Expand Using the Binomial Theorem ( square root of x+ square root of 6)^2

${(\sqrt{x} + \sqrt{6})}^{2}$

Step 1

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

$\sum_{k = 0}^{2} \frac{2!}{(2 - k)! k!} \cdot {(\sqrt{x})}^{2 - k} \cdot {(\sqrt{6})}^{k}$

Step 2

Expand the summation.

$\frac{2!}{(2 - 0)! 0!} \cdot {(\sqrt{x})}^{2 - 0} \cdot {(\sqrt{6})}^{0} + \frac{2!}{(2 - 1)! 1!} \cdot {(\sqrt{x})}^{2 - 1} \cdot {(\sqrt{6})}^{1} + \frac{2!}{(2 - 2)! 2!} \cdot {(\sqrt{x})}^{2 - 2} \cdot {(\sqrt{6})}^{2}$

Step 3

Simplify the exponents for each term of the expansion.

$1 \cdot {(\sqrt{x})}^{2} \cdot {(\sqrt{6})}^{0} + 2 \cdot {(\sqrt{x})}^{1} \cdot {(\sqrt{6})}^{1} + 1 \cdot {(\sqrt{x})}^{0} \cdot {(\sqrt{6})}^{2}$

Step 4

Simplify each term.

Multiply ${(\sqrt{x})}^{2}$ by $1$ .

${(\sqrt{x})}^{2} \cdot {(\sqrt{6})}^{0} + 2 \cdot {(\sqrt{x})}^{1} \cdot {(\sqrt{6})}^{1} + 1 \cdot {(\sqrt{x})}^{0} \cdot {(\sqrt{6})}^{2}$

Rewrite ${\sqrt{x}}^{2}$ as $x$ .

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{x}$ as $x^{\frac{1}{2}}$ .

${(x^{\frac{1}{2}})}^{2} \cdot {(\sqrt{6})}^{0} + 2 \cdot {(\sqrt{x})}^{1} \cdot {(\sqrt{6})}^{1} + 1 \cdot {(\sqrt{x})}^{0} \cdot {(\sqrt{6})}^{2}$

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

$x^{\frac{1}{2} \cdot 2} \cdot {(\sqrt{6})}^{0} + 2 \cdot {(\sqrt{x})}^{1} \cdot {(\sqrt{6})}^{1} + 1 \cdot {(\sqrt{x})}^{0} \cdot {(\sqrt{6})}^{2}$

Combine $\frac{1}{2}$ and $2$ .

$x^{\frac{2}{2}} \cdot {(\sqrt{6})}^{0} + 2 \cdot {(\sqrt{x})}^{1} \cdot {(\sqrt{6})}^{1} + 1 \cdot {(\sqrt{x})}^{0} \cdot {(\sqrt{6})}^{2}$

Cancel the common factor of $2$ .

Cancel the common factor.

$x^{\frac{2}{2}} \cdot {(\sqrt{6})}^{0} + 2 \cdot {(\sqrt{x})}^{1} \cdot {(\sqrt{6})}^{1} + 1 \cdot {(\sqrt{x})}^{0} \cdot {(\sqrt{6})}^{2}$

Rewrite the expression.

$x^{1} \cdot {(\sqrt{6})}^{0} + 2 \cdot {(\sqrt{x})}^{1} \cdot {(\sqrt{6})}^{1} + 1 \cdot {(\sqrt{x})}^{0} \cdot {(\sqrt{6})}^{2}$

Simplify.

$x \cdot {(\sqrt{6})}^{0} + 2 \cdot {(\sqrt{x})}^{1} \cdot {(\sqrt{6})}^{1} + 1 \cdot {(\sqrt{x})}^{0} \cdot {(\sqrt{6})}^{2}$

Anything raised to $0$ is $1$ .

$x \cdot 1 + 2 \cdot {(\sqrt{x})}^{1} \cdot {(\sqrt{6})}^{1} + 1 \cdot {(\sqrt{x})}^{0} \cdot {(\sqrt{6})}^{2}$

Multiply $x$ by $1$ .

$x + 2 \cdot {(\sqrt{x})}^{1} \cdot {(\sqrt{6})}^{1} + 1 \cdot {(\sqrt{x})}^{0} \cdot {(\sqrt{6})}^{2}$

Simplify.

$x + 2 \cdot \sqrt{x} \cdot {(\sqrt{6})}^{1} + 1 \cdot {(\sqrt{x})}^{0} \cdot {(\sqrt{6})}^{2}$

Simplify.

$x + 2 \sqrt{x} \cdot \sqrt{6} + 1 \cdot {(\sqrt{x})}^{0} \cdot {(\sqrt{6})}^{2}$

Combine using the product rule for radicals.

$x + 2 \sqrt{6 x} + 1 \cdot {(\sqrt{x})}^{0} \cdot {(\sqrt{6})}^{2}$

Multiply ${(\sqrt{x})}^{0}$ by $1$ .

$x + 2 \sqrt{6 x} + {(\sqrt{x})}^{0} \cdot {(\sqrt{6})}^{2}$

Anything raised to $0$ is $1$ .

$x + 2 \sqrt{6 x} + 1 \cdot {(\sqrt{6})}^{2}$

Multiply ${(\sqrt{6})}^{2}$ by $1$ .

$x + 2 \sqrt{6 x} + {(\sqrt{6})}^{2}$

Rewrite ${\sqrt{6}}^{2}$ as $6$ .

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{6}$ as $6^{\frac{1}{2}}$ .

$x + 2 \sqrt{6 x} + {(6^{\frac{1}{2}})}^{2}$

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

$x + 2 \sqrt{6 x} + 6^{\frac{1}{2} \cdot 2}$

Combine $\frac{1}{2}$ and $2$ .

$x + 2 \sqrt{6 x} + 6^{\frac{2}{2}}$

Cancel the common factor of $2$ .

Cancel the common factor.

$x + 2 \sqrt{6 x} + 6^{\frac{2}{2}}$

Rewrite the expression.

$x + 2 \sqrt{6 x} + 6^{1}$

Evaluate the exponent.

$x + 2 \sqrt{6 x} + 6$

Expand Using the Binomial Theorem ( square root of x+ square root of 6)^2

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