Finite Math Examples

$\frac{k (k + 1) (2 k + 1)}{6} + (k + 1) (k + 1)$

Step 1

Expand $(k + 1) (k + 1)$ using the FOIL Method.

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Step 1.1

Apply the distributive property.

$\frac{k (k + 1) (2 k + 1)}{6} + k (k + 1) + 1 (k + 1)$

Step 1.2

Apply the distributive property.

$\frac{k (k + 1) (2 k + 1)}{6} + k \cdot k + k \cdot 1 + 1 (k + 1)$

Step 1.3

Apply the distributive property.

$\frac{k (k + 1) (2 k + 1)}{6} + k \cdot k + k \cdot 1 + 1 k + 1 \cdot 1$

Step 2

Simplify and combine like terms.

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Step 2.1

Simplify each term.

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Step 2.1.1

Multiply $k$ by $k$ .

$\frac{k (k + 1) (2 k + 1)}{6} + k^{2} + k \cdot 1 + 1 k + 1 \cdot 1$

Step 2.1.2

Multiply $k$ by $1$ .

$\frac{k (k + 1) (2 k + 1)}{6} + k^{2} + k + 1 k + 1 \cdot 1$

Step 2.1.3

Multiply $k$ by $1$ .

$\frac{k (k + 1) (2 k + 1)}{6} + k^{2} + k + k + 1 \cdot 1$

Step 2.1.4

Multiply $1$ by $1$ .

$\frac{k (k + 1) (2 k + 1)}{6} + k^{2} + k + k + 1$

Step 2.2

Add $k$ and $k$ .

$\frac{k (k + 1) (2 k + 1)}{6} + k^{2} + 2 k + 1$

Step 3

To write $k^{2}$ as a fraction with a common denominator, multiply by $\frac{6}{6}$ .

$\frac{k (k + 1) (2 k + 1)}{6} + k^{2} \cdot \frac{6}{6} + 2 k + 1$

Step 4

Combine $k^{2}$ and $\frac{6}{6}$ .

$\frac{k (k + 1) (2 k + 1)}{6} + \frac{k^{2} \cdot 6}{6} + 2 k + 1$

Step 5

Combine the numerators over the common denominator.

$\frac{k (k + 1) (2 k + 1) + k^{2} \cdot 6}{6} + 2 k + 1$

Step 6

Reorder terms.

$2 k + \frac{k (k + 1) (2 k + 1) + k^{2} \cdot 6}{6} + 1$

Step 7

Factor $\frac{1}{6}$ out of each term.

$\frac{1}{6} (12 k + (k (k + 1)) (2 k + 1) + k^{2} \cdot 6 + 6)$

Step 8

Apply the distributive property.

$\frac{1}{6} \cdot (12 k + (k \cdot k + k \cdot 1) (2 k + 1) + k^{2} \cdot 6 + 6)$

Step 9

Multiply $k$ by $k$ .

$\frac{1}{6} \cdot (12 k + (k^{2} + k \cdot 1) (2 k + 1) + k^{2} \cdot 6 + 6)$

Step 10

Multiply $k$ by $1$ .

$\frac{1}{6} \cdot (12 k + (k^{2} + k) (2 k + 1) + k^{2} \cdot 6 + 6)$

Step 11

Expand $(k^{2} + k) (2 k + 1)$ using the FOIL Method.

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Step 11.1

Apply the distributive property.

$\frac{1}{6} \cdot (12 k + k^{2} (2 k + 1) + k (2 k + 1) + k^{2} \cdot 6 + 6)$

Step 11.2

Apply the distributive property.

$\frac{1}{6} \cdot (12 k + k^{2} (2 k) + k^{2} \cdot 1 + k (2 k + 1) + k^{2} \cdot 6 + 6)$

Step 11.3

Apply the distributive property.

$\frac{1}{6} \cdot (12 k + k^{2} (2 k) + k^{2} \cdot 1 + k (2 k) + k \cdot 1 + k^{2} \cdot 6 + 6)$

Step 12

Simplify and combine like terms.

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Step 12.1

Simplify each term.

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Step 12.1.1

Rewrite using the commutative property of multiplication.

$\frac{1}{6} \cdot (12 k + 2 k^{2} k + k^{2} \cdot 1 + k (2 k) + k \cdot 1 + k^{2} \cdot 6 + 6)$

Step 12.1.2

Multiply $k^{2}$ by $k$ by adding the exponents.

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Step 12.1.2.1

Move $k$ .

$\frac{1}{6} \cdot (12 k + 2 (k \cdot k^{2}) + k^{2} \cdot 1 + k (2 k) + k \cdot 1 + k^{2} \cdot 6 + 6)$

Step 12.1.2.2

Multiply $k$ by $k^{2}$ .

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Step 12.1.2.2.1

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

$\frac{1}{6} \cdot (12 k + 2 (k^{1} k^{2}) + k^{2} \cdot 1 + k (2 k) + k \cdot 1 + k^{2} \cdot 6 + 6)$

Step 12.1.2.2.2

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

$\frac{1}{6} \cdot (12 k + 2 k^{1 + 2} + k^{2} \cdot 1 + k (2 k) + k \cdot 1 + k^{2} \cdot 6 + 6)$

Step 12.1.2.3

Add $1$ and $2$ .

$\frac{1}{6} \cdot (12 k + 2 k^{3} + k^{2} \cdot 1 + k (2 k) + k \cdot 1 + k^{2} \cdot 6 + 6)$

Step 12.1.3

Multiply $k^{2}$ by $1$ .

$\frac{1}{6} \cdot (12 k + 2 k^{3} + k^{2} + k (2 k) + k \cdot 1 + k^{2} \cdot 6 + 6)$

Step 12.1.4

Rewrite using the commutative property of multiplication.

$\frac{1}{6} \cdot (12 k + 2 k^{3} + k^{2} + 2 k \cdot k + k \cdot 1 + k^{2} \cdot 6 + 6)$

Step 12.1.5

Multiply $k$ by $k$ by adding the exponents.

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Step 12.1.5.1

Move $k$ .

$\frac{1}{6} \cdot (12 k + 2 k^{3} + k^{2} + 2 (k \cdot k) + k \cdot 1 + k^{2} \cdot 6 + 6)$

Step 12.1.5.2

Multiply $k$ by $k$ .

$\frac{1}{6} \cdot (12 k + 2 k^{3} + k^{2} + 2 k^{2} + k \cdot 1 + k^{2} \cdot 6 + 6)$

Step 12.1.6

Multiply $k$ by $1$ .

$\frac{1}{6} \cdot (12 k + 2 k^{3} + k^{2} + 2 k^{2} + k + k^{2} \cdot 6 + 6)$

Step 12.2

Add $k^{2}$ and $2 k^{2}$ .

$\frac{1}{6} \cdot (12 k + 2 k^{3} + 3 k^{2} + k + k^{2} \cdot 6 + 6)$

Step 13

Move $6$ to the left of $k^{2}$ .

$\frac{1}{6} \cdot (12 k + 2 k^{3} + 3 k^{2} + k + 6 k^{2} + 6)$

Step 14

Add $12 k$ and $k$ .

$\frac{1}{6} \cdot (2 k^{3} + 3 k^{2} + 13 k + 6 k^{2} + 6)$

Step 15

Add $3 k^{2}$ and $6 k^{2}$ .

$\frac{1}{6} \cdot (2 k^{3} + 9 k^{2} + 13 k + 6)$

Step 16

Factor.

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Step 16.1

Rewrite $2 k^{3} + 9 k^{2} + 13 k + 6$ in a factored form.

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Step 16.1.1

Factor $2 k^{3} + 9 k^{2} + 13 k + 6$ using the rational roots test.

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Step 16.1.1.1

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.

$p = \pm 1, \pm 6, \pm 2, \pm 3$

$q = \pm 1, \pm 2$

Step 16.1.1.2

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

$\pm 1, \pm 0.5, \pm 6, \pm 3, \pm 2, \pm 1.5$

Step 16.1.1.3

Substitute $- 1$ and simplify the expression. In this case, the expression is equal to $0$ so $- 1$ is a root of the polynomial.

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Step 16.1.1.3.1

Substitute $- 1$ into the polynomial.

$2 {(- 1)}^{3} + 9 {(- 1)}^{2} + 13 \cdot - 1 + 6$

Step 16.1.1.3.2

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

$2 \cdot - 1 + 9 {(- 1)}^{2} + 13 \cdot - 1 + 6$

Step 16.1.1.3.3

Multiply $2$ by $- 1$ .

$- 2 + 9 {(- 1)}^{2} + 13 \cdot - 1 + 6$

Step 16.1.1.3.4

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

$- 2 + 9 \cdot 1 + 13 \cdot - 1 + 6$

Step 16.1.1.3.5

Multiply $9$ by $1$ .

$- 2 + 9 + 13 \cdot - 1 + 6$

Step 16.1.1.3.6

Add $- 2$ and $9$ .

$7 + 13 \cdot - 1 + 6$

Step 16.1.1.3.7

Multiply $13$ by $- 1$ .

$7 - 13 + 6$

Step 16.1.1.3.8

Subtract $13$ from $7$ .

$- 6 + 6$

Step 16.1.1.3.9

Add $- 6$ and $6$ .

$0$

Step 16.1.1.4

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

$\frac{2 k^{3} + 9 k^{2} + 13 k + 6}{k + 1}$

Step 16.1.1.5

Divide $2 k^{3} + 9 k^{2} + 13 k + 6$ by $k + 1$ .

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Step 16.1.1.5.1

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

$k$

$1$

$2 k^{3}$

$9 k^{2}$

$13 k$

$6$

Step 16.1.1.5.2

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

					$2 k^{2}$
	$k$	+	$1$		$2 k^{3}$	+	$9 k^{2}$	+	$13 k$	+	$6$

Step 16.1.1.5.3

Multiply the new quotient term by the divisor.

				$2 k^{2}$
$k$	+	$1$		$2 k^{3}$	+	$9 k^{2}$	+	$13 k$	+	$6$
			+	$2 k^{3}$	+	$2 k^{2}$

Step 16.1.1.5.4

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

				$2 k^{2}$
$k$	+	$1$		$2 k^{3}$	+	$9 k^{2}$	+	$13 k$	+	$6$
			-	$2 k^{3}$	-	$2 k^{2}$

Step 16.1.1.5.5

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

				$2 k^{2}$
$k$	+	$1$		$2 k^{3}$	+	$9 k^{2}$	+	$13 k$	+	$6$
			-	$2 k^{3}$	-	$2 k^{2}$
					+	$7 k^{2}$

Step 16.1.1.5.6

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

				$2 k^{2}$
$k$	+	$1$		$2 k^{3}$	+	$9 k^{2}$	+	$13 k$	+	$6$
			-	$2 k^{3}$	-	$2 k^{2}$
					+	$7 k^{2}$	+	$13 k$

Step 16.1.1.5.7

Divide the highest order term in the dividend $7 k^{2}$ by the highest order term in divisor $k$ .

				$2 k^{2}$	+	$7 k$
$k$	+	$1$		$2 k^{3}$	+	$9 k^{2}$	+	$13 k$	+	$6$
			-	$2 k^{3}$	-	$2 k^{2}$
					+	$7 k^{2}$	+	$13 k$

Step 16.1.1.5.8

Multiply the new quotient term by the divisor.

				$2 k^{2}$	+	$7 k$
$k$	+	$1$		$2 k^{3}$	+	$9 k^{2}$	+	$13 k$	+	$6$
			-	$2 k^{3}$	-	$2 k^{2}$
					+	$7 k^{2}$	+	$13 k$
					+	$7 k^{2}$	+	$7 k$

Step 16.1.1.5.9

The expression needs to be subtracted from the dividend, so change all the signs in $7 k^{2} + 7 k$

				$2 k^{2}$	+	$7 k$
$k$	+	$1$		$2 k^{3}$	+	$9 k^{2}$	+	$13 k$	+	$6$
			-	$2 k^{3}$	-	$2 k^{2}$
					+	$7 k^{2}$	+	$13 k$
					-	$7 k^{2}$	-	$7 k$

Step 16.1.1.5.10

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

				$2 k^{2}$	+	$7 k$
$k$	+	$1$		$2 k^{3}$	+	$9 k^{2}$	+	$13 k$	+	$6$
			-	$2 k^{3}$	-	$2 k^{2}$
					+	$7 k^{2}$	+	$13 k$
					-	$7 k^{2}$	-	$7 k$
							+	$6 k$

Step 16.1.1.5.11

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

				$2 k^{2}$	+	$7 k$
$k$	+	$1$		$2 k^{3}$	+	$9 k^{2}$	+	$13 k$	+	$6$
			-	$2 k^{3}$	-	$2 k^{2}$
					+	$7 k^{2}$	+	$13 k$
					-	$7 k^{2}$	-	$7 k$
							+	$6 k$	+	$6$

Step 16.1.1.5.12

Divide the highest order term in the dividend $6 k$ by the highest order term in divisor $k$ .

				$2 k^{2}$	+	$7 k$	+	$6$
$k$	+	$1$		$2 k^{3}$	+	$9 k^{2}$	+	$13 k$	+	$6$
			-	$2 k^{3}$	-	$2 k^{2}$
					+	$7 k^{2}$	+	$13 k$
					-	$7 k^{2}$	-	$7 k$
							+	$6 k$	+	$6$

Step 16.1.1.5.13

Multiply the new quotient term by the divisor.

				$2 k^{2}$	+	$7 k$	+	$6$
$k$	+	$1$		$2 k^{3}$	+	$9 k^{2}$	+	$13 k$	+	$6$
			-	$2 k^{3}$	-	$2 k^{2}$
					+	$7 k^{2}$	+	$13 k$
					-	$7 k^{2}$	-	$7 k$
							+	$6 k$	+	$6$
							+	$6 k$	+	$6$

Step 16.1.1.5.14

The expression needs to be subtracted from the dividend, so change all the signs in $6 k + 6$

				$2 k^{2}$	+	$7 k$	+	$6$
$k$	+	$1$		$2 k^{3}$	+	$9 k^{2}$	+	$13 k$	+	$6$
			-	$2 k^{3}$	-	$2 k^{2}$
					+	$7 k^{2}$	+	$13 k$
					-	$7 k^{2}$	-	$7 k$
							+	$6 k$	+	$6$
							-	$6 k$	-	$6$

Step 16.1.1.5.15

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

				$2 k^{2}$	+	$7 k$	+	$6$
$k$	+	$1$		$2 k^{3}$	+	$9 k^{2}$	+	$13 k$	+	$6$
			-	$2 k^{3}$	-	$2 k^{2}$
					+	$7 k^{2}$	+	$13 k$
					-	$7 k^{2}$	-	$7 k$
							+	$6 k$	+	$6$
							-	$6 k$	-	$6$
										$0$

Step 16.1.1.5.16

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

$2 k^{2} + 7 k + 6$

Step 16.1.1.6

Write $2 k^{3} + 9 k^{2} + 13 k + 6$ as a set of factors.

$\frac{1}{6} \cdot ((k + 1) (2 k^{2} + 7 k + 6))$

Step 16.1.2

Factor by grouping.

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Step 16.1.2.1

Factor by grouping.

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Step 16.1.2.1.1

For a polynomial of the form $a x^{2} + b x + c$ , rewrite the middle term as a sum of two terms whose product is $a \cdot c = 2 \cdot 6 = 12$ and whose sum is $b = 7$ .

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Step 16.1.2.1.1.1

Factor $7$ out of $7 k$ .

$\frac{1}{6} \cdot ((k + 1) (2 k^{2} + 7 (k) + 6))$

Step 16.1.2.1.1.2

Rewrite $7$ as $3$ plus $4$

$\frac{1}{6} \cdot ((k + 1) (2 k^{2} + (3 + 4) k + 6))$

Step 16.1.2.1.1.3

Apply the distributive property.

$\frac{1}{6} \cdot ((k + 1) (2 k^{2} + 3 k + 4 k + 6))$

Step 16.1.2.1.2

Factor out the greatest common factor from each group.

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Step 16.1.2.1.2.1

Group the first two terms and the last two terms.

$\frac{1}{6} \cdot ((k + 1) ((2 k^{2} + 3 k) + 4 k + 6))$

Step 16.1.2.1.2.2

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

$\frac{1}{6} \cdot ((k + 1) (k (2 k + 3) + 2 (2 k + 3)))$

Step 16.1.2.1.3

Factor the polynomial by factoring out the greatest common factor, $2 k + 3$ .

$\frac{1}{6} \cdot ((k + 1) ((2 k + 3) (k + 2)))$

Step 16.1.2.2

Remove unnecessary parentheses.

$\frac{1}{6} \cdot ((k + 1) (2 k + 3) (k + 2))$

Step 16.2

Remove unnecessary parentheses.

$\frac{1}{6} \cdot (k + 1) (2 k + 3) (k + 2)$