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Chapter 7: Problem 43

Use Pascal's triangle to simplify the indicated expression. $$ (2-\sqrt{3})^{5} $$

Short Answer

Expert verified
Using Pascal's triangle and the Binomial Theorem, the simplified expression for \((2-\sqrt{3})^{5}\) is \(1082 - 1403\sqrt{3}\).

Step by step solution

01

Refer to Pascal's Triangle

Refer to the rows in Pascal's triangle to find the coefficients in the 5th row (where \(n=5\)). The first row has one term (starting at row 0), and each successive row begins and ends with 1. So, the 5th row is 1, 5, 10, 10, 5, 1.
02

Apply the Binomial Theorem

Apply the Binomial Theorem to expand the expression \((2-\sqrt{3})^{5}\), using the coefficients from Pascal's triangle and the given values of \(a\) and \(b\). The Binomial Theorem gives: \((2-\sqrt{3})^{5} = \binom{5}{0}(2)^{5}(-\sqrt{3})^{0} + \binom{5}{1}(2)^{4}(-\sqrt{3})^{1} + \binom{5}{2}(2)^{3}(-\sqrt{3})^{2} + \binom{5}{3}(2)^{2}(-\sqrt{3})^{3} + \binom{5}{4}(2)^{1}(-\sqrt{3})^{4} + \binom{5}{5}(2)^{0}(-\sqrt{3})^{5}\)
03

Simplify the terms

Simplify each term in the expression by calculating the binomial coefficients, powers of \(a=2\), and powers of \(b=-\sqrt{3}\). The simplified terms are as follows: 1. \(\binom{5}{0}(2)^{5}(-\sqrt{3})^{0} = 1 \cdot 32 \cdot 1 = 32\) 2. \(\binom{5}{1}(2)^{4}(-\sqrt{3})^{1} = 5 \cdot 16 \cdot (-\sqrt{3}) = -80\sqrt{3}\) 3. \(\binom{5}{2}(2)^{3}(-\sqrt{3})^{2} = 10 \cdot 8 \cdot 3 = 240\) 4. \(\binom{5}{3}(2)^{2}(-\sqrt{3})^{3} = 10 \cdot 4 \cdot (-27\sqrt{3}) = -1080\sqrt{3}\) 5. \(\binom{5}{4}(2)^{1}(-\sqrt{3})^{4} = 5 \cdot 2 \cdot 81 = 810\) 6. \(\binom{5}{5}(2)^{0}(-\sqrt{3})^{5} = 1 \cdot 1 \cdot (-243\sqrt{3}) = -243\sqrt{3}\)
04

Add the simplified terms

Add the simplified terms together to obtain the final simplified expression: \((2-\sqrt{3})^5 = 32 - 80\sqrt{3} + 240 - 1080\sqrt{3} + 810 - 243\sqrt{3}\) Combine the terms and get: \((2-\sqrt{3})^5 = (32 + 240 + 810) + (-80\sqrt{3} - 1080\sqrt{3} - 243\sqrt{3})\) \((2-\sqrt{3})^5 =1082 - 1403\sqrt{3}\)

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