Expand Using the Binomial Theorem (3-8w^2)^2

${(3 - 8 w^{2})}^{2}$

Step 1

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

$\sum_{k = 0}^{2} \frac{2!}{(2 - k)! k!} \cdot {(3)}^{2 - k} \cdot {(- 8 w^{2})}^{k}$

Step 2

Expand the summation.

$\frac{2!}{(2 - 0)! 0!} \cdot {(3)}^{2 - 0} \cdot {(- 8 w^{2})}^{0} + \frac{2!}{(2 - 1)! 1!} \cdot {(3)}^{2 - 1} \cdot {(- 8 w^{2})}^{1} + \frac{2!}{(2 - 2)! 2!} \cdot {(3)}^{2 - 2} \cdot {(- 8 w^{2})}^{2}$

Step 3

Simplify the exponents for each term of the expansion.

$1 \cdot {(3)}^{2} \cdot {(- 8 w^{2})}^{0} + 2 \cdot {(3)}^{1} \cdot {(- 8 w^{2})}^{1} + 1 \cdot {(3)}^{0} \cdot {(- 8 w^{2})}^{2}$

Step 4

Simplify each term.

Multiply ${(3)}^{2}$ by $1$ .

${(3)}^{2} \cdot {(- 8 w^{2})}^{0} + 2 \cdot {(3)}^{1} \cdot {(- 8 w^{2})}^{1} + 1 \cdot {(3)}^{0} \cdot {(- 8 w^{2})}^{2}$

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

$9 \cdot {(- 8 w^{2})}^{0} + 2 \cdot {(3)}^{1} \cdot {(- 8 w^{2})}^{1} + 1 \cdot {(3)}^{0} \cdot {(- 8 w^{2})}^{2}$

Apply the product rule to $- 8 w^{2}$ .

$9 \cdot ({(- 8)}^{0} {(w^{2})}^{0}) + 2 \cdot {(3)}^{1} \cdot {(- 8 w^{2})}^{1} + 1 \cdot {(3)}^{0} \cdot {(- 8 w^{2})}^{2}$

Anything raised to $0$ is $1$ .

$9 \cdot (1 {(w^{2})}^{0}) + 2 \cdot {(3)}^{1} \cdot {(- 8 w^{2})}^{1} + 1 \cdot {(3)}^{0} \cdot {(- 8 w^{2})}^{2}$

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

$9 \cdot {(w^{2})}^{0} + 2 \cdot {(3)}^{1} \cdot {(- 8 w^{2})}^{1} + 1 \cdot {(3)}^{0} \cdot {(- 8 w^{2})}^{2}$

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

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

$9 \cdot w^{2 \cdot 0} + 2 \cdot {(3)}^{1} \cdot {(- 8 w^{2})}^{1} + 1 \cdot {(3)}^{0} \cdot {(- 8 w^{2})}^{2}$

Multiply $2$ by $0$ .

$9 \cdot w^{0} + 2 \cdot {(3)}^{1} \cdot {(- 8 w^{2})}^{1} + 1 \cdot {(3)}^{0} \cdot {(- 8 w^{2})}^{2}$

Anything raised to $0$ is $1$ .

$9 \cdot 1 + 2 \cdot {(3)}^{1} \cdot {(- 8 w^{2})}^{1} + 1 \cdot {(3)}^{0} \cdot {(- 8 w^{2})}^{2}$

Multiply $9$ by $1$ .

$9 + 2 \cdot {(3)}^{1} \cdot {(- 8 w^{2})}^{1} + 1 \cdot {(3)}^{0} \cdot {(- 8 w^{2})}^{2}$

Evaluate the exponent.

$9 + 2 \cdot 3 \cdot {(- 8 w^{2})}^{1} + 1 \cdot {(3)}^{0} \cdot {(- 8 w^{2})}^{2}$

Multiply $2$ by $3$ .

$9 + 6 \cdot {(- 8 w^{2})}^{1} + 1 \cdot {(3)}^{0} \cdot {(- 8 w^{2})}^{2}$

Simplify.

$9 + 6 \cdot (- 8 w^{2}) + 1 \cdot {(3)}^{0} \cdot {(- 8 w^{2})}^{2}$

Multiply $- 8$ by $6$ .

$9 - 48 w^{2} + 1 \cdot {(3)}^{0} \cdot {(- 8 w^{2})}^{2}$

Multiply ${(3)}^{0}$ by $1$ .

$9 - 48 w^{2} + {(3)}^{0} \cdot {(- 8 w^{2})}^{2}$

Anything raised to $0$ is $1$ .

$9 - 48 w^{2} + 1 \cdot {(- 8 w^{2})}^{2}$

Multiply ${(- 8 w^{2})}^{2}$ by $1$ .

$9 - 48 w^{2} + {(- 8 w^{2})}^{2}$

Apply the product rule to $- 8 w^{2}$ .

$9 - 48 w^{2} + {(- 8)}^{2} {(w^{2})}^{2}$

Raise $- 8$ to the power of $2$ .

$9 - 48 w^{2} + 64 {(w^{2})}^{2}$

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

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

$9 - 48 w^{2} + 64 w^{2 \cdot 2}$

Multiply $2$ by $2$ .

$9 - 48 w^{2} + 64 w^{4}$

Expand Using the Binomial Theorem (3-8w^2)^2

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