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Chapter 3: Problem 13

Expand the indicated expression. $$ (2+\sqrt{3})^{4} $$

Short Answer

Expert verified
The expanded expression of \((2+\sqrt{3})^{4}\) is \(97 + 56\sqrt{3}\).

Step by step solution

01

Identify a and b values and simplify powers if possible

In this case, we have \(a=2\) and \(b=\sqrt{3}\) and the exponent \(n=4\).
02

Apply the binomial theorem and expand the expression

To expand the expression, we apply the binomial theorem to obtain the sum of all possible terms and their coefficients: \[ (2+\sqrt{3})^{4} = \sum_{k=0}^4 {4 \choose k} 2^{4-k} (\sqrt{3})^k \] Now, let's find each term in the sum:
03

Calculate the binomial coefficients

1. For k = 0: \[ {4 \choose 0} = \frac{4!}{0!(4-0)!} = 1 \] 2. For k = 1: \[ {4 \choose 1} = \frac{4!}{1!(4-1)!} = 4 \] 3. For k = 2: \[ {4 \choose 2} = \frac{4!}{2!(4-2)!} = 6 \] 4. For k = 3: \[ {4 \choose 3} = \frac{4!}{3!(4-3)!} = 4 \] 5. For k = 4: \[ {4 \choose 4} = \frac{4!}{4!(4-4)!} = 1 \]
04

Plug in the binomial coefficients and simplify the powers of a and b

Now substitute the values of the binomial coefficients and simplify the powers of a and b: 1. For k = 0: \[ {4 \choose 0} 2^{4-0} (\sqrt{3})^0 = 1 \cdot 2^4 \cdot 1 = 16 \] 2. For k = 1: \[ {4 \choose 1} 2^{4-1} (\sqrt{3})^1 = 4 \cdot 2^3 \cdot \sqrt{3} = 32\sqrt{3} \] 3. For k = 2: \[ {4 \choose 2} 2^{4-2} (\sqrt{3})^2 = 6 \cdot 2^2 \cdot 3 = 72 \] 4. For k = 3: \[ {4 \choose 3} 2^{4-3} (\sqrt{3})^3 = 4 \cdot 2^1 \cdot 3\sqrt{3} = 24\sqrt{3} \] 5. For k = 4: \[ {4 \choose 4} 2^{4-4} (\sqrt{3})^4 = 1 \cdot 1 \cdot 9 = 9 \]
05

Add all the terms together to obtain the final expression

Finally, add all the terms together to get the expanded expression: \[ (2+\sqrt{3})^{4} = 16 + 32\sqrt{3} + 72 + 24\sqrt{3} + 9 = 97 + 56\sqrt{3} \] Therefore, the expanded expression for \((2+\sqrt{3})^{4}\) is \(97 + 56\sqrt{3}\).

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