Finite Math Examples

$(2 a - 3 b) \cdot 8$

Step 1

Use the binomial expansion theorem to find each term. The binomial theorem states ${(a + b)}^{n} = \sum_{k = 0}^{n} n C k \cdot (a^{n - k} b^{k})$ .

$\sum_{k = 0}^{1} \frac{1!}{(1 - k)! k!} \cdot {(2 a)}^{1 - k} \cdot {(- 3 b)}^{k}$

Step 2

Expand the summation.

$\frac{1!}{(1 - 0)! 0!} \cdot {(2 a)}^{1 - 0} \cdot {(- 3 b)}^{0} + \frac{1!}{(1 - 1)! 1!} \cdot {(2 a)}^{1 - 1} \cdot {(- 3 b)}^{1}$

Step 3

Simplify the exponents for each term of the expansion.

$1 \cdot {(2 a)}^{1} \cdot {(- 3 b)}^{0} + 1 \cdot {(2 a)}^{0} \cdot {(- 3 b)}^{1}$

Step 4

Simplify each term.

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Step 4.1

Multiply ${(2 a)}^{1}$ by $1$ .

${(2 a)}^{1} \cdot {(- 3 b)}^{0} + 1 \cdot {(2 a)}^{0} \cdot {(- 3 b)}^{1}$

Step 4.2

Simplify.

$2 a \cdot {(- 3 b)}^{0} + 1 \cdot {(2 a)}^{0} \cdot {(- 3 b)}^{1}$

Step 4.3

Apply the product rule to $- 3 b$ .

$2 a \cdot ({(- 3)}^{0} b^{0}) + 1 \cdot {(2 a)}^{0} \cdot {(- 3 b)}^{1}$

Step 4.4

Rewrite using the commutative property of multiplication.

$2 \cdot {(- 3)}^{0} a b^{0} + 1 \cdot {(2 a)}^{0} \cdot {(- 3 b)}^{1}$

Step 4.5

Anything raised to $0$ is $1$ .

$2 \cdot 1 a b^{0} + 1 \cdot {(2 a)}^{0} \cdot {(- 3 b)}^{1}$

Step 4.6

Multiply $2$ by $1$ .

$2 a b^{0} + 1 \cdot {(2 a)}^{0} \cdot {(- 3 b)}^{1}$

Step 4.7

Anything raised to $0$ is $1$ .

$2 a \cdot 1 + 1 \cdot {(2 a)}^{0} \cdot {(- 3 b)}^{1}$

Step 4.8

Multiply $2$ by $1$ .

$2 a + 1 \cdot {(2 a)}^{0} \cdot {(- 3 b)}^{1}$

Step 4.9

Multiply ${(2 a)}^{0}$ by $1$ .

$2 a + {(2 a)}^{0} \cdot {(- 3 b)}^{1}$

Step 4.10

Apply the product rule to $2 a$ .

$2 a + 2^{0} a^{0} \cdot {(- 3 b)}^{1}$

Step 4.11

Anything raised to $0$ is $1$ .

$2 a + 1 a^{0} \cdot {(- 3 b)}^{1}$

Step 4.12

Multiply $a^{0}$ by $1$ .

$2 a + a^{0} \cdot {(- 3 b)}^{1}$

Step 4.13

Anything raised to $0$ is $1$ .

$2 a + 1 \cdot {(- 3 b)}^{1}$

Step 4.14

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

$2 a + {(- 3 b)}^{1}$

Step 4.15

Simplify.

$2 a - 3 b$