Expand Using the Binomial Theorem (2r-3s)^12

${(2 r - 3 s)}^{12}$

Step 1

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

$\sum_{k = 0}^{12} \frac{12!}{(12 - k)! k!} \cdot {(2 r)}^{12 - k} \cdot {(- 3 s)}^{k}$

Step 2

Expand the summation.

$\frac{12!}{(12 - 0)! 0!} \cdot {(2 r)}^{12 - 0} \cdot {(- 3 s)}^{0} + \frac{12!}{(12 - 1)! 1!} \cdot {(2 r)}^{12 - 1} \cdot {(- 3 s)}^{1} + \dots + \frac{12!}{(12 - 12)! 12!} \cdot {(2 r)}^{12 - 12} \cdot {(- 3 s)}^{12}$

Step 3

Simplify the exponents for each term of the expansion.

$1 \cdot {(2 r)}^{12} \cdot {(- 3 s)}^{0} + 12 \cdot {(2 r)}^{11} \cdot {(- 3 s)}^{1} + \dots + 1 \cdot {(2 r)}^{0} \cdot {(- 3 s)}^{12}$

Step 4

Simplify the polynomial result.

$4096 r^{12} - 73728 r^{11} s + 608256 r^{10} s^{2} - 3041280 r^{9} s^{3} + 10264320 r^{8} s^{4} - 24634368 r^{7} s^{5} + 43110144 r^{6} s^{6} - 55427328 r^{5} s^{7} + 51963120 r^{4} s^{8} - 34642080 r^{3} s^{9} + 15588936 r^{2} s^{10} - 4251528 r s^{11} + 531441 s^{12}$

Expand Using the Binomial Theorem (2r-3s)^12

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