2}$	-	$97 x$
					+	$15 x^{2}$	-	$3 x$

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

				$2 x^{2}$	-	$3 x$
$5 x$	-	$1$		$10 x^{3}$	-	$17 x^{2}$	-	$97 x$	+	$20$
			-	$10 x^{3}$	+	$2 x^{2}$
					-	$15 x^{2}$	-	$97 x$
					+	$15 x^{2}$	-	$3 x$
							-	$100 x$

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

				$2 x^{2}$	-	$3 x$
$5 x$	-	$1$		$10 x^{3}$	-	$17 x^{2}$	-	$97 x$	+	$20$
			-	$10 x^{3}$	+	$2 x^{2}$
					-	$15 x^{2}$	-	$97 x$
					+	$15 x^{2}$	-	$3 x$
							-	$100 x$	+	$20$

Divide the highest order term in the dividend $- 100 x$ by the highest order term in divisor $5 x$ .

				$2 x^{2}$	-	$3 x$	-	$20$
$5 x$	-	$1$		$10 x^{3}$	-	$17 x^{2}$	-	$97 x$	+	$20$
			-	$10 x^{3}$	+	$2 x^{2}$
					-	$15 x^{2}$	-	$97 x$
					+	$15 x^{2}$	-	$3 x$
							-	$100 x$	+	$20$

Multiply the new quotient term by the divisor.

				$2 x^{2}$	-	$3 x$	-	$20$
$5 x$	-	$1$		$10 x^{3}$	-	$17 x^{2}$	-	$97 x$	+	$20$
			-	$10 x^{3}$	+	$2 x^{2}$
					-	$15 x^{2}$	-	$97 x$
					+	$15 x^{2}$	-	$3 x$
							-	$100 x$	+	$20$
							-	$100 x$	+	$20$

The expression needs to be subtracted from the dividend, so change all the signs in $- 100 x + 20$

				$2 x^{2}$	-	$3 x$	-	$20$
$5 x$	-	$1$		$10 x^{3}$	-	$17 x^{2}$	-	$97 x$	+	$20$
			-	$10 x^{3}$	+	$2 x^{2}$
					-	$15 x^{2}$	-	$97 x$
					+	$15 x^{2}$	-	$3 x$
							-	$100 x$	+	$20$
							+	$100 x$	-	$20$

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

				$2 x^{2}$	-	$3 x$	-	$20$
$5 x$	-	$1$		$10 x^{3}$	-	$17 x^{2}$	-	$97 x$	+	$20$
			-	$10 x^{3}$	+	$2 x^{2}$
					-	$15 x^{2}$	-	$97 x$
					+	$15 x^{2}$	-	$3 x$
							-	$100 x$	+	$20$
							+	$100 x$	-	$20$
										$0$

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

$2 x^{2} - 3 x - 20$

Write $10 x^{3} - 17 x^{2} - 97 x + 20$ as a set of factors.

$(5 x - 1) (2 x^{2} - 3 x - 20) = 0$

Factor by grouping.

For a polynomial of the form $a x^{2} + b x + c$ , rewrite the middle term as a sum of two terms whose product is $a \cdot c = 2 \cdot - 20 = - 40$ and whose sum is $b = - 3$ .

Factor $- 3$ out of $- 3 x$ .

$(5 x - 1) (2 x^{2} - 3 x - 20) = 0$

Rewrite $- 3$ as $5$ plus $- 8$

$(5 x - 1) (2 x^{2} + (5 - 8) x - 20) = 0$

Apply the distributive property.

$(5 x - 1) (2 x^{2} + 5 x - 8 x - 20) = 0$

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

$(5 x - 1) ((2 x^{2} + 5 x) - 8 x - 20) = 0$

Factor out the greatest common factor (GCF) from each group.

$(5 x - 1) (x (2 x + 5) - 4 (2 x + 5)) = 0$

Factor the polynomial by factoring out the greatest common factor, $2 x + 5$ .

$(5 x - 1) ((2 x + 5) (x - 4)) = 0$

Remove unnecessary parentheses.

$(5 x - 1) (2 x + 5) (x - 4) = 0$

Step 9

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

$5 x - 1 = 0$

$2 x + 5 = 0$

$x - 4 = 0$

Step 10

Set $5 x - 1$ equal to $0$ and solve for $x$ .

Set $5 x - 1$ equal to $0$ .

$5 x - 1 = 0$

Solve $5 x - 1 = 0$ for $x$ .

Add $1$ to both sides of the equation.

$5 x = 1$

Divide each term in $5 x = 1$ by $5$ and simplify.

Divide each term in $5 x = 1$ by $5$ .

$\frac{5 x}{5} = \frac{1}{5}$

Simplify the left side.

Cancel the common factor of $5$ .

Cancel the common factor.

$\frac{5 x}{5} = \frac{1}{5}$

Divide $x$ by $1$ .

$x = \frac{1}{5}$

Step 11

Set $2 x + 5$ equal to $0$ and solve for $x$ .

Set $2 x + 5$ equal to $0$ .

$2 x + 5 = 0$

Solve $2 x + 5 = 0$ for $x$ .

Subtract $5$ from both sides of the equation.

$2 x = - 5$

Divide each term in $2 x = - 5$ by $2$ and simplify.

Divide each term in $2 x = - 5$ by $2$ .

$\frac{2 x}{2} = \frac{- 5}{2}$

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

$\frac{2 x}{2} = \frac{- 5}{2}$

Divide $x$ by $1$ .

$x = \frac{- 5}{2}$

Simplify the right side.

Move the negative in front of the fraction.

$x = - \frac{5}{2}$

Step 12

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.

$x = 4$

Step 13

The final solution is all the values that make $(5 x - 1) (2 x + 5) (x - 4) = 0$ true.

$x = \frac{1}{5}, - \frac{5}{2}, 4$

Step 14

Find the Roots/Zeros Using the Rational Roots Test f(x)=10x^3-17x^2-97x+20

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