Expand Using the Binomial Theorem (x+ square root of 3)^2

${(x + \sqrt{3})}^{2}$

Step 1

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

$\sum_{k = 0}^{2} \frac{2!}{(2 - k)! k!} \cdot {(x)}^{2 - k} \cdot {(\sqrt{3})}^{k}$

Step 2

Expand the summation.

$\frac{2!}{(2 - 0)! 0!} {(x)}^{2 - 0} \cdot {(\sqrt{3})}^{0} + \frac{2!}{(2 - 1)! 1!} {(x)}^{2 - 1} \cdot {(\sqrt{3})}^{1} + \frac{2!}{(2 - 2)! 2!} {(x)}^{2 - 2} \cdot {(\sqrt{3})}^{2}$

Step 3

Simplify the exponents for each term of the expansion.

$1 \cdot {(x)}^{2} \cdot {(\sqrt{3})}^{0} + 2 \cdot {(x)}^{1} \cdot {(\sqrt{3})}^{1} + 1 \cdot {(x)}^{0} \cdot {(\sqrt{3})}^{2}$

Step 4

Simplify each term.

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

${(x)}^{2} \cdot {(\sqrt{3})}^{0} + 2 \cdot {(x)}^{1} \cdot {(\sqrt{3})}^{1} + 1 \cdot {(x)}^{0} \cdot {(\sqrt{3})}^{2}$

Anything raised to $0$ is $1$ .

$x^{2} \cdot 1 + 2 \cdot {(x)}^{1} \cdot {(\sqrt{3})}^{1} + 1 \cdot {(x)}^{0} \cdot {(\sqrt{3})}^{2}$

Multiply $x^{2}$ by $1$ .

$x^{2} + 2 \cdot {(x)}^{1} \cdot {(\sqrt{3})}^{1} + 1 \cdot {(x)}^{0} \cdot {(\sqrt{3})}^{2}$

Simplify.

$x^{2} + 2 \cdot x \cdot {(\sqrt{3})}^{1} + 1 \cdot {(x)}^{0} \cdot {(\sqrt{3})}^{2}$

Simplify.

$x^{2} + 2 x \cdot \sqrt{3} + 1 \cdot {(x)}^{0} \cdot {(\sqrt{3})}^{2}$

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

$x^{2} + 2 x \sqrt{3} + {(x)}^{0} \cdot {(\sqrt{3})}^{2}$

Anything raised to $0$ is $1$ .

$x^{2} + 2 x \sqrt{3} + 1 \cdot {(\sqrt{3})}^{2}$

Multiply ${(\sqrt{3})}^{2}$ by $1$ .

$x^{2} + 2 x \sqrt{3} + {(\sqrt{3})}^{2}$

Rewrite ${\sqrt{3}}^{2}$ as $3$ .

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{3}$ as $3^{\frac{1}{2}}$ .

$x^{2} + 2 x \sqrt{3} + {(3^{\frac{1}{2}})}^{2}$

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

$x^{2} + 2 x \sqrt{3} + 3^{\frac{1}{2} \cdot 2}$

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

$x^{2} + 2 x \sqrt{3} + 3^{\frac{2}{2}}$

Cancel the common factor of $2$ .

Cancel the common factor.

$x^{2} + 2 x \sqrt{3} + 3^{\frac{2}{2}}$

Rewrite the expression.

$x^{2} + 2 x \sqrt{3} + 3^{1}$

Evaluate the exponent.

$x^{2} + 2 x \sqrt{3} + 3$

Expand Using the Binomial Theorem (x+ square root of 3)^2

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