Precalculus Examples

${(x^{2} + 2 y)}^{10}$

Step 1

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

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

Step 2

Expand the summation.

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

Step 3

Simplify the exponents for each term of the expansion.

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

Step 4

Simplify the polynomial result.

$x^{20} + 20 x^{18} y + 180 x^{16} y^{2} + 960 x^{14} y^{3} + 3360 x^{12} y^{4} + 8064 x^{10} y^{5} + 13440 x^{8} y^{6} + 15360 x^{6} y^{7} + 11520 x^{4} y^{8} + 5120 x^{2} y^{9} + 1024 y^{10}$