Expand Using the Binomial Theorem ( square root of x+ square root of 6)^5

${(\sqrt{x} + \sqrt{6})}^{5}$

Step 1

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

$\sum_{k = 0}^{5} \frac{5!}{(5 - k)! k!} \cdot {(\sqrt{x})}^{5 - k} \cdot {(\sqrt{6})}^{k}$

Step 2

Expand the summation.

$\frac{5!}{(5 - 0)! 0!} \cdot {(\sqrt{x})}^{5 - 0} \cdot {(\sqrt{6})}^{0} + \frac{5!}{(5 - 1)! 1!} \cdot {(\sqrt{x})}^{5 - 1} \cdot {(\sqrt{6})}^{1} + \dots + \frac{5!}{(5 - 5)! 5!} \cdot {(\sqrt{x})}^{5 - 5} \cdot {(\sqrt{6})}^{5}$

Step 3

Simplify the exponents for each term of the expansion.

$1 \cdot {(\sqrt{x})}^{5} \cdot {(\sqrt{6})}^{0} + 5 \cdot {(\sqrt{x})}^{4} \cdot {(\sqrt{6})}^{1} + \dots + 1 \cdot {(\sqrt{x})}^{0} \cdot {(\sqrt{6})}^{5}$

Step 4

Simplify the polynomial result.

$x^{2} \sqrt{x} + 5 x^{2} \sqrt{6} + 60 x \sqrt{x} + 60 x \sqrt{6} + 180 \sqrt{x} + 36 \sqrt{6}$

Expand Using the Binomial Theorem ( square root of x+ square root of 6)^5

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