Find the Roots/Zeros Using the Rational Roots Test f(x)=10x^3-33x^2+8x+3

$f (x) = 10 x^{3} - 33 x^{2} + 8 x + 3$

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.

$p = \pm 1, \pm 3$

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

Step 2

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

$\pm 1, \pm \frac{1}{2}, \pm \frac{1}{5}, \pm \frac{1}{10}, \pm 3, \pm \frac{3}{2}, \pm \frac{3}{5}, \pm \frac{3}{10}$

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.

$10 {(\frac{1}{2})}^{3} - 33 {(\frac{1}{2})}^{2} + 8 (\frac{1}{2}) + 3$

Step 4

Simplify the expression. In this case, the expression is equal to $0$ so $x = \frac{1}{2}$ is a root of the polynomial.

Simplify each term.

Apply the product rule to $\frac{1}{2}$ .

$10 \frac{1^{3}}{2^{3}} - 33 {(\frac{1}{2})}^{2} + 8 (\frac{1}{2}) + 3$

One to any power is one.

$10 \frac{1}{2^{3}} - 33 {(\frac{1}{2})}^{2} + 8 (\frac{1}{2}) + 3$

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

$10 (\frac{1}{8}) - 33 {(\frac{1}{2})}^{2} + 8 (\frac{1}{2}) + 3$

Cancel the common factor of $2$ .

Factor $2$ out of $10$ .

$2 (5) \frac{1}{8} - 33 {(\frac{1}{2})}^{2} + 8 (\frac{1}{2}) + 3$

Factor $2$ out of $8$ .

$2 \cdot 5 \frac{1}{2 \cdot 4} - 33 {(\frac{1}{2})}^{2} + 8 (\frac{1}{2}) + 3$

Cancel the common factor.

$2 \cdot 5 \frac{1}{2 \cdot 4} - 33 {(\frac{1}{2})}^{2} + 8 (\frac{1}{2}) + 3$

Rewrite the expression.

$5 (\frac{1}{4}) - 33 {(\frac{1}{2})}^{2} + 8 (\frac{1}{2}) + 3$

Combine $5$ and $\frac{1}{4}$ .

$\frac{5}{4} - 33 {(\frac{1}{2})}^{2} + 8 (\frac{1}{2}) + 3$

Apply the product rule to $\frac{1}{2}$ .

$\frac{5}{4} - 33 \frac{1^{2}}{2^{2}} + 8 (\frac{1}{2}) + 3$

One to any power is one.

$\frac{5}{4} - 33 \frac{1}{2^{2}} + 8 (\frac{1}{2}) + 3$

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

$\frac{5}{4} - 33 (\frac{1}{4}) + 8 (\frac{1}{2}) + 3$

Combine $- 33$ and $\frac{1}{4}$ .

$\frac{5}{4} + \frac{- 33}{4} + 8 (\frac{1}{2}) + 3$

Move the negative in front of the fraction.

$\frac{5}{4} - \frac{33}{4} + 8 (\frac{1}{2}) + 3$

Cancel the common factor of $2$ .

Factor $2$ out of $8$ .

$\frac{5}{4} - \frac{33}{4} + 2 (4) \frac{1}{2} + 3$

Cancel the common factor.

$\frac{5}{4} - \frac{33}{4} + 2 \cdot 4 \frac{1}{2} + 3$

Rewrite the expression.

$\frac{5}{4} - \frac{33}{4} + 4 + 3$

Combine fractions.

Combine the numerators over the common denominator.

$4 + 3 + \frac{5 - 33}{4}$

Subtract $33$ from $5$ .

$4 + 3 + \frac{- 28}{4}$

Find the common denominator.

Write $4$ as a fraction with denominator $1$ .

$\frac{4}{1} + 3 + \frac{- 28}{4}$

Multiply $\frac{4}{1}$ by $\frac{4}{4}$ .

$\frac{4}{1} \cdot \frac{4}{4} + 3 + \frac{- 28}{4}$

Multiply $\frac{4}{1}$ by $\frac{4}{4}$ .

$\frac{4 \cdot 4}{4} + 3 + \frac{- 28}{4}$

Write $3$ as a fraction with denominator $1$ .

$\frac{4 \cdot 4}{4} + \frac{3}{1} + \frac{- 28}{4}$

Multiply $\frac{3}{1}$ by $\frac{4}{4}$ .

$\frac{4 \cdot 4}{4} + \frac{3}{1} \cdot \frac{4}{4} + \frac{- 28}{4}$

Multiply $\frac{3}{1}$ by $\frac{4}{4}$ .

$\frac{4 \cdot 4}{4} + \frac{3 \cdot 4}{4} + \frac{- 28}{4}$

Combine the numerators over the common denominator.

$\frac{4 \cdot 4 + 3 \cdot 4 - 28}{4}$

Simplify each term.

Multiply $4$ by $4$ .

$\frac{16 + 3 \cdot 4 - 28}{4}$

Multiply $3$ by $4$ .

$\frac{16 + 12 - 28}{4}$

Simplify the expression.

Add $16$ and $12$ .

$\frac{28 - 28}{4}$

Subtract $28$ from $28$ .

$\frac{0}{4}$

Divide $0$ by $4$ .

$0$

Step 5

Since $\frac{1}{2}$ is a known root, divide the polynomial by $x - \frac{1}{2}$ to find the quotient polynomial. This polynomial can then be used to find the remaining roots.

$\frac{10 x^{3} - 33 x^{2} + 8 x + 3}{x - \frac{1}{2}}$

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.

$\frac{1}{2}$	$10$	$- 33$	$8$	$3$
	\xa0			\xa0

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

$\frac{1}{2}$	$10$	$- 33$	$8$	$3$
	\xa0			\xa0
	$10$

Multiply the newest entry in the result $(10)$ by the divisor $(\frac{1}{2})$ and place the result of $(5)$ under the next term in the dividend $(- 33)$ .

$\frac{1}{2}$	$10$	$- 33$	$8$	$3$
	\xa0	$5$		\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.

$\frac{1}{2}$	$10$	$- 33$	$8$	$3$
	\xa0	$5$		\xa0
	$10$	$- 28$

Multiply the newest entry in the result $(- 28)$ by the divisor $(\frac{1}{2})$ and place the result of $(- 14)$ under the next term in the dividend $(8)$ .

$\frac{1}{2}$	$10$	$- 33$	$8$	$3$
	\xa0	$5$	$- 14$	\xa0
	$10$	$- 28$

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

$\frac{1}{2}$	$10$	$- 33$	$8$	$3$
	\xa0	$5$	$- 14$	\xa0
	$10$	$- 28$	$- 6$

Multiply the newest entry in the result $(- 6)$ by the divisor $(\frac{1}{2})$ and place the result of $(- 3)$ under the next term in the dividend $(3)$ .

$\frac{1}{2}$	$10$	$- 33$	$8$	$3$
	\xa0	$5$	$- 14$	$- 3$
	$10$	$- 28$	$- 6$

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

$\frac{1}{2}$	$10$	$- 33$	$8$	$3$
	\xa0	$5$	$- 14$	$- 3$
	$10$	$- 28$	$- 6$	$0$

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

$10 x^{2} + - 28 x - 6$

Simplify the quotient polynomial.

$10 x^{2} - 28 x - 6$

Step 7

Factor $2$ out of $10 x^{2} - 28 x - 6$ .

Factor $2$ out of $10 x^{2}$ .

$(x - \frac{1}{2}) (2 (5 x^{2}) - 28 x - 6)$

Factor $2$ out of $- 28 x$ .

$(x - \frac{1}{2}) (2 (5 x^{2}) + 2 (- 14 x) - 6)$

Factor $2$ out of $- 6$ .

$(x - \frac{1}{2}) (2 (5 x^{2}) + 2 (- 14 x) + 2 (- 3))$

Factor $2$ out of $2 (5 x^{2}) + 2 (- 14 x)$ .

$(x - \frac{1}{2}) (2 (5 x^{2} - 14 x) + 2 (- 3))$

Factor $2$ out of $2 (5 x^{2} - 14 x) + 2 (- 3)$ .

$(x - \frac{1}{2}) (2 (5 x^{2} - 14 x - 3))$

Step 8

Factor.

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 $a \cdot c = 5 \cdot - 3 = - 15$ and whose sum is $b = - 14$ .

Factor $- 14$ out of $- 14 x$ .

$(x - \frac{1}{2}) (2 (5 x^{2} - 14 x - 3))$

Rewrite $- 14$ as $1$ plus $- 15$

$(x - \frac{1}{2}) (2 (5 x^{2} + (1 - 15) x - 3))$

Apply the distributive property.

$(x - \frac{1}{2}) (2 (5 x^{2} + 1 x - 15 x - 3))$

Multiply $x$ by $1$ .

$(x - \frac{1}{2}) (2 (5 x^{2} + x - 15 x - 3))$

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

$(x - \frac{1}{2}) (2 ((5 x^{2} + x) - 15 x - 3))$

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

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

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

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

Remove unnecessary parentheses.

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

Step 9

Factor the left side of the equation.

Factor $10 x^{3} - 33 x^{2} + 8 x + 3$ 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.

$p = \pm 1, \pm 3 q = \pm 1, \pm 10, \pm 2, \pm 5$

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

$\pm 1, \pm 0.1, \pm 0.5, \pm 0.2, \pm 3, \pm 0.3, \pm 1.5, \pm 0.6$

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.

$10 {(- 0.2)}^{3} - 33 {(- 0.2)}^{2} + 8 \cdot - 0.2 + 3$

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

$10 \cdot - 0.008 - 33 {(- 0.2)}^{2} + 8 \cdot - 0.2 + 3$

Multiply $10$ by $- 0.008$ .

$- 0.08 - 33 {(- 0.2)}^{2} + 8 \cdot - 0.2 + 3$

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

$- 0.08 - 33 \cdot 0.04 + 8 \cdot - 0.2 + 3$

Multiply $- 33$ by $0.04$ .

$- 0.08 - 1.32 + 8 \cdot - 0.2 + 3$

Subtract $1.32$ from $- 0.08$ .

$- 1.4 + 8 \cdot - 0.2 + 3$

Multiply $8$ by $- 0.2$ .

$- 1.4 - 1.6 + 3$

Subtract $1.6$ from $- 1.4$ .

$- 3 + 3$

Add $- 3$ and $3$ .

$0$

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.

$\frac{10 x^{3} - 33 x^{2} + 8 x + 3}{5 x + 1}$

Divide $10 x^{3} - 33 x^{2} + 8 x + 3$ 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 $0$ .

$5 x$

$1$

$10 x^{3}$

$33 x^{2}$

$8 x$

$3$

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

					$2 x^{2}$
	$5 x$	+	$1$		$10 x^{3}$	-	$33 x^{2}$	+	$8 x$	+	$3$

Multiply the new quotient term by the divisor.

				$2 x^{2}$
$5 x$	+	$1$		$10 x^{3}$	-	$33 x^{2}$	+	$8 x$	+	$3$
			+	$10 x^{3}$	+	$2 x^{2}$

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

				$2 x^{2}$
$5 x$	+	$1$		$10 x^{3}$	-	$33 x^{2}$	+	$8 x$	+	$3$
			-	$10 x^{3}$	-	$2 x^{2}$

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

				$2 x^{2}$
$5 x$	+	$1$		$10 x^{3}$	-	$33 x^{2}$	+	$8 x$	+	$3$
			-	$10 x^{3}$	-	$2 x^{2}$
					-	$35 x^{2}$

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

				$2 x^{2}$
$5 x$	+	$1$		$10 x^{3}$	-	$33 x^{2}$	+	$8 x$	+	$3$
			-	$10 x^{3}$	-	$2 x^{2}$
					-	$35 x^{2}$	+	$8 x$

Divide the highest order term in the dividend $- 35 x^{2}$ by the highest order term in divisor $5 x$ .

				$2 x^{2}$	-	$7 x$
$5 x$	+	$1$		$10 x^{3}$	-	$33 x^{2}$	+	$8 x$	+	$3$
			-	$10 x^{3}$	-	$2 x^{2}$
					-	$35 x^{2}$	+	$8 x$

Multiply the new quotient term by the divisor.

				$2 x^{2}$	-	$7 x$
$5 x$	+	$1$		$10 x^{3}$	-	$33 x^{2}$	+	$8 x$	+	$3$
			-	$10 x^{3}$	-	$2 x^{2}$
					-	$35 x^{2}$	+	$8 x$
					-	$35 x^{2}$	-	$7 x$

The expression needs to be subtracted from the dividend, so change all the signs in $- 35 x^{2} - 7 x$

				$2 x^{2}$	-	$7 x$
$5 x$	+	$1$		$10 x^{3}$	-	$33 x^{2}$	+	$8 x$	+	$3$
			-	$10 x^{3}$	-	$2 x^{2}$
					-	$35 x^{2}$	+	$8 x$
					+	$35 x^{2}$	+	$7 x$

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

				$2 x^{2}$	-	$7 x$
$5 x$	+	$1$		$10 x^{3}$	-	$33 x^{2}$	+	$8 x$	+	$3$
			-	$10 x^{3}$	-	$2 x^{2}$
					-	$35 x^{2}$	+	$8 x$
					+	$35 x^{2}$	+	$7 x$
							+	$15 x$

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

				$2 x^{2}$	-	$7 x$
$5 x$	+	$1$		$10 x^{3}$	-	$33 x^{2}$	+	$8 x$	+	$3$
			-	$10 x^{3}$	-	$2 x^{2}$
					-	$35 x^{2}$	+	$8 x$
					+	$35 x^{2}$	+	$7 x$
							+	$15 x$	+	$3$

Divide the highest order term in the dividend $15 x$ by the highest order term in divisor $5 x$ .

				$2 x^{2}$	-	$7 x$	+	$3$
$5 x$	+	$1$		$10 x^{3}$	-	$33 x^{2}$	+	$8 x$	+	$3$
			-	$10 x^{3}$	-	$2 x^{2}$
					-	$35 x^{2}$	+	$8 x$
					+	$35 x^{2}$	+	$7 x$
							+	$15 x$	+	$3$

Multiply the new quotient term by the divisor.

				$2 x^{2}$	-	$7 x$	+	$3$
$5 x$	+	$1$		$10 x^{3}$	-	$33 x^{2}$	+	$8 x$	+	$3$
			-	$10 x^{3}$	-	$2 x^{2}$
					-	$35 x^{2}$	+	$8 x$
					+	$35 x^{2}$	+	$7 x$
							+	$15 x$	+	$3$
							+	$15 x$	+	$3$

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

				$2 x^{2}$	-	$7 x$	+	$3$
$5 x$	+	$1$		$10 x^{3}$	-	$33 x^{2}$	+	$8 x$	+	$3$
			-	$10 x^{3}$	-	$2 x^{2}$
					-	$35 x^{2}$	+	$8 x$
					+	$35 x^{2}$	+	$7 x$
							+	$15 x$	+	$3$
							-	$15 x$	-	$3$

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

				$2 x^{2}$	-	$7 x$	+	$3$
$5 x$	+	$1$		$10 x^{3}$	-	$33 x^{2}$	+	$8 x$	+	$3$
			-	$10 x^{3}$	-	$2 x^{2}$
					-	$35 x^{2}$	+	$8 x$
					+	$35 x^{2}$	+	$7 x$
							+	$15 x$	+	$3$
							-	$15 x$	-	$3$
										$0$

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

$2 x^{2} - 7 x + 3$

Write $10 x^{3} - 33 x^{2} + 8 x + 3$ as a set of factors.

$(5 x + 1) (2 x^{2} - 7 x + 3) = 0$

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 $a \cdot c = 2 \cdot 3 = 6$ and whose sum is $b = - 7$ .

Factor $- 7$ out of $- 7 x$ .

$(5 x + 1) (2 x^{2} - 7 x + 3) = 0$

Rewrite $- 7$ as $- 1$ plus $- 6$

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

Apply the distributive property.

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

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

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

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

$(5 x + 1) (x (2 x - 1) - 3 (2 x - 1)) = 0$

Factor the polynomial by factoring out the greatest common factor, $2 x - 1$ .

$(5 x + 1) ((2 x - 1) (x - 3)) = 0$

Remove unnecessary parentheses.

$(5 x + 1) (2 x - 1) (x - 3) = 0$

Step 10

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

$5 x + 1 = 0$

$2 x - 1 = 0$

$x - 3 = 0$

Step 11

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 $x$ .

Subtract $1$ from 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 $1$ .

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

Simplify the right side.

Move the negative in front of the fraction.

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

Step 12

Set $2 x - 1$ equal to $0$ and solve for $x$ .

Set $2 x - 1$ equal to $0$ .

$2 x - 1 = 0$

Solve $2 x - 1 = 0$ for $x$ .

Add $1$ to both sides of the equation.

$2 x = 1$

Divide each term in $2 x = 1$ by $2$ and simplify.

Divide each term in $2 x = 1$ by $2$ .

$\frac{2 x}{2} = \frac{1}{2}$

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

$\frac{2 x}{2} = \frac{1}{2}$

Divide $x$ by $1$ .

$x = \frac{1}{2}$

Step 13

Set $x - 3$ equal to $0$ and solve for $x$ .

Set $x - 3$ equal to $0$ .

$x - 3 = 0$

Add $3$ to both sides of the equation.

$x = 3$

Step 14

The final solution is all the values that make $(5 x + 1) (2 x - 1) (x - 3) = 0$ true.

$x = - \frac{1}{5}, \frac{1}{2}, 3$

Step 15

Find the Roots/Zeros Using the Rational Roots Test f(x)=10x^3-33x^2+8x+3

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