Find the Roots/Zeros Using the Rational Roots Test f(t)=25t^3-65t^2+51t+13

$f (t) = 25 t^{3} - 65 t^{2} + 51 t + 13$

Step 1

Factor $25 t^{3} - 65 t^{2} + 51 t + 13$ 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.

$p = \pm 1, \pm 13 q = \pm 1, \pm 25, \pm 5$

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

$\pm 1, \pm 0.04, \pm 0.2, \pm 13, \pm 0.52, \pm 2.6$

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.

$25 {(- 0.2)}^{3} - 65 {(- 0.2)}^{2} + 51 \cdot - 0.2 + 13$

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

$25 \cdot - 0.008 - 65 {(- 0.2)}^{2} + 51 \cdot - 0.2 + 13$

Multiply $25$ by $- 0.008$ .

$- 0.2 - 65 {(- 0.2)}^{2} + 51 \cdot - 0.2 + 13$

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

$- 0.2 - 65 \cdot 0.04 + 51 \cdot - 0.2 + 13$

Multiply $- 65$ by $0.04$ .

$- 0.2 - 2.6 + 51 \cdot - 0.2 + 13$

Subtract $2.6$ from $- 0.2$ .

$- 2.8 + 51 \cdot - 0.2 + 13$

Multiply $51$ by $- 0.2$ .

$- 2.8 - 10.2 + 13$

Subtract $10.2$ from $- 2.8$ .

$- 13 + 13$

Add $- 13$ and $13$ .

$0$

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

$\frac{25 t^{3} - 65 t^{2} + 51 t + 13}{5 t + 1}$

Divide $25 t^{3} - 65 t^{2} + 51 t + 13$ by $5 t + 1$ .

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

$5 t$

$1$

$25 t^{3}$

$65 t^{2}$

$51 t$

$13$

Divide the highest order term in the dividend $25 t^{3}$ by the highest order term in divisor $5 t$ .

					$5 t^{2}$
	$5 t$	+	$1$		$25 t^{3}$	-	$65 t^{2}$	+	$51 t$	+	$13$

Multiply the new quotient term by the divisor.

				$5 t^{2}$
$5 t$	+	$1$		$25 t^{3}$	-	$65 t^{2}$	+	$51 t$	+	$13$
			+	$25 t^{3}$	+	$5 t^{2}$

The expression needs to be subtracted from the dividend, so change all the signs in $25 t^{3} + 5 t^{2}$

				$5 t^{2}$
$5 t$	+	$1$		$25 t^{3}$	-	$65 t^{2}$	+	$51 t$	+	$13$
			-	$25 t^{3}$	-	$5 t^{2}$

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

				$5 t^{2}$
$5 t$	+	$1$		$25 t^{3}$	-	$65 t^{2}$	+	$51 t$	+	$13$
			-	$25 t^{3}$	-	$5 t^{2}$
					-	$70 t^{2}$

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

				$5 t^{2}$
$5 t$	+	$1$		$25 t^{3}$	-	$65 t^{2}$	+	$51 t$	+	$13$
			-	$25 t^{3}$	-	$5 t^{2}$
					-	$70 t^{2}$	+	$51 t$

Divide the highest order term in the dividend $- 70 t^{2}$ by the highest order term in divisor $5 t$ .

				$5 t^{2}$	-	$14 t$
$5 t$	+	$1$		$25 t^{3}$	-	$65 t^{2}$	+	$51 t$	+	$13$
			-	$25 t^{3}$	-	$5 t^{2}$
					-	$70 t^{2}$	+	$51 t$

Multiply the new quotient term by the divisor.

				$5 t^{2}$	-	$14 t$
$5 t$	+	$1$		$25 t^{3}$	-	$65 t^{2}$	+	$51 t$	+	$13$
			-	$25 t^{3}$	-	$5 t^{2}$
					-	$70 t^{2}$	+	$51 t$
					-	$70 t^{2}$	-	$14 t$

The expression needs to be subtracted from the dividend, so change all the signs in $- 70 t^{2} - 14 t$

				$5 t^{2}$	-	$14 t$
$5 t$	+	$1$		$25 t^{3}$	-	$65 t^{2}$	+	$51 t$	+	$13$
			-	$25 t^{3}$	-	$5 t^{2}$
					-	$70 t^{2}$	+	$51 t$
					+	$70 t^{2}$	+	$14 t$

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

				$5 t^{2}$	-	$14 t$
$5 t$	+	$1$		$25 t^{3}$	-	$65 t^{2}$	+	$51 t$	+	$13$
			-	$25 t^{3}$	-	$5 t^{2}$
					-	$70 t^{2}$	+	$51 t$
					+	$70 t^{2}$	+	$14 t$
							+	$65 t$

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

				$5 t^{2}$	-	$14 t$
$5 t$	+	$1$		$25 t^{3}$	-	$65 t^{2}$	+	$51 t$	+	$13$
			-	$25 t^{3}$	-	$5 t^{2}$
					-	$70 t^{2}$	+	$51 t$
					+	$70 t^{2}$	+	$14 t$
							+	$65 t$	+	$13$

Divide the highest order term in the dividend $65 t$ by the highest order term in divisor $5 t$ .

				$5 t^{2}$	-	$14 t$	+	$13$
$5 t$	+	$1$		$25 t^{3}$	-	$65 t^{2}$	+	$51 t$	+	$13$
			-	$25 t^{3}$	-	$5 t^{2}$
					-	$70 t^{2}$	+	$51 t$
					+	$70 t^{2}$	+	$14 t$
							+	$65 t$	+	$13$

Multiply the new quotient term by the divisor.

				$5 t^{2}$	-	$14 t$	+	$13$
$5 t$	+	$1$		$25 t^{3}$	-	$65 t^{2}$	+	$51 t$	+	$13$
			-	$25 t^{3}$	-	$5 t^{2}$
					-	$70 t^{2}$	+	$51 t$
					+	$70 t^{2}$	+	$14 t$
							+	$65 t$	+	$13$
							+	$65 t$	+	$13$

The expression needs to be subtracted from the dividend, so change all the signs in $65 t + 13$

				$5 t^{2}$	-	$14 t$	+	$13$
$5 t$	+	$1$		$25 t^{3}$	-	$65 t^{2}$	+	$51 t$	+	$13$
			-	$25 t^{3}$	-	$5 t^{2}$
					-	$70 t^{2}$	+	$51 t$
					+	$70 t^{2}$	+	$14 t$
							+	$65 t$	+	$13$
							-	$65 t$	-	$13$

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

				$5 t^{2}$	-	$14 t$	+	$13$
$5 t$	+	$1$		$25 t^{3}$	-	$65 t^{2}$	+	$51 t$	+	$13$
			-	$25 t^{3}$	-	$5 t^{2}$
					-	$70 t^{2}$	+	$51 t$
					+	$70 t^{2}$	+	$14 t$
							+	$65 t$	+	$13$
							-	$65 t$	-	$13$
										$0$

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

$5 t^{2} - 14 t + 13$

Write $25 t^{3} - 65 t^{2} + 51 t + 13$ as a set of factors.

$(5 t + 1) (5 t^{2} - 14 t + 13) = 0$

Step 2

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

$5 t + 1 = 0$

$5 t^{2} - 14 t + 13 = 0$

Step 3

Set $5 t + 1$ equal to $0$ and solve for $t$ .

Set $5 t + 1$ equal to $0$ .

$5 t + 1 = 0$

Solve $5 t + 1 = 0$ for $t$ .

Subtract $1$ from both sides of the equation.

$5 t = - 1$

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

Divide each term in $5 t = - 1$ by $5$ .

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

Simplify the left side.

Cancel the common factor of $5$ .

Cancel the common factor.

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

Divide $t$ by $1$ .

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

Simplify the right side.

Move the negative in front of the fraction.

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

Step 4

Set $5 t^{2} - 14 t + 13$ equal to $0$ and solve for $t$ .

Set $5 t^{2} - 14 t + 13$ equal to $0$ .

$5 t^{2} - 14 t + 13 = 0$

Solve $5 t^{2} - 14 t + 13 = 0$ for $t$ .

Use the quadratic formula to find the solutions.

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

Substitute the values $a = 5$ , $b = - 14$ , and $c = 13$ into the quadratic formula and solve for $t$ .

$\frac{14 \pm \sqrt{{(- 14)}^{2} - 4 \cdot (5 \cdot 13)}}{2 \cdot 5}$

Simplify.

Simplify the numerator.

Raise $- 14$ to the power of $2$ .

$t = \frac{14 \pm \sqrt{196 - 4 \cdot 5 \cdot 13}}{2 \cdot 5}$

Multiply $- 4 \cdot 5 \cdot 13$ .

Multiply $- 4$ by $5$ .

$t = \frac{14 \pm \sqrt{196 - 20 \cdot 13}}{2 \cdot 5}$

Multiply $- 20$ by $13$ .

$t = \frac{14 \pm \sqrt{196 - 260}}{2 \cdot 5}$

Subtract $260$ from $196$ .

$t = \frac{14 \pm \sqrt{- 64}}{2 \cdot 5}$

Rewrite $- 64$ as $- 1 (64)$ .

$t = \frac{14 \pm \sqrt{- 1 \cdot 64}}{2 \cdot 5}$

Rewrite $\sqrt{- 1 (64)}$ as $\sqrt{- 1} \cdot \sqrt{64}$ .

$t = \frac{14 \pm \sqrt{- 1} \cdot \sqrt{64}}{2 \cdot 5}$

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

$t = \frac{14 \pm i \cdot \sqrt{64}}{2 \cdot 5}$

Rewrite $64$ as $8^{2}$ .

$t = \frac{14 \pm i \cdot \sqrt{8^{2}}}{2 \cdot 5}$

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

$t = \frac{14 \pm i \cdot 8}{2 \cdot 5}$

Move $8$ to the left of $i$ .

$t = \frac{14 \pm 8 i}{2 \cdot 5}$

Multiply $2$ by $5$ .

$t = \frac{14 \pm 8 i}{10}$

Simplify $\frac{14 \pm 8 i}{10}$ .

$t = \frac{7 \pm 4 i}{5}$

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

Simplify the numerator.

Raise $- 14$ to the power of $2$ .

$t = \frac{14 \pm \sqrt{196 - 4 \cdot 5 \cdot 13}}{2 \cdot 5}$

Multiply $- 4 \cdot 5 \cdot 13$ .

Multiply $- 4$ by $5$ .

$t = \frac{14 \pm \sqrt{196 - 20 \cdot 13}}{2 \cdot 5}$

Multiply $- 20$ by $13$ .

$t = \frac{14 \pm \sqrt{196 - 260}}{2 \cdot 5}$

Subtract $260$ from $196$ .

$t = \frac{14 \pm \sqrt{- 64}}{2 \cdot 5}$

Rewrite $- 64$ as $- 1 (64)$ .

$t = \frac{14 \pm \sqrt{- 1 \cdot 64}}{2 \cdot 5}$

Rewrite $\sqrt{- 1 (64)}$ as $\sqrt{- 1} \cdot \sqrt{64}$ .

$t = \frac{14 \pm \sqrt{- 1} \cdot \sqrt{64}}{2 \cdot 5}$

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

$t = \frac{14 \pm i \cdot \sqrt{64}}{2 \cdot 5}$

Rewrite $64$ as $8^{2}$ .

$t = \frac{14 \pm i \cdot \sqrt{8^{2}}}{2 \cdot 5}$

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

$t = \frac{14 \pm i \cdot 8}{2 \cdot 5}$

Move $8$ to the left of $i$ .

$t = \frac{14 \pm 8 i}{2 \cdot 5}$

Multiply $2$ by $5$ .

$t = \frac{14 \pm 8 i}{10}$

Simplify $\frac{14 \pm 8 i}{10}$ .

$t = \frac{7 \pm 4 i}{5}$

Change the $\pm$ to $+$ .

$t = \frac{7 + 4 i}{5}$

Split the fraction $\frac{7 + 4 i}{5}$ into two fractions.

$t = \frac{7}{5} + \frac{4 i}{5}$

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

Simplify the numerator.

Raise $- 14$ to the power of $2$ .

$t = \frac{14 \pm \sqrt{196 - 4 \cdot 5 \cdot 13}}{2 \cdot 5}$

Multiply $- 4 \cdot 5 \cdot 13$ .

Multiply $- 4$ by $5$ .

$t = \frac{14 \pm \sqrt{196 - 20 \cdot 13}}{2 \cdot 5}$

Multiply $- 20$ by $13$ .

$t = \frac{14 \pm \sqrt{196 - 260}}{2 \cdot 5}$

Subtract $260$ from $196$ .

$t = \frac{14 \pm \sqrt{- 64}}{2 \cdot 5}$

Rewrite $- 64$ as $- 1 (64)$ .

$t = \frac{14 \pm \sqrt{- 1 \cdot 64}}{2 \cdot 5}$

Rewrite $\sqrt{- 1 (64)}$ as $\sqrt{- 1} \cdot \sqrt{64}$ .

$t = \frac{14 \pm \sqrt{- 1} \cdot \sqrt{64}}{2 \cdot 5}$

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

$t = \frac{14 \pm i \cdot \sqrt{64}}{2 \cdot 5}$

Rewrite $64$ as $8^{2}$ .

$t = \frac{14 \pm i \cdot \sqrt{8^{2}}}{2 \cdot 5}$

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

$t = \frac{14 \pm i \cdot 8}{2 \cdot 5}$

Move $8$ to the left of $i$ .

$t = \frac{14 \pm 8 i}{2 \cdot 5}$

Multiply $2$ by $5$ .

$t = \frac{14 \pm 8 i}{10}$

Simplify $\frac{14 \pm 8 i}{10}$ .

$t = \frac{7 \pm 4 i}{5}$

Change the $\pm$ to $-$ .

$t = \frac{7 - 4 i}{5}$

Split the fraction $\frac{7 - 4 i}{5}$ into two fractions.

$t = \frac{7}{5} + \frac{- 4 i}{5}$

Move the negative in front of the fraction.

$t = \frac{7}{5} - \frac{4 i}{5}$

The final answer is the combination of both solutions.

$t = \frac{7}{5} + \frac{4 i}{5}, \frac{7}{5} - \frac{4 i}{5}$

Step 5

The final solution is all the values that make $(5 t + 1) (5 t^{2} - 14 t + 13) = 0$ true.

$t = - \frac{1}{5}, \frac{7}{5} + \frac{4 i}{5}, \frac{7}{5} - \frac{4 i}{5}$

Step 6

Find the Roots/Zeros Using the Rational Roots Test f(t)=25t^3-65t^2+51t+13

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