Expand Using the Binomial Theorem 3(a-4b)^2

$3 {(a - 4 b)}^{2}$

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

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

Expand the summation.

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

Simplify the exponents for each term of the expansion.

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

Simplify each term.

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

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

Apply the product rule to $- 4 b$ .

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

Rewrite using the commutative property of multiplication.

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

Anything raised to $0$ is $1$ .

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

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

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

Anything raised to $0$ is $1$ .

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

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

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

Simplify.

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

Simplify.

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

Rewrite using the commutative property of multiplication.

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

Multiply $2$ by $- 4$ .

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

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

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

Anything raised to $0$ is $1$ .

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

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

$a^{2} - 8 a b + {(- 4 b)}^{2}$

Apply the product rule to $- 4 b$ .

$a^{2} - 8 a b + {(- 4)}^{2} b^{2}$

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

$a^{2} - 8 a b + 16 b^{2}$

Expand Using the Binomial Theorem 3(a-4b)^2

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