Expand Using the Binomial Theorem (2p-3q)^11

${(2 p - 3 q)}^{11}$

Step 1

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

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

Step 2

Expand the summation.

$\frac{11!}{(11 - 0)! 0!} \cdot {(2 p)}^{11 - 0} \cdot {(- 3 q)}^{0} + \frac{11!}{(11 - 1)! 1!} \cdot {(2 p)}^{11 - 1} \cdot {(- 3 q)}^{1} + \dots + \frac{11!}{(11 - 11)! 11!} \cdot {(2 p)}^{11 - 11} \cdot {(- 3 q)}^{11}$

Step 3

Simplify the exponents for each term of the expansion.

$1 \cdot {(2 p)}^{11} \cdot {(- 3 q)}^{0} + 11 \cdot {(2 p)}^{10} \cdot {(- 3 q)}^{1} + \dots + 1 \cdot {(2 p)}^{0} \cdot {(- 3 q)}^{11}$

Step 4

Simplify the polynomial result.

$2048 p^{11} - 33792 p^{10} q + 253440 p^{9} q^{2} - 1140480 p^{8} q^{3} + 3421440 p^{7} q^{4} - 7185024 p^{6} q^{5} + 10777536 p^{5} q^{6} - 11547360 p^{4} q^{7} + 8660520 p^{3} q^{8} - 4330260 p^{2} q^{9} + 1299078 p q^{10} - 177147 q^{11}$

Expand Using the Binomial Theorem (2p-3q)^11

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