Find the Roots/Zeros Using the Rational Roots Test f(t)=4t^3-10t^2+28t+17

$f (t) = 4 t^{3} - 10 t^{2} + 28 t + 17$

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.

$p = \pm 1, \pm 17$

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

Step 2

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

$\pm 1, \pm \frac{1}{2}, \pm \frac{1}{4}, \pm 17, \pm \frac{17}{2}, \pm \frac{17}{4}$

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.

$4 {(- \frac{1}{2})}^{3} - 10 {(- \frac{1}{2})}^{2} + 28 (- \frac{1}{2}) + 17$

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 $- \frac{1}{2}$ .

$4 ({(- 1)}^{3} {(\frac{1}{2})}^{3}) - 10 {(- \frac{1}{2})}^{2} + 28 (- \frac{1}{2}) + 17$

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

$4 ({(- 1)}^{3} \frac{1^{3}}{2^{3}}) - 10 {(- \frac{1}{2})}^{2} + 28 (- \frac{1}{2}) + 17$

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

$4 (- \frac{1^{3}}{2^{3}}) - 10 {(- \frac{1}{2})}^{2} + 28 (- \frac{1}{2}) + 17$

One to any power is one.

$4 (- \frac{1}{2^{3}}) - 10 {(- \frac{1}{2})}^{2} + 28 (- \frac{1}{2}) + 17$

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

$4 (- \frac{1}{8}) - 10 {(- \frac{1}{2})}^{2} + 28 (- \frac{1}{2}) + 17$

Cancel the common factor of $4$ .

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

$4 (\frac{- 1}{8}) - 10 {(- \frac{1}{2})}^{2} + 28 (- \frac{1}{2}) + 17$

Factor $4$ out of $8$ .

$4 \frac{- 1}{4 (2)} - 10 {(- \frac{1}{2})}^{2} + 28 (- \frac{1}{2}) + 17$

Cancel the common factor.

$4 \frac{- 1}{4 \cdot 2} - 10 {(- \frac{1}{2})}^{2} + 28 (- \frac{1}{2}) + 17$

Rewrite the expression.

$\frac{- 1}{2} - 10 {(- \frac{1}{2})}^{2} + 28 (- \frac{1}{2}) + 17$

Move the negative in front of the fraction.

$- \frac{1}{2} - 10 {(- \frac{1}{2})}^{2} + 28 (- \frac{1}{2}) + 17$

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

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

$- \frac{1}{2} - 10 ({(- 1)}^{2} {(\frac{1}{2})}^{2}) + 28 (- \frac{1}{2}) + 17$

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

$- \frac{1}{2} - 10 ({(- 1)}^{2} \frac{1^{2}}{2^{2}}) + 28 (- \frac{1}{2}) + 17$

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

$- \frac{1}{2} - 10 (1 \frac{1^{2}}{2^{2}}) + 28 (- \frac{1}{2}) + 17$

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

$- \frac{1}{2} - 10 \frac{1^{2}}{2^{2}} + 28 (- \frac{1}{2}) + 17$

One to any power is one.

$- \frac{1}{2} - 10 \frac{1}{2^{2}} + 28 (- \frac{1}{2}) + 17$

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

$- \frac{1}{2} - 10 (\frac{1}{4}) + 28 (- \frac{1}{2}) + 17$

Cancel the common factor of $2$ .

Factor $2$ out of $- 10$ .

$- \frac{1}{2} + 2 (- 5) \frac{1}{4} + 28 (- \frac{1}{2}) + 17$

Factor $2$ out of $4$ .

$- \frac{1}{2} + 2 \cdot - 5 \frac{1}{2 \cdot 2} + 28 (- \frac{1}{2}) + 17$

Cancel the common factor.

$- \frac{1}{2} + 2 \cdot - 5 \frac{1}{2 \cdot 2} + 28 (- \frac{1}{2}) + 17$

Rewrite the expression.

$- \frac{1}{2} - 5 (\frac{1}{2}) + 28 (- \frac{1}{2}) + 17$

Combine $- 5$ and $\frac{1}{2}$ .

$- \frac{1}{2} + \frac{- 5}{2} + 28 (- \frac{1}{2}) + 17$

Move the negative in front of the fraction.

$- \frac{1}{2} - \frac{5}{2} + 28 (- \frac{1}{2}) + 17$

Cancel the common factor of $2$ .

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

$- \frac{1}{2} - \frac{5}{2} + 28 (\frac{- 1}{2}) + 17$

Factor $2$ out of $28$ .

$- \frac{1}{2} - \frac{5}{2} + 2 (14) \frac{- 1}{2} + 17$

Cancel the common factor.

$- \frac{1}{2} - \frac{5}{2} + 2 \cdot 14 \frac{- 1}{2} + 17$

Rewrite the expression.

$- \frac{1}{2} - \frac{5}{2} + 14 \cdot - 1 + 17$

Multiply $14$ by $- 1$ .

$- \frac{1}{2} - \frac{5}{2} - 14 + 17$

Combine fractions.

Combine the numerators over the common denominator.

$- 14 + 17 + \frac{- 1 - 5}{2}$

Subtract $5$ from $- 1$ .

$- 14 + 17 + \frac{- 6}{2}$

Find the common denominator.

Write $- 14$ as a fraction with denominator $1$ .

$\frac{- 14}{1} + 17 + \frac{- 6}{2}$

Multiply $\frac{- 14}{1}$ by $\frac{2}{2}$ .

$\frac{- 14}{1} \cdot \frac{2}{2} + 17 + \frac{- 6}{2}$

Multiply $\frac{- 14}{1}$ by $\frac{2}{2}$ .

$\frac{- 14 \cdot 2}{2} + 17 + \frac{- 6}{2}$

Write $17$ as a fraction with denominator $1$ .

$\frac{- 14 \cdot 2}{2} + \frac{17}{1} + \frac{- 6}{2}$

Multiply $\frac{17}{1}$ by $\frac{2}{2}$ .

$\frac{- 14 \cdot 2}{2} + \frac{17}{1} \cdot \frac{2}{2} + \frac{- 6}{2}$

Multiply $\frac{17}{1}$ by $\frac{2}{2}$ .

$\frac{- 14 \cdot 2}{2} + \frac{17 \cdot 2}{2} + \frac{- 6}{2}$

Combine the numerators over the common denominator.

$\frac{- 14 \cdot 2 + 17 \cdot 2 - 6}{2}$

Simplify each term.

Multiply $- 14$ by $2$ .

$\frac{- 28 + 17 \cdot 2 - 6}{2}$

Multiply $17$ by $2$ .

$\frac{- 28 + 34 - 6}{2}$

Simplify the expression.

Add $- 28$ and $34$ .

$\frac{6 - 6}{2}$

Subtract $6$ from $6$ .

$\frac{0}{2}$

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.

$\frac{4 t^{3} - 10 t^{2} + 28 t + 17}{t + \frac{1}{2}}$

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.

$- \frac{1}{2}$	$4$	$- 10$	$28$	$17$
	\xa0			\xa0

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

$- \frac{1}{2}$	$4$	$- 10$	$28$	$17$
	\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 $(- 10)$ .

$- \frac{1}{2}$	$4$	$- 10$	$28$	$17$
	\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.

$- \frac{1}{2}$	$4$	$- 10$	$28$	$17$
	\xa0	$- 2$		\xa0
	$4$	$- 12$

Multiply the newest entry in the result $(- 12)$ by the divisor $(- \frac{1}{2})$ and place the result of $(6)$ under the next term in the dividend $(28)$ .

$- \frac{1}{2}$	$4$	$- 10$	$28$	$17$
	\xa0	$- 2$	$6$	\xa0
	$4$	$- 12$

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

$- \frac{1}{2}$	$4$	$- 10$	$28$	$17$
	\xa0	$- 2$	$6$	\xa0
	$4$	$- 12$	$34$

Multiply the newest entry in the result $(34)$ by the divisor $(- \frac{1}{2})$ and place the result of $(- 17)$ under the next term in the dividend $(17)$ .

$- \frac{1}{2}$	$4$	$- 10$	$28$	$17$
	\xa0	$- 2$	$6$	$- 17$
	$4$	$- 12$	$34$

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

$- \frac{1}{2}$	$4$	$- 10$	$28$	$17$
	\xa0	$- 2$	$6$	$- 17$
	$4$	$- 12$	$34$	$0$

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

$4 t^{2} + - 12 t + 34$

Simplify the quotient polynomial.

$4 t^{2} - 12 t + 34$

Step 7

Factor $2$ out of $4 t^{2} - 12 t + 34$ .

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

$(t + \frac{1}{2}) (2 (2 t^{2}) - 12 t + 34)$

Factor $2$ out of $- 12 t$ .

$(t + \frac{1}{2}) (2 (2 t^{2}) + 2 (- 6 t) + 34)$

Factor $2$ out of $34$ .

$(t + \frac{1}{2}) (2 (2 t^{2}) + 2 (- 6 t) + 2 (17))$

Factor $2$ out of $2 (2 t^{2}) + 2 (- 6 t)$ .

$(t + \frac{1}{2}) (2 (2 t^{2} - 6 t) + 2 (17))$

Factor $2$ out of $2 (2 t^{2} - 6 t) + 2 (17)$ .

$(t + \frac{1}{2}) (2 (2 t^{2} - 6 t + 17))$

Step 8

Factor $4 t^{3} - 10 t^{2} + 28 t + 17$ 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.

$p = \pm 1, \pm 17 q = \pm 1, \pm 4, \pm 2$

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

$\pm 1, \pm 0.25, \pm 0.5, \pm 17, \pm 4.25, \pm 8.5$

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.

$4 {(- 0.5)}^{3} - 10 {(- 0.5)}^{2} + 28 \cdot - 0.5 + 17$

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

$4 \cdot - 0.125 - 10 {(- 0.5)}^{2} + 28 \cdot - 0.5 + 17$

Multiply $4$ by $- 0.125$ .

$- 0.5 - 10 {(- 0.5)}^{2} + 28 \cdot - 0.5 + 17$

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

$- 0.5 - 10 \cdot 0.25 + 28 \cdot - 0.5 + 17$

Multiply $- 10$ by $0.25$ .

$- 0.5 - 2.5 + 28 \cdot - 0.5 + 17$

Subtract $2.5$ from $- 0.5$ .

$- 3 + 28 \cdot - 0.5 + 17$

Multiply $28$ by $- 0.5$ .

$- 3 - 14 + 17$

Subtract $14$ from $- 3$ .

$- 17 + 17$

Add $- 17$ and $17$ .

$0$

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.

$\frac{4 t^{3} - 10 t^{2} + 28 t + 17}{2 t + 1}$

Divide $4 t^{3} - 10 t^{2} + 28 t + 17$ 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 $0$ .

$2 t$

$1$

$4 t^{3}$

$10 t^{2}$

$28 t$

$17$

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

					$2 t^{2}$
	$2 t$	+	$1$		$4 t^{3}$	-	$10 t^{2}$	+	$28 t$	+	$17$

Multiply the new quotient term by the divisor.

				$2 t^{2}$
$2 t$	+	$1$		$4 t^{3}$	-	$10 t^{2}$	+	$28 t$	+	$17$
			+	$4 t^{3}$	+	$2 t^{2}$

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

				$2 t^{2}$
$2 t$	+	$1$		$4 t^{3}$	-	$10 t^{2}$	+	$28 t$	+	$17$
			-	$4 t^{3}$	-	$2 t^{2}$

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

				$2 t^{2}$
$2 t$	+	$1$		$4 t^{3}$	-	$10 t^{2}$	+	$28 t$	+	$17$
			-	$4 t^{3}$	-	$2 t^{2}$
					-	$12 t^{2}$

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

				$2 t^{2}$
$2 t$	+	$1$		$4 t^{3}$	-	$10 t^{2}$	+	$28 t$	+	$17$
			-	$4 t^{3}$	-	$2 t^{2}$
					-	$12 t^{2}$	+	$28 t$

Divide the highest order term in the dividend $- 12 t^{2}$ by the highest order term in divisor $2 t$ .

				$2 t^{2}$	-	$6 t$
$2 t$	+	$1$		$4 t^{3}$	-	$10 t^{2}$	+	$28 t$	+	$17$
			-	$4 t^{3}$	-	$2 t^{2}$
					-	$12 t^{2}$	+	$28 t$

Multiply the new quotient term by the divisor.

				$2 t^{2}$	-	$6 t$
$2 t$	+	$1$		$4 t^{3}$	-	$10 t^{2}$	+	$28 t$	+	$17$
			-	$4 t^{3}$	-	$2 t^{2}$
					-	$12 t^{2}$	+	$28 t$
					-	$12 t^{2}$	-	$6 t$

The expression needs to be subtracted from the dividend, so change all the signs in $- 12 t^{2} - 6 t$

				$2 t^{2}$	-	$6 t$
$2 t$	+	$1$		$4 t^{3}$	-	$10 t^{2}$	+	$28 t$	+	$17$
			-	$4 t^{3}$	-	$2 t^{2}$
					-	$12 t^{2}$	+	$28 t$
					+	$12 t^{2}$	+	$6 t$

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

				$2 t^{2}$	-	$6 t$
$2 t$	+	$1$		$4 t^{3}$	-	$10 t^{2}$	+	$28 t$	+	$17$
			-	$4 t^{3}$	-	$2 t^{2}$
					-	$12 t^{2}$	+	$28 t$
					+	$12 t^{2}$	+	$6 t$
							+	$34 t$

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

				$2 t^{2}$	-	$6 t$
$2 t$	+	$1$		$4 t^{3}$	-	$10 t^{2}$	+	$28 t$	+	$17$
			-	$4 t^{3}$	-	$2 t^{2}$
					-	$12 t^{2}$	+	$28 t$
					+	$12 t^{2}$	+	$6 t$
							+	$34 t$	+	$17$

Divide the highest order term in the dividend $34 t$ by the highest order term in divisor $2 t$ .

				$2 t^{2}$	-	$6 t$	+	$17$
$2 t$	+	$1$		$4 t^{3}$	-	$10 t^{2}$	+	$28 t$	+	$17$
			-	$4 t^{3}$	-	$2 t^{2}$
					-	$12 t^{2}$	+	$28 t$
					+	$12 t^{2}$	+	$6 t$
							+	$34 t$	+	$17$

Multiply the new quotient term by the divisor.

				$2 t^{2}$	-	$6 t$	+	$17$
$2 t$	+	$1$		$4 t^{3}$	-	$10 t^{2}$	+	$28 t$	+	$17$
			-	$4 t^{3}$	-	$2 t^{2}$
					-	$12 t^{2}$	+	$28 t$
					+	$12 t^{2}$	+	$6 t$
							+	$34 t$	+	$17$
							+	$34 t$	+	$17$

The expression needs to be subtracted from the dividend, so change all the signs in $34 t + 17$

				$2 t^{2}$	-	$6 t$	+	$17$
$2 t$	+	$1$		$4 t^{3}$	-	$10 t^{2}$	+	$28 t$	+	$17$
			-	$4 t^{3}$	-	$2 t^{2}$
					-	$12 t^{2}$	+	$28 t$
					+	$12 t^{2}$	+	$6 t$
							+	$34 t$	+	$17$
							-	$34 t$	-	$17$

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

				$2 t^{2}$	-	$6 t$	+	$17$
$2 t$	+	$1$		$4 t^{3}$	-	$10 t^{2}$	+	$28 t$	+	$17$
			-	$4 t^{3}$	-	$2 t^{2}$
					-	$12 t^{2}$	+	$28 t$
					+	$12 t^{2}$	+	$6 t$
							+	$34 t$	+	$17$
							-	$34 t$	-	$17$
										$0$

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

$2 t^{2} - 6 t + 17$

Write $4 t^{3} - 10 t^{2} + 28 t + 17$ as a set of factors.

$(2 t + 1) (2 t^{2} - 6 t + 17) = 0$

Step 9

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

$2 t + 1 = 0$

$2 t^{2} - 6 t + 17 = 0$

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 $2 t^{2} - 6 t + 17$ equal to $0$ and solve for $t$ .

Set $2 t^{2} - 6 t + 17$ equal to $0$ .

$2 t^{2} - 6 t + 17 = 0$

Solve $2 t^{2} - 6 t + 17 = 0$ for $t$ .

Use the quadratic formula to find the solutions.

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

Substitute the values $a = 2$ , $b = - 6$ , and $c = 17$ into the quadratic formula and solve for $t$ .

$\frac{6 \pm \sqrt{{(- 6)}^{2} - 4 \cdot (2 \cdot 17)}}{2 \cdot 2}$

Simplify.

Simplify the numerator.

Raise $- 6$ to the power of $2$ .

$t = \frac{6 \pm \sqrt{36 - 4 \cdot 2 \cdot 17}}{2 \cdot 2}$

Multiply $- 4 \cdot 2 \cdot 17$ .

Multiply $- 4$ by $2$ .

$t = \frac{6 \pm \sqrt{36 - 8 \cdot 17}}{2 \cdot 2}$

Multiply $- 8$ by $17$ .

$t = \frac{6 \pm \sqrt{36 - 136}}{2 \cdot 2}$

Subtract $136$ from $36$ .

$t = \frac{6 \pm \sqrt{- 100}}{2 \cdot 2}$

Rewrite $- 100$ as $- 1 (100)$ .

$t = \frac{6 \pm \sqrt{- 1 \cdot 100}}{2 \cdot 2}$

Rewrite $\sqrt{- 1 (100)}$ as $\sqrt{- 1} \cdot \sqrt{100}$ .

$t = \frac{6 \pm \sqrt{- 1} \cdot \sqrt{100}}{2 \cdot 2}$

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

$t = \frac{6 \pm i \cdot \sqrt{100}}{2 \cdot 2}$

Rewrite $100$ as $10^{2}$ .

$t = \frac{6 \pm i \cdot \sqrt{10^{2}}}{2 \cdot 2}$

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

$t = \frac{6 \pm i \cdot 10}{2 \cdot 2}$

Move $10$ to the left of $i$ .

$t = \frac{6 \pm 10 i}{2 \cdot 2}$

Multiply $2$ by $2$ .

$t = \frac{6 \pm 10 i}{4}$

Simplify $\frac{6 \pm 10 i}{4}$ .

$t = \frac{3 \pm 5 i}{2}$

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

Simplify the numerator.

Raise $- 6$ to the power of $2$ .

$t = \frac{6 \pm \sqrt{36 - 4 \cdot 2 \cdot 17}}{2 \cdot 2}$

Multiply $- 4 \cdot 2 \cdot 17$ .

Multiply $- 4$ by $2$ .

$t = \frac{6 \pm \sqrt{36 - 8 \cdot 17}}{2 \cdot 2}$

Multiply $- 8$ by $17$ .

$t = \frac{6 \pm \sqrt{36 - 136}}{2 \cdot 2}$

Subtract $136$ from $36$ .

$t = \frac{6 \pm \sqrt{- 100}}{2 \cdot 2}$

Rewrite $- 100$ as $- 1 (100)$ .

$t = \frac{6 \pm \sqrt{- 1 \cdot 100}}{2 \cdot 2}$

Rewrite $\sqrt{- 1 (100)}$ as $\sqrt{- 1} \cdot \sqrt{100}$ .

$t = \frac{6 \pm \sqrt{- 1} \cdot \sqrt{100}}{2 \cdot 2}$

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

$t = \frac{6 \pm i \cdot \sqrt{100}}{2 \cdot 2}$

Rewrite $100$ as $10^{2}$ .

$t = \frac{6 \pm i \cdot \sqrt{10^{2}}}{2 \cdot 2}$

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

$t = \frac{6 \pm i \cdot 10}{2 \cdot 2}$

Move $10$ to the left of $i$ .

$t = \frac{6 \pm 10 i}{2 \cdot 2}$

Multiply $2$ by $2$ .

$t = \frac{6 \pm 10 i}{4}$

Simplify $\frac{6 \pm 10 i}{4}$ .

$t = \frac{3 \pm 5 i}{2}$

Change the $\pm$ to $+$ .

$t = \frac{3 + 5 i}{2}$

Split the fraction $\frac{3 + 5 i}{2}$ into two fractions.

$t = \frac{3}{2} + \frac{5 i}{2}$

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

Simplify the numerator.

Raise $- 6$ to the power of $2$ .

$t = \frac{6 \pm \sqrt{36 - 4 \cdot 2 \cdot 17}}{2 \cdot 2}$

Multiply $- 4 \cdot 2 \cdot 17$ .

Multiply $- 4$ by $2$ .

$t = \frac{6 \pm \sqrt{36 - 8 \cdot 17}}{2 \cdot 2}$

Multiply $- 8$ by $17$ .

$t = \frac{6 \pm \sqrt{36 - 136}}{2 \cdot 2}$

Subtract $136$ from $36$ .

$t = \frac{6 \pm \sqrt{- 100}}{2 \cdot 2}$

Rewrite $- 100$ as $- 1 (100)$ .

$t = \frac{6 \pm \sqrt{- 1 \cdot 100}}{2 \cdot 2}$

Rewrite $\sqrt{- 1 (100)}$ as $\sqrt{- 1} \cdot \sqrt{100}$ .

$t = \frac{6 \pm \sqrt{- 1} \cdot \sqrt{100}}{2 \cdot 2}$

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

$t = \frac{6 \pm i \cdot \sqrt{100}}{2 \cdot 2}$

Rewrite $100$ as $10^{2}$ .

$t = \frac{6 \pm i \cdot \sqrt{10^{2}}}{2 \cdot 2}$

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

$t = \frac{6 \pm i \cdot 10}{2 \cdot 2}$

Move $10$ to the left of $i$ .

$t = \frac{6 \pm 10 i}{2 \cdot 2}$

Multiply $2$ by $2$ .

$t = \frac{6 \pm 10 i}{4}$

Simplify $\frac{6 \pm 10 i}{4}$ .

$t = \frac{3 \pm 5 i}{2}$

Change the $\pm$ to $-$ .

$t = \frac{3 - 5 i}{2}$

Split the fraction $\frac{3 - 5 i}{2}$ into two fractions.

$t = \frac{3}{2} + \frac{- 5 i}{2}$

Move the negative in front of the fraction.

$t = \frac{3}{2} - \frac{5 i}{2}$

The final answer is the combination of both solutions.

$t = \frac{3}{2} + \frac{5 i}{2}, \frac{3}{2} - \frac{5 i}{2}$

Step 12

The final solution is all the values that make $(2 t + 1) (2 t^{2} - 6 t + 17) = 0$ true.

$t = - \frac{1}{2}, \frac{3}{2} + \frac{5 i}{2}, \frac{3}{2} - \frac{5 i}{2}$

Step 13

Find the Roots/Zeros Using the Rational Roots Test f(t)=4t^3-10t^2+28t+17

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