Finite Math Examples

$t^{4} - 2 t^{3} + 2 t^{2} - 10 t - 15$

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.

$p = \pm 1, \pm 3, \pm 5, \pm 15$

$q = \pm 1$

Step 2

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

$\pm 1, \pm 3, \pm 5, \pm 15$

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.

${(- 1)}^{4} - 2 {(- 1)}^{3} + 2 {(- 1)}^{2} - 10 \cdot - 1 - 15$

Step 4

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

Tap for more steps...

Step 4.1

Simplify each term.

Tap for more steps...

Step 4.1.1

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

$1 - 2 {(- 1)}^{3} + 2 {(- 1)}^{2} - 10 \cdot - 1 - 15$

Step 4.1.2

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

$1 - 2 \cdot - 1 + 2 {(- 1)}^{2} - 10 \cdot - 1 - 15$

Step 4.1.3

Multiply $- 2$ by $- 1$ .

$1 + 2 + 2 {(- 1)}^{2} - 10 \cdot - 1 - 15$

Step 4.1.4

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

$1 + 2 + 2 \cdot 1 - 10 \cdot - 1 - 15$

Step 4.1.5

Multiply $2$ by $1$ .

$1 + 2 + 2 - 10 \cdot - 1 - 15$

Step 4.1.6

Multiply $- 10$ by $- 1$ .

$1 + 2 + 2 + 10 - 15$

Step 4.2

Simplify by adding and subtracting.

Tap for more steps...

Step 4.2.1

Add $1$ and $2$ .

$3 + 2 + 10 - 15$

Step 4.2.2

Add $3$ and $2$ .

$5 + 10 - 15$

Step 4.2.3

Add $5$ and $10$ .

$15 - 15$

Step 4.2.4

Subtract $15$ from $15$ .

$0$

Step 5

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.

$\frac{t^{4} - 2 t^{3} + 2 t^{2} - 10 t - 15}{t + 1}$

Step 6

Next, find the roots of the remaining polynomial. The order of the polynomial has been reduced by $1$ .

Tap for more steps...

Step 6.1

Place the numbers representing the divisor and the dividend into a division-like configuration.

$- 1$	$1$	$- 2$	$2$	$- 10$	$- 15$

Step 6.2

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

$- 1$	$1$	$- 2$	$2$	$- 10$	$- 15$

	$1$

Step 6.3

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 $(- 2)$ .

$- 1$	$1$	$- 2$	$2$	$- 10$	$- 15$
		$- 1$
	$1$

Step 6.4

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

$- 1$	$1$	$- 2$	$2$	$- 10$	$- 15$
		$- 1$
	$1$	$- 3$

Step 6.5

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

$- 1$	$1$	$- 2$	$2$	$- 10$	$- 15$
		$- 1$	$3$
	$1$	$- 3$

Step 6.6

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

$- 1$	$1$	$- 2$	$2$	$- 10$	$- 15$
		$- 1$	$3$
	$1$	$- 3$	$5$

Step 6.7

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 $(- 10)$ .

$- 1$	$1$	$- 2$	$2$	$- 10$	$- 15$
		$- 1$	$3$	$- 5$
	$1$	$- 3$	$5$

Step 6.8

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

$- 1$	$1$	$- 2$	$2$	$- 10$	$- 15$
		$- 1$	$3$	$- 5$
	$1$	$- 3$	$5$	$- 15$

Step 6.9

Multiply the newest entry in the result $(- 15)$ by the divisor $(- 1)$ and place the result of $(15)$ under the next term in the dividend $(- 15)$ .

$- 1$	$1$	$- 2$	$2$	$- 10$	$- 15$
		$- 1$	$3$	$- 5$	$15$
	$1$	$- 3$	$5$	$- 15$

Step 6.10

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

$- 1$	$1$	$- 2$	$2$	$- 10$	$- 15$
		$- 1$	$3$	$- 5$	$15$
	$1$	$- 3$	$5$	$- 15$	$0$

Step 6.11

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

$1 t^{3} + - 3 t^{2} + (5) t - 15$

Step 6.12

Simplify the quotient polynomial.

$t^{3} - 3 t^{2} + 5 t - 15$

Step 7

Factor out the greatest common factor from each group.

Tap for more steps...

Step 7.1

Group the first two terms and the last two terms.

$(t + 1) (t^{3} - 3 t^{2}) + 5 t - 15$

Step 7.2

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

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

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

Step 8

Factor the polynomial by factoring out the greatest common factor, $t - 3$ .

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

Step 9

Factor the left side of the equation.

Tap for more steps...

Step 9.1

Regroup terms.

$- 2 t^{3} - 10 t + t^{4} + 2 t^{2} - 15 = 0$

Step 9.2

Factor $- 2 t$ out of $- 2 t^{3} - 10 t$ .

Tap for more steps...

Step 9.2.1

Factor $- 2 t$ out of $- 2 t^{3}$ .

$- 2 t \cdot t^{2} - 10 t + t^{4} + 2 t^{2} - 15 = 0$

Step 9.2.2

Factor $- 2 t$ out of $- 10 t$ .

$- 2 t \cdot t^{2} - 2 t \cdot 5 + t^{4} + 2 t^{2} - 15 = 0$

Step 9.2.3

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

$- 2 t (t^{2} + 5) + t^{4} + 2 t^{2} - 15 = 0$

Step 9.3

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

$- 2 t (t^{2} + 5) + {(t^{2})}^{2} + 2 t^{2} - 15 = 0$

Step 9.4

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

$- 2 t (t^{2} + 5) + u^{2} + 2 u - 15 = 0$

Step 9.5

Factor $u^{2} + 2 u - 15$ using the AC method.

Tap for more steps...

Step 9.5.1

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 $- 15$ and whose sum is $2$ .

$- 3, 5$

Step 9.5.2

Write the factored form using these integers.

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

Step 9.6

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

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

Step 9.7

Factor $t^{2} + 5$ out of $- 2 t (t^{2} + 5) + (t^{2} - 3) (t^{2} + 5)$ .

Tap for more steps...

Step 9.7.1

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

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

Step 9.7.2

Factor $t^{2} + 5$ out of $(t^{2} - 3) (t^{2} + 5)$ .

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

Step 9.7.3

Factor $t^{2} + 5$ out of $(t^{2} + 5) (- 2 t) + (t^{2} + 5) (t^{2} - 3)$ .

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

Step 9.8

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

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

Step 9.9

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

Tap for more steps...

Step 9.9.1

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 $- 3$ and whose sum is $- 2$ .

$- 3, 1$

Step 9.9.2

Write the factored form using these integers.

$(t^{2} + 5) ((u - 3) (u + 1)) = 0$

Step 9.10

Factor.

Tap for more steps...

Step 9.10.1

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

$(t^{2} + 5) ((t - 3) (t + 1)) = 0$

Step 9.10.2

Remove unnecessary parentheses.

$(t^{2} + 5) (t - 3) (t + 1) = 0$

Step 10

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

$t^{2} + 5 = 0$

$t - 3 = 0$

$t + 1 = 0$

Step 11

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

Tap for more steps...

Step 11.1

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

$t^{2} + 5 = 0$

Step 11.2

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

Tap for more steps...

Step 11.2.1

Subtract $5$ from both sides of the equation.

$t^{2} = - 5$

Step 11.2.2

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

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

Step 11.2.3

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

Tap for more steps...

Step 11.2.3.1

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

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

Step 11.2.3.2

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

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

Step 11.2.3.3

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

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

Step 11.2.4

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

Tap for more steps...

Step 11.2.4.1

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

$t = i \sqrt{5}$

Step 11.2.4.2

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

$t = - i \sqrt{5}$

Step 11.2.4.3

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

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

Step 12

Set $t - 3$ equal to $0$ and solve for $t$ .

Tap for more steps...

Step 12.1

Set $t - 3$ equal to $0$ .

$t - 3 = 0$

Step 12.2

Add $3$ to both sides of the equation.

$t = 3$

Step 13

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

Tap for more steps...

Step 13.1

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

$t + 1 = 0$

Step 13.2

Subtract $1$ from both sides of the equation.

$t = - 1$

Step 14

The final solution is all the values that make $(t^{2} + 5) (t - 3) (t + 1) = 0$ true.

$t = i \sqrt{5}, - i \sqrt{5}, 3, - 1$

Step 15