Expand Using the Binomial Theorem (2u^3+4v^2)^10

${(2 u^{3} + 4 v^{2})}^{10}$

Step 1

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

$\sum_{k = 0}^{10} \frac{10!}{(10 - k)! k!} \cdot {(2 u^{3})}^{10 - k} \cdot {(4 v^{2})}^{k}$

Step 2

Expand the summation.

$\frac{10!}{(10 - 0)! 0!} \cdot {(2 u^{3})}^{10 - 0} \cdot {(4 v^{2})}^{0} + \frac{10!}{(10 - 1)! 1!} \cdot {(2 u^{3})}^{10 - 1} \cdot {(4 v^{2})}^{1} + \dots + \frac{10!}{(10 - 10)! 10!} \cdot {(2 u^{3})}^{10 - 10} \cdot {(4 v^{2})}^{10}$

Step 3

Simplify the exponents for each term of the expansion.

$1 \cdot {(2 u^{3})}^{10} \cdot {(4 v^{2})}^{0} + 10 \cdot {(2 u^{3})}^{9} \cdot {(4 v^{2})}^{1} + \dots + 1 \cdot {(2 u^{3})}^{0} \cdot {(4 v^{2})}^{10}$

Step 4

Simplify the polynomial result.

$1024 u^{30} + 20480 u^{27} v^{2} + 184320 u^{24} v^{4} + 983040 u^{21} v^{6} + 3440640 u^{18} v^{8} + 8257536 u^{15} v^{10} + 13762560 u^{12} v^{12} + 15728640 u^{9} v^{14} + 11796480 u^{6} v^{16} + 5242880 u^{3} v^{18} + 1048576 v^{20}$

Expand Using the Binomial Theorem (2u^3+4v^2)^10

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