Find the Roots/Zeros Using the Rational Roots Test 4t^3-2t^2+8t+5

$4 t^{3} - 2 t^{2} + 8 t + 5$

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.

$p = \pm 1, \pm 5$

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

Step 2

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

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

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.

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

Step 4

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

Simplify each term.

Use the power rule ${(a b)}^{n} = a^{n} b^{n}$ to distribute the exponent.

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

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

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

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

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

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

One to any power is one.

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

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

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

Cancel the common factor of $4$ .

Move the leading negative in $- \frac{1}{8}$ into the numerator.

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

Factor $4$ out of $8$ .

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

Cancel the common factor.

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

Rewrite the expression.

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

Move the negative in front of the fraction.

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

Use the power rule ${(a b)}^{n} = a^{n} b^{n}$ to distribute the exponent.

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

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

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

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

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

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

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

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

One to any power is one.

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

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

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

Cancel the common factor of $2$ .

Factor $2$ out of $- 2$ .

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

Factor $2$ out of $4$ .

$- \frac{1}{2} + 2 \cdot - 1 \frac{1}{2 \cdot 2} + 8 (- \frac{1}{2}) + 5$

Cancel the common factor.

$- \frac{1}{2} + 2 \cdot - 1 \frac{1}{2 \cdot 2} + 8 (- \frac{1}{2}) + 5$

Rewrite the expression.

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

Rewrite $- 1 (\frac{1}{2})$ as $- (\frac{1}{2})$ .

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

Cancel the common factor of $2$ .

Move the leading negative in $- \frac{1}{2}$ into the numerator.

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

Factor $2$ out of $8$ .

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

Cancel the common factor.

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

Rewrite the expression.

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

Multiply $4$ by $- 1$ .

$- \frac{1}{2} - \frac{1}{2} - 4 + 5$

Combine fractions.

Combine the numerators over the common denominator.

$- 4 + 5 + \frac{- 1 - 1}{2}$

Subtract $1$ from $- 1$ .

$- 4 + 5 + \frac{- 2}{2}$

Find the common denominator.

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

$\frac{- 4}{1} + 5 + \frac{- 2}{2}$

Multiply $\frac{- 4}{1}$ by $\frac{2}{2}$ .

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

Multiply $\frac{- 4}{1}$ by $\frac{2}{2}$ .

$\frac{- 4 \cdot 2}{2} + 5 + \frac{- 2}{2}$

Write $5$ as a fraction with denominator $1$ .

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

Multiply $\frac{5}{1}$ by $\frac{2}{2}$ .

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

Multiply $\frac{5}{1}$ by $\frac{2}{2}$ .

$\frac{- 4 \cdot 2}{2} + \frac{5 \cdot 2}{2} + \frac{- 2}{2}$

Combine the numerators over the common denominator.

$\frac{- 4 \cdot 2 + 5 \cdot 2 - 2}{2}$

Simplify each term.

Multiply $- 4$ by $2$ .

$\frac{- 8 + 5 \cdot 2 - 2}{2}$

Multiply $5$ by $2$ .

$\frac{- 8 + 10 - 2}{2}$

Simplify the expression.

Add $- 8$ and $10$ .

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

Subtract $2$ from $2$ .

$\frac{0}{2}$

Divide $0$ by $2$ .

$0$

Step 5

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

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

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.

$- \frac{1}{2}$	$4$	$- 2$	$8$	$5$
	\xa0			\xa0

The first number in the dividend $(4)$ is put into the first position of the result area (below the horizontal line).

$- \frac{1}{2}$	$4$	$- 2$	$8$	$5$
	\xa0			\xa0
	$4$

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

$- \frac{1}{2}$	$4$	$- 2$	$8$	$5$
	\xa0	$- 2$		\xa0
	$4$

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

$- \frac{1}{2}$	$4$	$- 2$	$8$	$5$
	\xa0	$- 2$		\xa0
	$4$	$- 4$

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

$- \frac{1}{2}$	$4$	$- 2$	$8$	$5$
	\xa0	$- 2$	$2$	\xa0
	$4$	$- 4$

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

$- \frac{1}{2}$	$4$	$- 2$	$8$	$5$
	\xa0	$- 2$	$2$	\xa0
	$4$	$- 4$	$10$

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 $(5)$ .

$- \frac{1}{2}$	$4$	$- 2$	$8$	$5$
	\xa0	$- 2$	$2$	$- 5$
	$4$	$- 4$	$10$

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

$- \frac{1}{2}$	$4$	$- 2$	$8$	$5$
	\xa0	$- 2$	$2$	$- 5$
	$4$	$- 4$	$10$	$0$

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

$4 t^{2} + - 4 t + 10$

Simplify the quotient polynomial.

$4 t^{2} - 4 t + 10$

Step 7

Factor $2$ out of $4 t^{2} - 4 t + 10$ .

Factor $2$ out of $4 t^{2}$ .

$(t + \frac{1}{2}) (2 (2 t^{2}) - 4 t + 10)$

Factor $2$ out of $- 4 t$ .

$(t + \frac{1}{2}) (2 (2 t^{2}) + 2 (- 2 t) + 10)$

Factor $2$ out of $10$ .

$(t + \frac{1}{2}) (2 (2 t^{2}) + 2 (- 2 t) + 2 (5))$

Factor $2$ out of $2 (2 t^{2}) + 2 (- 2 t)$ .

$(t + \frac{1}{2}) (2 (2 t^{2} - 2 t) + 2 (5))$

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

$(t + \frac{1}{2}) (2 (2 t^{2} - 2 t + 5))$

Step 8

Factor $4 t^{3} - 2 t^{2} + 8 t + 5$ 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.

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

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

$\pm 1, \pm 0.25, \pm 0.5, \pm 5, \pm 1.25, \pm 2.5$

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.

$4 {(- 0.5)}^{3} - 2 {(- 0.5)}^{2} + 8 \cdot - 0.5 + 5$

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

$4 \cdot - 0.125 - 2 {(- 0.5)}^{2} + 8 \cdot - 0.5 + 5$

Multiply $4$ by $- 0.125$ .

$- 0.5 - 2 {(- 0.5)}^{2} + 8 \cdot - 0.5 + 5$

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

$- 0.5 - 2 \cdot 0.25 + 8 \cdot - 0.5 + 5$

Multiply $- 2$ by $0.25$ .

$- 0.5 - 0.5 + 8 \cdot - 0.5 + 5$

Subtract $0.5$ from $- 0.5$ .

$- 1 + 8 \cdot - 0.5 + 5$

Multiply $8$ by $- 0.5$ .

$- 1 - 4 + 5$

Subtract $4$ from $- 1$ .

$- 5 + 5$

Add $- 5$ and $5$ .

$0$

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

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

Divide $4 t^{3} - 2 t^{2} + 8 t + 5$ by $2 t + 1$ .

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

$2 t$

$1$

$4 t^{3}$

$2 t^{2}$

$8 t$

$5$

Divide the highest order term in the dividend $4 t^{3}$ by the highest order term in divisor $2 t$ .

					$2 t^{2}$
	$2 t$	+	$1$		$4 t^{3}$	-	$2 t^{2}$	+	$8 t$	+	$5$

Multiply the new quotient term by the divisor.

				$2 t^{2}$
$2 t$	+	$1$		$4 t^{3}$	-	$2 t^{2}$	+	$8 t$	+	$5$
			+	$4 t^{3}$	+	$2 t^{2}$

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

				$2 t^{2}$
$2 t$	+	$1$		$4 t^{3}$	-	$2 t^{2}$	+	$8 t$	+	$5$
			-	$4 t^{3}$	-	$2 t^{2}$

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

				$2 t^{2}$
$2 t$	+	$1$		$4 t^{3}$	-	$2 t^{2}$	+	$8 t$	+	$5$
			-	$4 t^{3}$	-	$2 t^{2}$
					-	$4 t^{2}$

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

				$2 t^{2}$
$2 t$	+	$1$		$4 t^{3}$	-	$2 t^{2}$	+	$8 t$	+	$5$
			-	$4 t^{3}$	-	$2 t^{2}$
					-	$4 t^{2}$	+	$8 t$

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

				$2 t^{2}$	-	$2 t$
$2 t$	+	$1$		$4 t^{3}$	-	$2 t^{2}$	+	$8 t$	+	$5$
			-	$4 t^{3}$	-	$2 t^{2}$
					-	$4 t^{2}$	+	$8 t$

Multiply the new quotient term by the divisor.

				$2 t^{2}$	-	$2 t$
$2 t$	+	$1$		$4 t^{3}$	-	$2 t^{2}$	+	$8 t$	+	$5$
			-	$4 t^{3}$	-	$2 t^{2}$
					-	$4 t^{2}$	+	$8 t$
					-	$4 t^{2}$	-	$2 t$

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

				$2 t^{2}$	-	$2 t$
$2 t$	+	$1$		$4 t^{3}$	-	$2 t^{2}$	+	$8 t$	+	$5$
			-	$4 t^{3}$	-	$2 t^{2}$
					-	$4 t^{2}$	+	$8 t$
					+	$4 t^{2}$	+	$2 t$

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

				$2 t^{2}$	-	$2 t$
$2 t$	+	$1$		$4 t^{3}$	-	$2 t^{2}$	+	$8 t$	+	$5$
			-	$4 t^{3}$	-	$2 t^{2}$
					-	$4 t^{2}$	+	$8 t$
					+	$4 t^{2}$	+	$2 t$
							+	$10 t$

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

				$2 t^{2}$	-	$2 t$
$2 t$	+	$1$		$4 t^{3}$	-	$2 t^{2}$	+	$8 t$	+	$5$
			-	$4 t^{3}$	-	$2 t^{2}$
					-	$4 t^{2}$	+	$8 t$
					+	$4 t^{2}$	+	$2 t$
							+	$10 t$	+	$5$

Divide the highest order term in the dividend $10 t$ by the highest order term in divisor $2 t$ .

				$2 t^{2}$	-	$2 t$	+	$5$
$2 t$	+	$1$		$4 t^{3}$	-	$2 t^{2}$	+	$8 t$	+	$5$
			-	$4 t^{3}$	-	$2 t^{2}$
					-	$4 t^{2}$	+	$8 t$
					+	$4 t^{2}$	+	$2 t$
							+	$10 t$	+	$5$

Multiply the new quotient term by the divisor.

				$2 t^{2}$	-	$2 t$	+	$5$
$2 t$	+	$1$		$4 t^{3}$	-	$2 t^{2}$	+	$8 t$	+	$5$
			-	$4 t^{3}$	-	$2 t^{2}$
					-	$4 t^{2}$	+	$8 t$
					+	$4 t^{2}$	+	$2 t$
							+	$10 t$	+	$5$
							+	$10 t$	+	$5$

The expression needs to be subtracted from the dividend, so change all the signs in $10 t + 5$

				$2 t^{2}$	-	$2 t$	+	$5$
$2 t$	+	$1$		$4 t^{3}$	-	$2 t^{2}$	+	$8 t$	+	$5$
			-	$4 t^{3}$	-	$2 t^{2}$
					-	$4 t^{2}$	+	$8 t$
					+	$4 t^{2}$	+	$2 t$
							+	$10 t$	+	$5$
							-	$10 t$	-	$5$

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

				$2 t^{2}$	-	$2 t$	+	$5$
$2 t$	+	$1$		$4 t^{3}$	-	$2 t^{2}$	+	$8 t$	+	$5$
			-	$4 t^{3}$	-	$2 t^{2}$
					-	$4 t^{2}$	+	$8 t$
					+	$4 t^{2}$	+	$2 t$
							+	$10 t$	+	$5$
							-	$10 t$	-	$5$
										$0$

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

$2 t^{2} - 2 t + 5$

Write $4 t^{3} - 2 t^{2} + 8 t + 5$ as a set of factors.

$(2 t + 1) (2 t^{2} - 2 t + 5) = 0$

Step 9

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

$2 t + 1 = 0$

$2 t^{2} - 2 t + 5 = 0$

Step 10

Set $2 t + 1$ equal to $0$ and solve for $t$ .

Set $2 t + 1$ equal to $0$ .

$2 t + 1 = 0$

Solve $2 t + 1 = 0$ for $t$ .

Subtract $1$ from both sides of the equation.

$2 t = - 1$

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

Divide each term in $2 t = - 1$ by $2$ .

$\frac{2 t}{2} = \frac{- 1}{2}$

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

$\frac{2 t}{2} = \frac{- 1}{2}$

Divide $t$ by $1$ .

$t = \frac{- 1}{2}$

Simplify the right side.

Move the negative in front of the fraction.

$t = - \frac{1}{2}$

Step 11

Set $2 t^{2} - 2 t + 5$ equal to $0$ and solve for $t$ .

Set $2 t^{2} - 2 t + 5$ equal to $0$ .

$2 t^{2} - 2 t + 5 = 0$

Solve $2 t^{2} - 2 t + 5 = 0$ for $t$ .

Use the quadratic formula to find the solutions.

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

Substitute the values $a = 2$ , $b = - 2$ , and $c = 5$ into the quadratic formula and solve for $t$ .

$\frac{2 \pm \sqrt{{(- 2)}^{2} - 4 \cdot (2 \cdot 5)}}{2 \cdot 2}$

Simplify.

Simplify the numerator.

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

$t = \frac{2 \pm \sqrt{4 - 4 \cdot 2 \cdot 5}}{2 \cdot 2}$

Multiply $- 4 \cdot 2 \cdot 5$ .

Multiply $- 4$ by $2$ .

$t = \frac{2 \pm \sqrt{4 - 8 \cdot 5}}{2 \cdot 2}$

Multiply $- 8$ by $5$ .

$t = \frac{2 \pm \sqrt{4 - 40}}{2 \cdot 2}$

Subtract $40$ from $4$ .

$t = \frac{2 \pm \sqrt{- 36}}{2 \cdot 2}$

Rewrite $- 36$ as $- 1 (36)$ .

$t = \frac{2 \pm \sqrt{- 1 \cdot 36}}{2 \cdot 2}$

Rewrite $\sqrt{- 1 (36)}$ as $\sqrt{- 1} \cdot \sqrt{36}$ .

$t = \frac{2 \pm \sqrt{- 1} \cdot \sqrt{36}}{2 \cdot 2}$

Rewrite $\sqrt{- 1}$ as $i$ .

$t = \frac{2 \pm i \cdot \sqrt{36}}{2 \cdot 2}$

Rewrite $36$ as $6^{2}$ .

$t = \frac{2 \pm i \cdot \sqrt{6^{2}}}{2 \cdot 2}$

Pull terms out from under the radical, assuming positive real numbers.

$t = \frac{2 \pm i \cdot 6}{2 \cdot 2}$

Move $6$ to the left of $i$ .

$t = \frac{2 \pm 6 i}{2 \cdot 2}$

Multiply $2$ by $2$ .

$t = \frac{2 \pm 6 i}{4}$

Simplify $\frac{2 \pm 6 i}{4}$ .

$t = \frac{1 \pm 3 i}{2}$

The final answer is the combination of both solutions.

$t = \frac{1}{2} + \frac{3 i}{2}, \frac{1}{2} - \frac{3 i}{2}$

Step 12

The final solution is all the values that make $(2 t + 1) (2 t^{2} - 2 t + 5) = 0$ true.

$t = - \frac{1}{2}, \frac{1}{2} + \frac{3 i}{2}, \frac{1}{2} - \frac{3 i}{2}$

Step 13

Find the Roots/Zeros Using the Rational Roots Test 4t^3-2t^2+8t+5

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