Precalculus Examples

${(x^{2} + y^{5})}^{8}$

Step 1

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

$\sum_{k = 0}^{8} \frac{8!}{(8 - k)! k!} \cdot {(x^{2})}^{8 - k} \cdot {(y^{5})}^{k}$

Step 2

Expand the summation.

$\frac{8!}{(8 - 0)! 0!} \cdot {(x^{2})}^{8 - 0} \cdot {(y^{5})}^{0} + \frac{8!}{(8 - 1)! 1!} \cdot {(x^{2})}^{8 - 1} \cdot {(y^{5})}^{1} + \dots + \frac{8!}{(8 - 8)! 8!} \cdot {(x^{2})}^{8 - 8} \cdot {(y^{5})}^{8}$

Step 3

Simplify the exponents for each term of the expansion.

$1 \cdot {(x^{2})}^{8} \cdot {(y^{5})}^{0} + 8 \cdot {(x^{2})}^{7} \cdot {(y^{5})}^{1} + \dots + 1 \cdot {(x^{2})}^{0} \cdot {(y^{5})}^{8}$

Step 4

Simplify the polynomial result.

$x^{16} + 8 x^{14} y^{5} + 28 x^{12} y^{10} + 56 x^{10} y^{15} + 70 x^{8} y^{20} + 56 x^{6} y^{25} + 28 x^{4} y^{30} + 8 x^{2} y^{35} + y^{40}$