Expand using the Binomial Theorem (x+2)^10

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

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

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

Expand the summation.

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

Simplify the exponents for each term of the expansion.

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

Simplify the polynomial result.

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

Expand using the Binomial Theorem (x+2)^10

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