Expand Using the Binomial Theorem (3a-11b)^2

${(3 a - 11 b)}^{2}$

Step 1

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

$\sum_{k = 0}^{2} \frac{2!}{(2 - k)! k!} \cdot {(3 a)}^{2 - k} \cdot {(- 11 b)}^{k}$

Step 2

Expand the summation.

$\frac{2!}{(2 - 0)! 0!} \cdot {(3 a)}^{2 - 0} \cdot {(- 11 b)}^{0} + \frac{2!}{(2 - 1)! 1!} \cdot {(3 a)}^{2 - 1} \cdot {(- 11 b)}^{1} + \frac{2!}{(2 - 2)! 2!} \cdot {(3 a)}^{2 - 2} \cdot {(- 11 b)}^{2}$

Step 3

Simplify the exponents for each term of the expansion.

$1 \cdot {(3 a)}^{2} \cdot {(- 11 b)}^{0} + 2 \cdot {(3 a)}^{1} \cdot {(- 11 b)}^{1} + 1 \cdot {(3 a)}^{0} \cdot {(- 11 b)}^{2}$

Step 4

Simplify each term.

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

${(3 a)}^{2} \cdot {(- 11 b)}^{0} + 2 \cdot {(3 a)}^{1} \cdot {(- 11 b)}^{1} + 1 \cdot {(3 a)}^{0} \cdot {(- 11 b)}^{2}$

Apply the product rule to $3 a$ .

$3^{2} a^{2} \cdot {(- 11 b)}^{0} + 2 \cdot {(3 a)}^{1} \cdot {(- 11 b)}^{1} + 1 \cdot {(3 a)}^{0} \cdot {(- 11 b)}^{2}$

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

$9 a^{2} \cdot {(- 11 b)}^{0} + 2 \cdot {(3 a)}^{1} \cdot {(- 11 b)}^{1} + 1 \cdot {(3 a)}^{0} \cdot {(- 11 b)}^{2}$

Apply the product rule to $- 11 b$ .

$9 a^{2} \cdot ({(- 11)}^{0} b^{0}) + 2 \cdot {(3 a)}^{1} \cdot {(- 11 b)}^{1} + 1 \cdot {(3 a)}^{0} \cdot {(- 11 b)}^{2}$

Rewrite using the commutative property of multiplication.

$9 \cdot {(- 11)}^{0} a^{2} b^{0} + 2 \cdot {(3 a)}^{1} \cdot {(- 11 b)}^{1} + 1 \cdot {(3 a)}^{0} \cdot {(- 11 b)}^{2}$

Anything raised to $0$ is $1$ .

$9 \cdot 1 a^{2} b^{0} + 2 \cdot {(3 a)}^{1} \cdot {(- 11 b)}^{1} + 1 \cdot {(3 a)}^{0} \cdot {(- 11 b)}^{2}$

Multiply $9$ by $1$ .

$9 a^{2} b^{0} + 2 \cdot {(3 a)}^{1} \cdot {(- 11 b)}^{1} + 1 \cdot {(3 a)}^{0} \cdot {(- 11 b)}^{2}$

Anything raised to $0$ is $1$ .

$9 a^{2} \cdot 1 + 2 \cdot {(3 a)}^{1} \cdot {(- 11 b)}^{1} + 1 \cdot {(3 a)}^{0} \cdot {(- 11 b)}^{2}$

Multiply $9$ by $1$ .

$9 a^{2} + 2 \cdot {(3 a)}^{1} \cdot {(- 11 b)}^{1} + 1 \cdot {(3 a)}^{0} \cdot {(- 11 b)}^{2}$

Simplify.

$9 a^{2} + 2 \cdot (3 a) \cdot {(- 11 b)}^{1} + 1 \cdot {(3 a)}^{0} \cdot {(- 11 b)}^{2}$

Multiply $3$ by $2$ .

$9 a^{2} + 6 a \cdot {(- 11 b)}^{1} + 1 \cdot {(3 a)}^{0} \cdot {(- 11 b)}^{2}$

Simplify.

$9 a^{2} + 6 a \cdot (- 11 b) + 1 \cdot {(3 a)}^{0} \cdot {(- 11 b)}^{2}$

Rewrite using the commutative property of multiplication.

$9 a^{2} + 6 \cdot - 11 a b + 1 \cdot {(3 a)}^{0} \cdot {(- 11 b)}^{2}$

Multiply $6$ by $- 11$ .

$9 a^{2} - 66 a b + 1 \cdot {(3 a)}^{0} \cdot {(- 11 b)}^{2}$

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

$9 a^{2} - 66 a b + {(3 a)}^{0} \cdot {(- 11 b)}^{2}$

Apply the product rule to $3 a$ .

$9 a^{2} - 66 a b + 3^{0} a^{0} \cdot {(- 11 b)}^{2}$

Anything raised to $0$ is $1$ .

$9 a^{2} - 66 a b + 1 a^{0} \cdot {(- 11 b)}^{2}$

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

$9 a^{2} - 66 a b + a^{0} \cdot {(- 11 b)}^{2}$

Anything raised to $0$ is $1$ .

$9 a^{2} - 66 a b + 1 \cdot {(- 11 b)}^{2}$

Multiply ${(- 11 b)}^{2}$ by $1$ .

$9 a^{2} - 66 a b + {(- 11 b)}^{2}$

Apply the product rule to $- 11 b$ .

$9 a^{2} - 66 a b + {(- 11)}^{2} b^{2}$

Raise $- 11$ to the power of $2$ .

$9 a^{2} - 66 a b + 121 b^{2}$

Expand Using the Binomial Theorem (3a-11b)^2

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