Precalculus Examples

${(a - 2 b^{2})}^{4}$

Step 1

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

$\sum_{k = 0}^{4} \frac{4!}{(4 - k)! k!} \cdot {(a)}^{4 - k} \cdot {(- 2 b^{2})}^{k}$

Step 2

Expand the summation.

$\frac{4!}{(4 - 0)! 0!} {(a)}^{4 - 0} \cdot {(- 2 b^{2})}^{0} + \frac{4!}{(4 - 1)! 1!} {(a)}^{4 - 1} \cdot {(- 2 b^{2})}^{1} + \frac{4!}{(4 - 2)! 2!} {(a)}^{4 - 2} \cdot {(- 2 b^{2})}^{2} + \frac{4!}{(4 - 3)! 3!} {(a)}^{4 - 3} \cdot {(- 2 b^{2})}^{3} + \frac{4!}{(4 - 4)! 4!} {(a)}^{4 - 4} \cdot {(- 2 b^{2})}^{4}$

Step 3

Simplify the exponents for each term of the expansion.

$1 \cdot {(a)}^{4} \cdot {(- 2 b^{2})}^{0} + 4 \cdot {(a)}^{3} \cdot {(- 2 b^{2})}^{1} + 6 \cdot {(a)}^{2} \cdot {(- 2 b^{2})}^{2} + 4 \cdot {(a)}^{1} \cdot {(- 2 b^{2})}^{3} + 1 \cdot {(a)}^{0} \cdot {(- 2 b^{2})}^{4}$

Step 4

Simplify each term.

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Step 4.1

Multiply ${(a)}^{4}$ by $1$ .

${(a)}^{4} \cdot {(- 2 b^{2})}^{0} + 4 \cdot {(a)}^{3} \cdot {(- 2 b^{2})}^{1} + 6 \cdot {(a)}^{2} \cdot {(- 2 b^{2})}^{2} + 4 \cdot {(a)}^{1} \cdot {(- 2 b^{2})}^{3} + 1 \cdot {(a)}^{0} \cdot {(- 2 b^{2})}^{4}$

Step 4.2

Apply the product rule to $- 2 b^{2}$ .

$a^{4} \cdot ({(- 2)}^{0} {(b^{2})}^{0}) + 4 \cdot {(a)}^{3} \cdot {(- 2 b^{2})}^{1} + 6 \cdot {(a)}^{2} \cdot {(- 2 b^{2})}^{2} + 4 \cdot {(a)}^{1} \cdot {(- 2 b^{2})}^{3} + 1 \cdot {(a)}^{0} \cdot {(- 2 b^{2})}^{4}$

Step 4.3

Rewrite using the commutative property of multiplication.

${(- 2)}^{0} a^{4} {(b^{2})}^{0} + 4 \cdot {(a)}^{3} \cdot {(- 2 b^{2})}^{1} + 6 \cdot {(a)}^{2} \cdot {(- 2 b^{2})}^{2} + 4 \cdot {(a)}^{1} \cdot {(- 2 b^{2})}^{3} + 1 \cdot {(a)}^{0} \cdot {(- 2 b^{2})}^{4}$

Step 4.4

Anything raised to $0$ is $1$ .

$1 a^{4} {(b^{2})}^{0} + 4 \cdot {(a)}^{3} \cdot {(- 2 b^{2})}^{1} + 6 \cdot {(a)}^{2} \cdot {(- 2 b^{2})}^{2} + 4 \cdot {(a)}^{1} \cdot {(- 2 b^{2})}^{3} + 1 \cdot {(a)}^{0} \cdot {(- 2 b^{2})}^{4}$

Step 4.5

Multiply $a^{4}$ by $1$ .

$a^{4} {(b^{2})}^{0} + 4 \cdot {(a)}^{3} \cdot {(- 2 b^{2})}^{1} + 6 \cdot {(a)}^{2} \cdot {(- 2 b^{2})}^{2} + 4 \cdot {(a)}^{1} \cdot {(- 2 b^{2})}^{3} + 1 \cdot {(a)}^{0} \cdot {(- 2 b^{2})}^{4}$

Step 4.6

Multiply the exponents in ${(b^{2})}^{0}$ .

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Step 4.6.1

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

$a^{4} b^{2 \cdot 0} + 4 \cdot {(a)}^{3} \cdot {(- 2 b^{2})}^{1} + 6 \cdot {(a)}^{2} \cdot {(- 2 b^{2})}^{2} + 4 \cdot {(a)}^{1} \cdot {(- 2 b^{2})}^{3} + 1 \cdot {(a)}^{0} \cdot {(- 2 b^{2})}^{4}$

Step 4.6.2

Multiply $2$ by $0$ .

$a^{4} b^{0} + 4 \cdot {(a)}^{3} \cdot {(- 2 b^{2})}^{1} + 6 \cdot {(a)}^{2} \cdot {(- 2 b^{2})}^{2} + 4 \cdot {(a)}^{1} \cdot {(- 2 b^{2})}^{3} + 1 \cdot {(a)}^{0} \cdot {(- 2 b^{2})}^{4}$

Step 4.7

Anything raised to $0$ is $1$ .

$a^{4} \cdot 1 + 4 \cdot {(a)}^{3} \cdot {(- 2 b^{2})}^{1} + 6 \cdot {(a)}^{2} \cdot {(- 2 b^{2})}^{2} + 4 \cdot {(a)}^{1} \cdot {(- 2 b^{2})}^{3} + 1 \cdot {(a)}^{0} \cdot {(- 2 b^{2})}^{4}$

Step 4.8

Multiply $a^{4}$ by $1$ .

$a^{4} + 4 \cdot {(a)}^{3} \cdot {(- 2 b^{2})}^{1} + 6 \cdot {(a)}^{2} \cdot {(- 2 b^{2})}^{2} + 4 \cdot {(a)}^{1} \cdot {(- 2 b^{2})}^{3} + 1 \cdot {(a)}^{0} \cdot {(- 2 b^{2})}^{4}$

Step 4.9

Simplify.

$a^{4} + 4 a^{3} \cdot (- 2 b^{2}) + 6 \cdot {(a)}^{2} \cdot {(- 2 b^{2})}^{2} + 4 \cdot {(a)}^{1} \cdot {(- 2 b^{2})}^{3} + 1 \cdot {(a)}^{0} \cdot {(- 2 b^{2})}^{4}$

Step 4.10

Rewrite using the commutative property of multiplication.

$a^{4} + 4 \cdot - 2 a^{3} b^{2} + 6 \cdot {(a)}^{2} \cdot {(- 2 b^{2})}^{2} + 4 \cdot {(a)}^{1} \cdot {(- 2 b^{2})}^{3} + 1 \cdot {(a)}^{0} \cdot {(- 2 b^{2})}^{4}$

Step 4.11

Multiply $4$ by $- 2$ .

$a^{4} - 8 a^{3} b^{2} + 6 \cdot {(a)}^{2} \cdot {(- 2 b^{2})}^{2} + 4 \cdot {(a)}^{1} \cdot {(- 2 b^{2})}^{3} + 1 \cdot {(a)}^{0} \cdot {(- 2 b^{2})}^{4}$

Step 4.12

Apply the product rule to $- 2 b^{2}$ .

$a^{4} - 8 a^{3} b^{2} + 6 a^{2} \cdot ({(- 2)}^{2} {(b^{2})}^{2}) + 4 \cdot {(a)}^{1} \cdot {(- 2 b^{2})}^{3} + 1 \cdot {(a)}^{0} \cdot {(- 2 b^{2})}^{4}$

Step 4.13

Rewrite using the commutative property of multiplication.

$a^{4} - 8 a^{3} b^{2} + 6 \cdot {(- 2)}^{2} a^{2} {(b^{2})}^{2} + 4 \cdot {(a)}^{1} \cdot {(- 2 b^{2})}^{3} + 1 \cdot {(a)}^{0} \cdot {(- 2 b^{2})}^{4}$

Step 4.14

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

$a^{4} - 8 a^{3} b^{2} + 6 \cdot 4 a^{2} {(b^{2})}^{2} + 4 \cdot {(a)}^{1} \cdot {(- 2 b^{2})}^{3} + 1 \cdot {(a)}^{0} \cdot {(- 2 b^{2})}^{4}$

Step 4.15

Multiply $6$ by $4$ .

$a^{4} - 8 a^{3} b^{2} + 24 a^{2} {(b^{2})}^{2} + 4 \cdot {(a)}^{1} \cdot {(- 2 b^{2})}^{3} + 1 \cdot {(a)}^{0} \cdot {(- 2 b^{2})}^{4}$

Step 4.16

Multiply the exponents in ${(b^{2})}^{2}$ .

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Step 4.16.1

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

$a^{4} - 8 a^{3} b^{2} + 24 a^{2} b^{2 \cdot 2} + 4 \cdot {(a)}^{1} \cdot {(- 2 b^{2})}^{3} + 1 \cdot {(a)}^{0} \cdot {(- 2 b^{2})}^{4}$

Step 4.16.2

Multiply $2$ by $2$ .

$a^{4} - 8 a^{3} b^{2} + 24 a^{2} b^{4} + 4 \cdot {(a)}^{1} \cdot {(- 2 b^{2})}^{3} + 1 \cdot {(a)}^{0} \cdot {(- 2 b^{2})}^{4}$

Step 4.17

Simplify.

$a^{4} - 8 a^{3} b^{2} + 24 a^{2} b^{4} + 4 \cdot a \cdot {(- 2 b^{2})}^{3} + 1 \cdot {(a)}^{0} \cdot {(- 2 b^{2})}^{4}$

Step 4.18

Apply the product rule to $- 2 b^{2}$ .

$a^{4} - 8 a^{3} b^{2} + 24 a^{2} b^{4} + 4 a \cdot ({(- 2)}^{3} {(b^{2})}^{3}) + 1 \cdot {(a)}^{0} \cdot {(- 2 b^{2})}^{4}$

Step 4.19

Rewrite using the commutative property of multiplication.

$a^{4} - 8 a^{3} b^{2} + 24 a^{2} b^{4} + 4 \cdot {(- 2)}^{3} a {(b^{2})}^{3} + 1 \cdot {(a)}^{0} \cdot {(- 2 b^{2})}^{4}$

Step 4.20

Raise $- 2$ to the power of $3$ .

$a^{4} - 8 a^{3} b^{2} + 24 a^{2} b^{4} + 4 \cdot - 8 a {(b^{2})}^{3} + 1 \cdot {(a)}^{0} \cdot {(- 2 b^{2})}^{4}$

Step 4.21

Multiply $4$ by $- 8$ .

$a^{4} - 8 a^{3} b^{2} + 24 a^{2} b^{4} - 32 a {(b^{2})}^{3} + 1 \cdot {(a)}^{0} \cdot {(- 2 b^{2})}^{4}$

Step 4.22

Multiply the exponents in ${(b^{2})}^{3}$ .

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Step 4.22.1

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

$a^{4} - 8 a^{3} b^{2} + 24 a^{2} b^{4} - 32 a b^{2 \cdot 3} + 1 \cdot {(a)}^{0} \cdot {(- 2 b^{2})}^{4}$

Step 4.22.2

Multiply $2$ by $3$ .

$a^{4} - 8 a^{3} b^{2} + 24 a^{2} b^{4} - 32 a b^{6} + 1 \cdot {(a)}^{0} \cdot {(- 2 b^{2})}^{4}$

Step 4.23

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

$a^{4} - 8 a^{3} b^{2} + 24 a^{2} b^{4} - 32 a b^{6} + {(a)}^{0} \cdot {(- 2 b^{2})}^{4}$

Step 4.24

Anything raised to $0$ is $1$ .

$a^{4} - 8 a^{3} b^{2} + 24 a^{2} b^{4} - 32 a b^{6} + 1 \cdot {(- 2 b^{2})}^{4}$

Step 4.25

Multiply ${(- 2 b^{2})}^{4}$ by $1$ .

$a^{4} - 8 a^{3} b^{2} + 24 a^{2} b^{4} - 32 a b^{6} + {(- 2 b^{2})}^{4}$

Step 4.26

Apply the product rule to $- 2 b^{2}$ .

$a^{4} - 8 a^{3} b^{2} + 24 a^{2} b^{4} - 32 a b^{6} + {(- 2)}^{4} {(b^{2})}^{4}$

Step 4.27

Raise $- 2$ to the power of $4$ .

$a^{4} - 8 a^{3} b^{2} + 24 a^{2} b^{4} - 32 a b^{6} + 16 {(b^{2})}^{4}$

Step 4.28

Multiply the exponents in ${(b^{2})}^{4}$ .

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Step 4.28.1

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

$a^{4} - 8 a^{3} b^{2} + 24 a^{2} b^{4} - 32 a b^{6} + 16 b^{2 \cdot 4}$

Step 4.28.2

Multiply $2$ by $4$ .

$a^{4} - 8 a^{3} b^{2} + 24 a^{2} b^{4} - 32 a b^{6} + 16 b^{8}$