Expand Using the Binomial Theorem (-3u^2+v)^5

${(- 3 u^{2} + v)}^{5}$

Step 1

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

$\sum_{k = 0}^{5} \frac{5!}{(5 - k)! k!} \cdot {(- 3 u^{2})}^{5 - k} \cdot {(v)}^{k}$

Step 2

Expand the summation.

$\frac{5!}{(5 - 0)! 0!} \cdot {(- 3 u^{2})}^{5 - 0} \cdot {(v)}^{0} + \frac{5!}{(5 - 1)! 1!} \cdot {(- 3 u^{2})}^{5 - 1} \cdot {(v)}^{1} + \dots + \frac{5!}{(5 - 5)! 5!} \cdot {(- 3 u^{2})}^{5 - 5} \cdot {(v)}^{5}$

Step 3

Simplify the exponents for each term of the expansion.

$1 \cdot {(- 3 u^{2})}^{5} \cdot {(v)}^{0} + 5 \cdot {(- 3 u^{2})}^{4} \cdot {(v)}^{1} + \dots + 1 \cdot {(- 3 u^{2})}^{0} \cdot {(v)}^{5}$

Step 4

Simplify the polynomial result.

$- 243 u^{10} + 405 u^{8} v - 270 u^{6} v^{2} + 90 u^{4} v^{3} - 15 u^{2} v^{4} + v^{5}$

Expand Using the Binomial Theorem (-3u^2+v)^5

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