Find the Roots/Zeros Using the Rational Roots Test t^4-t^3+3t^2-5t-10

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

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.

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

$q = \pm 1$

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

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

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.

${(- 1)}^{4} - {(- 1)}^{3} + 3 {(- 1)}^{2} - 5 \cdot - 1 - 10$

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

Simplify each term.

Raise $- 1$ to the power of $4$ .

$1 - {(- 1)}^{3} + 3 {(- 1)}^{2} - 5 \cdot - 1 - 10$

Multiply $- 1$ by ${(- 1)}^{3}$ by adding the exponents.

Multiply $- 1$ by ${(- 1)}^{3}$ .

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

$1 + {(- 1)}^{1} {(- 1)}^{3} + 3 {(- 1)}^{2} - 5 \cdot - 1 - 10$

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$1 + {(- 1)}^{1 + 3} + 3 {(- 1)}^{2} - 5 \cdot - 1 - 10$

Add $1$ and $3$ .

$1 + {(- 1)}^{4} + 3 {(- 1)}^{2} - 5 \cdot - 1 - 10$

Raise $- 1$ to the power of $4$ .

$1 + 1 + 3 {(- 1)}^{2} - 5 \cdot - 1 - 10$

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

$1 + 1 + 3 \cdot 1 - 5 \cdot - 1 - 10$

Multiply $3$ by $1$ .

$1 + 1 + 3 - 5 \cdot - 1 - 10$

Multiply $- 5$ by $- 1$ .

$1 + 1 + 3 + 5 - 10$

Simplify by adding and subtracting.

Add $1$ and $1$ .

$2 + 3 + 5 - 10$

Add $2$ and $3$ .

$5 + 5 - 10$

Add $5$ and $5$ .

$10 - 10$

Subtract $10$ from $10$ .

$0$

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

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

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.

$- 1$	$1$	$- 1$	$3$	$- 5$	$- 10$
	\xa0				\xa0

The first number in the dividend $(1)$ is put into the first position of the result area (below the horizontal line).

$- 1$	$1$	$- 1$	$3$	$- 5$	$- 10$
	\xa0				\xa0
	$1$

Multiply the newest entry in the result $(1)$ by the divisor $(- 1)$ and place the result of $(- 1)$ under the next term in the dividend $(- 1)$ .

$- 1$	$1$	$- 1$	$3$	$- 5$	$- 10$
	\xa0	$- 1$			\xa0
	$1$

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

$- 1$	$1$	$- 1$	$3$	$- 5$	$- 10$
	\xa0	$- 1$			\xa0
	$1$	$- 2$

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

$- 1$	$1$	$- 1$	$3$	$- 5$	$- 10$
	\xa0	$- 1$	$2$		\xa0
	$1$	$- 2$

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

$- 1$	$1$	$- 1$	$3$	$- 5$	$- 10$
	\xa0	$- 1$	$2$		\xa0
	$1$	$- 2$	$5$

Multiply the newest entry in the result $(5)$ by the divisor $(- 1)$ and place the result of $(- 5)$ under the next term in the dividend $(- 5)$ .

$- 1$	$1$	$- 1$	$3$	$- 5$	$- 10$
	\xa0	$- 1$	$2$	$- 5$	\xa0
	$1$	$- 2$	$5$

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

$- 1$	$1$	$- 1$	$3$	$- 5$	$- 10$
	\xa0	$- 1$	$2$	$- 5$	\xa0
	$1$	$- 2$	$5$	$- 10$

Multiply the newest entry in the result $(- 10)$ by the divisor $(- 1)$ and place the result of $(10)$ under the next term in the dividend $(- 10)$ .

$- 1$	$1$	$- 1$	$3$	$- 5$	$- 10$
	\xa0	$- 1$	$2$	$- 5$	$10$
	$1$	$- 2$	$5$	$- 10$

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

$- 1$	$1$	$- 1$	$3$	$- 5$	$- 10$
	\xa0	$- 1$	$2$	$- 5$	$10$
	$1$	$- 2$	$5$	$- 10$	$0$

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

$1 t^{3} + - 2 t^{2} + (5) t - 10$

Simplify the quotient polynomial.

$t^{3} - 2 t^{2} + 5 t - 10$

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

$(t + 1) (t^{3} - 2 t^{2}) + 5 t - 10$

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

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

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

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

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

Factor the left side of the equation.

Regroup terms.

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

Factor $- t$ out of $- t^{3} - 5 t$ .

Factor $- t$ out of $- t^{3}$ .

$- t \cdot t^{2} - 5 t + t^{4} + 3 t^{2} - 10 = 0$

Factor $- t$ out of $- 5 t$ .

$- t \cdot t^{2} - t \cdot 5 + t^{4} + 3 t^{2} - 10 = 0$

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

$- t (t^{2} + 5) + t^{4} + 3 t^{2} - 10 = 0$

Rewrite $t^{4}$ as ${(t^{2})}^{2}$ .

$- t (t^{2} + 5) + {(t^{2})}^{2} + 3 t^{2} - 10 = 0$

Let $u = t^{2}$ . Substitute $u$ for all occurrences of $t^{2}$ .

$- t (t^{2} + 5) + u^{2} + 3 u - 10 = 0$

Factor $u^{2} + 3 u - 10$ using the AC method.

Consider the form $x^{2} + b x + c$ . Find a pair of integers whose product is $c$ and whose sum is $b$ . In this case, whose product is $- 10$ and whose sum is $3$ .

$- 2, 5$

Write the factored form using these integers.

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

Replace all occurrences of $u$ with $t^{2}$ .

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

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

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

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

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

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

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

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

Let $u = t$ . Substitute $u$ for all occurrences of $t$ .

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

Factor $- u + u^{2} - 2$ using the AC method.

Consider the form $x^{2} + b x + c$ . Find a pair of integers whose product is $c$ and whose sum is $b$ . In this case, whose product is $- 2$ and whose sum is $- 1$ .

$- 2, 1$

Write the factored form using these integers.

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

Factor.

Replace all occurrences of $u$ with $t$ .

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

Remove unnecessary parentheses.

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

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

$t^{2} + 5 = 0$

$t - 2 = 0$

$t + 1 = 0$

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

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

$t^{2} + 5 = 0$

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

Subtract $5$ from both sides of the equation.

$t^{2} = - 5$

Take the square root of both sides of the equation to eliminate the exponent on the left side.

$t = \pm \sqrt{- 5}$

Simplify $\pm \sqrt{- 5}$ .

Rewrite $- 5$ as $- 1 (5)$ .

$t = \pm \sqrt{- 1 (5)}$

Rewrite $\sqrt{- 1 (5)}$ as $\sqrt{- 1} \cdot \sqrt{5}$ .

$t = \pm \sqrt{- 1} \cdot \sqrt{5}$

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

$t = \pm i \sqrt{5}$

The complete solution is the result of both the positive and negative portions of the solution.

First, use the positive value of the $\pm$ to find the first solution.

$t = i \sqrt{5}$

Next, use the negative value of the $\pm$ to find the second solution.

$t = - i \sqrt{5}$

The complete solution is the result of both the positive and negative portions of the solution.

$t = i \sqrt{5}, - i \sqrt{5}$

Set $t - 2$ equal to $0$ and solve for $t$ .

Set $t - 2$ equal to $0$ .

$t - 2 = 0$

Add $2$ to both sides of the equation.

$t = 2$

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

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

$t + 1 = 0$

Subtract $1$ from both sides of the equation.

$t = - 1$

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

$t = i \sqrt{5}, - i \sqrt{5}, 2, - 1$

Find the Roots/Zeros Using the Rational Roots Test t^4-t^3+3t^2-5t-10

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