Expand Using the Binomial Theorem (x-2y)^12

${(x - 2 y)}^{12}$

Step 1

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

$\sum_{k = 0}^{12} \frac{12!}{(12 - k)! k!} \cdot {(x)}^{12 - k} \cdot {(- 2 y)}^{k}$

Step 2

Expand the summation.

$\frac{12!}{(12 - 0)! 0!} {(x)}^{12 - 0} \cdot {(- 2 y)}^{0} + \frac{12!}{(12 - 1)! 1!} {(x)}^{12 - 1} \cdot {(- 2 y)}^{1} + \dots + \frac{12!}{(12 - 12)! 12!} {(x)}^{12 - 12} \cdot {(- 2 y)}^{12}$

Step 3

Simplify the exponents for each term of the expansion.

$1 \cdot {(x)}^{12} \cdot {(- 2 y)}^{0} + 12 \cdot {(x)}^{11} \cdot {(- 2 y)}^{1} + \dots + 1 \cdot {(x)}^{0} \cdot {(- 2 y)}^{12}$

Step 4

Simplify the polynomial result.

$x^{12} - 24 x^{11} y + 264 x^{10} y^{2} - 1760 x^{9} y^{3} + 7920 x^{8} y^{4} - 25344 x^{7} y^{5} + 59136 x^{6} y^{6} - 101376 x^{5} y^{7} + 126720 x^{4} y^{8} - 112640 x^{3} y^{9} + 67584 x^{2} y^{10} - 24576 x y^{11} + 4096 y^{12}$

Expand Using the Binomial Theorem (x-2y)^12

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