Find the Roots/Zeros Using the Rational Roots Test 8x^3+50x^2-41x+7

$8 x^{3} + 50 x^{2} - 41 x + 7$

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.

$p = \pm 1, \pm 7$

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

Step 2

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

$\pm 1, \pm \frac{1}{2}, \pm \frac{1}{4}, \pm \frac{1}{8}, \pm 7, \pm \frac{7}{2}, \pm \frac{7}{4}, \pm \frac{7}{8}$

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.

$8 {(\frac{1}{2})}^{3} + 50 {(\frac{1}{2})}^{2} - 41 (\frac{1}{2}) + 7$

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 $\frac{1}{2}$ .

$8 \frac{1^{3}}{2^{3}} + 50 {(\frac{1}{2})}^{2} - 41 (\frac{1}{2}) + 7$

One to any power is one.

$8 \frac{1}{2^{3}} + 50 {(\frac{1}{2})}^{2} - 41 (\frac{1}{2}) + 7$

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

$8 (\frac{1}{8}) + 50 {(\frac{1}{2})}^{2} - 41 (\frac{1}{2}) + 7$

Cancel the common factor of $8$ .

Cancel the common factor.

$8 (\frac{1}{8}) + 50 {(\frac{1}{2})}^{2} - 41 (\frac{1}{2}) + 7$

Rewrite the expression.

$1 + 50 {(\frac{1}{2})}^{2} - 41 (\frac{1}{2}) + 7$

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

$1 + 50 \frac{1^{2}}{2^{2}} - 41 (\frac{1}{2}) + 7$

One to any power is one.

$1 + 50 \frac{1}{2^{2}} - 41 (\frac{1}{2}) + 7$

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

$1 + 50 (\frac{1}{4}) - 41 (\frac{1}{2}) + 7$

Cancel the common factor of $2$ .

Factor $2$ out of $50$ .

$1 + 2 (25) \frac{1}{4} - 41 (\frac{1}{2}) + 7$

Factor $2$ out of $4$ .

$1 + 2 \cdot 25 \frac{1}{2 \cdot 2} - 41 (\frac{1}{2}) + 7$

Cancel the common factor.

$1 + 2 \cdot 25 \frac{1}{2 \cdot 2} - 41 (\frac{1}{2}) + 7$

Rewrite the expression.

$1 + 25 (\frac{1}{2}) - 41 (\frac{1}{2}) + 7$

Combine $25$ and $\frac{1}{2}$ .

$1 + \frac{25}{2} - 41 (\frac{1}{2}) + 7$

Combine $- 41$ and $\frac{1}{2}$ .

$1 + \frac{25}{2} + \frac{- 41}{2} + 7$

Move the negative in front of the fraction.

$1 + \frac{25}{2} - \frac{41}{2} + 7$

Combine fractions.

Combine the numerators over the common denominator.

$1 + 7 + \frac{25 - 41}{2}$

Subtract $41$ from $25$ .

$1 + 7 + \frac{- 16}{2}$

Find the common denominator.

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

$\frac{1}{1} + 7 + \frac{- 16}{2}$

Multiply $\frac{1}{1}$ by $\frac{2}{2}$ .

$\frac{1}{1} \cdot \frac{2}{2} + 7 + \frac{- 16}{2}$

Multiply $\frac{1}{1}$ by $\frac{2}{2}$ .

$\frac{2}{2} + 7 + \frac{- 16}{2}$

Write $7$ as a fraction with denominator $1$ .

$\frac{2}{2} + \frac{7}{1} + \frac{- 16}{2}$

Multiply $\frac{7}{1}$ by $\frac{2}{2}$ .

$\frac{2}{2} + \frac{7}{1} \cdot \frac{2}{2} + \frac{- 16}{2}$

Multiply $\frac{7}{1}$ by $\frac{2}{2}$ .

$\frac{2}{2} + \frac{7 \cdot 2}{2} + \frac{- 16}{2}$

Combine the numerators over the common denominator.

$\frac{2 + 7 \cdot 2 - 16}{2}$

Simplify the expression.

Multiply $7$ by $2$ .

$\frac{2 + 14 - 16}{2}$

Add $2$ and $14$ .

$\frac{16 - 16}{2}$

Subtract $16$ from $16$ .

$\frac{0}{2}$

Divide $0$ by $2$ .

$0$

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.

$\frac{8 x^{3} + 50 x^{2} - 41 x + 7}{x - \frac{1}{2}}$

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.

$\frac{1}{2}$	$8$	$50$	$- 41$	$7$
	\xa0			\xa0

The first number in the dividend $(8)$ is put into the first position of the result area (below the horizontal line).

$\frac{1}{2}$	$8$	$50$	$- 41$	$7$
	\xa0			\xa0
	$8$

Multiply the newest entry in the result $(8)$ by the divisor $(\frac{1}{2})$ and place the result of $(4)$ under the next term in the dividend $(50)$ .

$\frac{1}{2}$	$8$	$50$	$- 41$	$7$
	\xa0	$4$		\xa0
	$8$

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

$\frac{1}{2}$	$8$	$50$	$- 41$	$7$
	\xa0	$4$		\xa0
	$8$	$54$

Multiply the newest entry in the result $(54)$ by the divisor $(\frac{1}{2})$ and place the result of $(27)$ under the next term in the dividend $(- 41)$ .

$\frac{1}{2}$	$8$	$50$	$- 41$	$7$
	\xa0	$4$	$27$	\xa0
	$8$	$54$

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

$\frac{1}{2}$	$8$	$50$	$- 41$	$7$
	\xa0	$4$	$27$	\xa0
	$8$	$54$	$- 14$

Multiply the newest entry in the result $(- 14)$ by the divisor $(\frac{1}{2})$ and place the result of $(- 7)$ under the next term in the dividend $(7)$ .

$\frac{1}{2}$	$8$	$50$	$- 41$	$7$
	\xa0	$4$	$27$	$- 7$
	$8$	$54$	$- 14$

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

$\frac{1}{2}$	$8$	$50$	$- 41$	$7$
	\xa0	$4$	$27$	$- 7$
	$8$	$54$	$- 14$	$0$

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

$8 x^{2} + 54 x - 14$

Step 7

Factor $2$ out of $8 x^{2} + 54 x - 14$ .

Factor $2$ out of $8 x^{2}$ .

$(x - \frac{1}{2}) (2 (4 x^{2}) + 54 x - 14)$

Factor $2$ out of $54 x$ .

$(x - \frac{1}{2}) (2 (4 x^{2}) + 2 (27 x) - 14)$

Factor $2$ out of $- 14$ .

$(x - \frac{1}{2}) (2 (4 x^{2}) + 2 (27 x) + 2 (- 7))$

Factor $2$ out of $2 (4 x^{2}) + 2 (27 x)$ .

$(x - \frac{1}{2}) (2 (4 x^{2} + 27 x) + 2 (- 7))$

Factor $2$ out of $2 (4 x^{2} + 27 x) + 2 (- 7)$ .

$(x - \frac{1}{2}) (2 (4 x^{2} + 27 x - 7))$

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 $a \cdot c = 4 \cdot - 7 = - 28$ and whose sum is $b = 27$ .

Factor $27$ out of $27 x$ .

$(x - \frac{1}{2}) (2 (4 x^{2} + 27 (x) - 7))$

Rewrite $27$ as $- 1$ plus $28$

$(x - \frac{1}{2}) (2 (4 x^{2} + (- 1 + 28) x - 7))$

Apply the distributive property.

$(x - \frac{1}{2}) (2 (4 x^{2} - 1 x + 28 x - 7))$

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

$(x - \frac{1}{2}) (2 ((4 x^{2} - 1 x) + 28 x - 7))$

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

$(x - \frac{1}{2}) (2 (x (4 x - 1) + 7 (4 x - 1)))$

Factor the polynomial by factoring out the greatest common factor, $4 x - 1$ .

$(x - \frac{1}{2}) (2 ((4 x - 1) (x + 7)))$

Remove unnecessary parentheses.

$(x - \frac{1}{2}) (2 (4 x - 1) (x + 7))$

Step 9

Factor the left side of the equation.

Factor $8 x^{3} + 50 x^{2} - 41 x + 7$ 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.

$p = \pm 1, \pm 7 q = \pm 1, \pm 8, \pm 2, \pm 4$

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

$\pm 1, \pm 0.125, \pm 0.5, \pm 0.25, \pm 7, \pm 0.875, \pm 3.5, \pm 1.75$

Substitute $0.5$ and simplify the expression. In this case, the expression is equal to $0$ so $0.5$ is a root of the polynomial.

Substitute $0.5$ into the polynomial.

$8 \cdot {0.5}^{3} + 50 \cdot {0.5}^{2} - 41 \cdot 0.5 + 7$

Raise $0.5$ to the power of $3$ .

$8 \cdot 0.125 + 50 \cdot {0.5}^{2} - 41 \cdot 0.5 + 7$

Multiply $8$ by $0.125$ .

$1 + 50 \cdot {0.5}^{2} - 41 \cdot 0.5 + 7$

Raise $0.5$ to the power of $2$ .

$1 + 50 \cdot 0.25 - 41 \cdot 0.5 + 7$

Multiply $50$ by $0.25$ .

$1 + 12.5 - 41 \cdot 0.5 + 7$

Add $1$ and $12.5$ .

$13.5 - 41 \cdot 0.5 + 7$

Multiply $- 41$ by $0.5$ .

$13.5 - 20.5 + 7$

Subtract $20.5$ from $13.5$ .

$- 7 + 7$

Add $- 7$ and $7$ .

$0$

Since $0.5$ is a known root, divide the polynomial by $2 x - 1$ to find the quotient polynomial. This polynomial can then be used to find the remaining roots.

$\frac{8 x^{3} + 50 x^{2} - 41 x + 7}{2 x - 1}$

Divide $8 x^{3} + 50 x^{2} - 41 x + 7$ by $2 x - 1$ .

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

$2 x$

$1$

$8 x^{3}$

$50 x^{2}$

$41 x$

$7$

Divide the highest order term in the dividend $8 x^{3}$ by the highest order term in divisor $2 x$ .

					$4 x^{2}$
	$2 x$	-	$1$		$8 x^{3}$	+	$50 x^{2}$	-	$41 x$	+	$7$

Multiply the new quotient term by the divisor.

				$4 x^{2}$
$2 x$	-	$1$		$8 x^{3}$	+	$50 x^{2}$	-	$41 x$	+	$7$
			+	$8 x^{3}$	-	$4 x^{2}$

The expression needs to be subtracted from the dividend, so change all the signs in $8 x^{3} - 4 x^{2}$

				$4 x^{2}$
$2 x$	-	$1$		$8 x^{3}$	+	$50 x^{2}$	-	$41 x$	+	$7$
			-	$8 x^{3}$	+	$4 x^{2}$

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

				$4 x^{2}$
$2 x$	-	$1$		$8 x^{3}$	+	$50 x^{2}$	-	$41 x$	+	$7$
			-	$8 x^{3}$	+	$4 x^{2}$
					+	$54 x^{2}$

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

				$4 x^{2}$
$2 x$	-	$1$		$8 x^{3}$	+	$50 x^{2}$	-	$41 x$	+	$7$
			-	$8 x^{3}$	+	$4 x^{2}$
					+	$54 x^{2}$	-	$41 x$

Divide the highest order term in the dividend $54 x^{2}$ by the highest order term in divisor $2 x$ .

				$4 x^{2}$	+	$27 x$
$2 x$	-	$1$		$8 x^{3}$	+	$50 x^{2}$	-	$41 x$	+	$7$
			-	$8 x^{3}$	+	$4 x^{2}$
					+	$54 x^{2}$	-	$41 x$

Multiply the new quotient term by the divisor.

				$4 x^{2}$	+	$27 x$
$2 x$	-	$1$		$8 x^{3}$	+	$50 x^{2}$	-	$41 x$	+	$7$
			-	$8 x^{3}$	+	$4 x^{2}$
					+	$54 x^{2}$	-	$41 x$
					+	$54 x^{2}$	-	$27 x$

The expression needs to be subtracted from the dividend, so change all the signs in $54 x^{2} - 27 x$

				$4 x^{2}$	+	$27 x$
$2 x$	-	$1$		$8 x^{3}$	+	$50 x^{2}$	-	$41 x$	+	$7$
			-	$8 x^{3}$	+	$4 x^{2}$
					+	$54 x^{2}$	-	$41 x$
					-	$54 x^{2}$	+	$27 x$

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

				$4 x^{2}$	+	$27 x$
$2 x$	-	$1$		$8 x^{3}$	+	$50 x^{2}$	-	$41 x$	+	$7$
			-	$8 x^{3}$	+	$4 x^{2}$
					+	$54 x^{2}$	-	$41 x$
					-	$54 x^{2}$	+	$27 x$
							-	$14 x$

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

				$4 x^{2}$	+	$27 x$
$2 x$	-	$1$		$8 x^{3}$	+	$50 x^{2}$	-	$41 x$	+	$7$
			-	$8 x^{3}$	+	$4 x^{2}$
					+	$54 x^{2}$	-	$41 x$
					-	$54 x^{2}$	+	$27 x$
							-	$14 x$	+	$7$

Divide the highest order term in the dividend $- 14 x$ by the highest order term in divisor $2 x$ .

				$4 x^{2}$	+	$27 x$	-	$7$
$2 x$	-	$1$		$8 x^{3}$	+	$50 x^{2}$	-	$41 x$	+	$7$
			-	$8 x^{3}$	+	$4 x^{2}$
					+	$54 x^{2}$	-	$41 x$
					-	$54 x^{2}$	+	$27 x$
							-	$14 x$	+	$7$

Multiply the new quotient term by the divisor.

				$4 x^{2}$	+	$27 x$	-	$7$
$2 x$	-	$1$		$8 x^{3}$	+	$50 x^{2}$	-	$41 x$	+	$7$
			-	$8 x^{3}$	+	$4 x^{2}$
					+	$54 x^{2}$	-	$41 x$
					-	$54 x^{2}$	+	$27 x$
							-	$14 x$	+	$7$
							-	$14 x$	+	$7$

The expression needs to be subtracted from the dividend, so change all the signs in $- 14 x + 7$

				$4 x^{2}$	+	$27 x$	-	$7$
$2 x$	-	$1$		$8 x^{3}$	+	$50 x^{2}$	-	$41 x$	+	$7$
			-	$8 x^{3}$	+	$4 x^{2}$
					+	$54 x^{2}$	-	$41 x$
					-	$54 x^{2}$	+	$27 x$
							-	$14 x$	+	$7$
							+	$14 x$	-	$7$

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

				$4 x^{2}$	+	$27 x$	-	$7$
$2 x$	-	$1$		$8 x^{3}$	+	$50 x^{2}$	-	$41 x$	+	$7$
			-	$8 x^{3}$	+	$4 x^{2}$
					+	$54 x^{2}$	-	$41 x$
					-	$54 x^{2}$	+	$27 x$
							-	$14 x$	+	$7$
							+	$14 x$	-	$7$
										$0$

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

$4 x^{2} + 27 x - 7$

Write $8 x^{3} + 50 x^{2} - 41 x + 7$ as a set of factors.

$(2 x - 1) (4 x^{2} + 27 x - 7) = 0$

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 $a \cdot c = 4 \cdot - 7 = - 28$ and whose sum is $b = 27$ .

Factor $27$ out of $27 x$ .

$(2 x - 1) (4 x^{2} + 27 (x) - 7) = 0$

Rewrite $27$ as $- 1$ plus $28$

$(2 x - 1) (4 x^{2} + (- 1 + 28) x - 7) = 0$

Apply the distributive property.

$(2 x - 1) (4 x^{2} - 1 x + 28 x - 7) = 0$

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

$(2 x - 1) ((4 x^{2} - 1 x) + 28 x - 7) = 0$

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

$(2 x - 1) (x (4 x - 1) + 7 (4 x - 1)) = 0$

Factor the polynomial by factoring out the greatest common factor, $4 x - 1$ .

$(2 x - 1) ((4 x - 1) (x + 7)) = 0$

Remove unnecessary parentheses.

$(2 x - 1) (4 x - 1) (x + 7) = 0$

Step 10

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

$2 x - 1 = 0$

$4 x - 1 = 0$

$x + 7 = 0$

Step 11

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 12

Set $4 x - 1$ equal to $0$ and solve for $x$ .

Set $4 x - 1$ equal to $0$ .

$4 x - 1 = 0$

Solve $4 x - 1 = 0$ for $x$ .

Add $1$ to both sides of the equation.

$4 x = 1$

Divide each term in $4 x = 1$ by $4$ and simplify.

Divide each term in $4 x = 1$ by $4$ .

$\frac{4 x}{4} = \frac{1}{4}$

Simplify the left side.

Cancel the common factor of $4$ .

Cancel the common factor.

$\frac{4 x}{4} = \frac{1}{4}$

Divide $x$ by $1$ .

$x = \frac{1}{4}$

Step 13

Set $x + 7$ equal to $0$ and solve for $x$ .

Set $x + 7$ equal to $0$ .

$x + 7 = 0$

Subtract $7$ from both sides of the equation.

$x = - 7$

Step 14

The final solution is all the values that make $(2 x - 1) (4 x - 1) (x + 7) = 0$ true.

$x = \frac{1}{2}, \frac{1}{4}, - 7$

Step 15

Find the Roots/Zeros Using the Rational Roots Test 8x^3+50x^2-41x+7

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