Precalculus Examples

${(2 x - 3 y)}^{10}$

Step 1

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

$\sum_{k = 0}^{10} \frac{10!}{(10 - k)! k!} \cdot {(2 x)}^{10 - k} \cdot {(- 3 y)}^{k}$

Step 2

Expand the summation.

$\frac{10!}{(10 - 0)! 0!} \cdot {(2 x)}^{10 - 0} \cdot {(- 3 y)}^{0} + \frac{10!}{(10 - 1)! 1!} \cdot {(2 x)}^{10 - 1} \cdot {(- 3 y)}^{1} + \dots + \frac{10!}{(10 - 10)! 10!} \cdot {(2 x)}^{10 - 10} \cdot {(- 3 y)}^{10}$

Step 3

Simplify the exponents for each term of the expansion.

$1 \cdot {(2 x)}^{10} \cdot {(- 3 y)}^{0} + 10 \cdot {(2 x)}^{9} \cdot {(- 3 y)}^{1} + \dots + 1 \cdot {(2 x)}^{0} \cdot {(- 3 y)}^{10}$

Step 4

Simplify the polynomial result.

$1024 x^{10} - 15360 x^{9} y + 103680 x^{8} y^{2} - 414720 x^{7} y^{3} + 1088640 x^{6} y^{4} - 1959552 x^{5} y^{5} + 2449440 x^{4} y^{6} - 2099520 x^{3} y^{7} + 1180980 x^{2} y^{8} - 393660 x y^{9} + 59049 y^{10}$