Problem 35 - Contemporary Precalculus

Question

Expand and (where possible) simplify the expression. $$(1+\sqrt{3})^{4}+(1-\sqrt{3})^{4}$$

Step-by-Step Solution

Answer

Answer: The simplified form of the expression is \(200\).

Step 1: Expand the expression using the binomial theorem

The binomial theorem states that for any positive integer n, and any real numbers a and b: $$(a + b)^{n} = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^{k}$$ Expanding our expression using this formula, we get: \([(1+\sqrt{3})^{4}+(1-\sqrt{3})^{4}]= [\sum_{k=0}^{4} \binom{4}{k} 1^{4-k} (\sqrt{3})^{k}]+\left[\sum_{k=0}^{4} \binom{4}{k} 1^{4-k} (-\sqrt{3})^{k}\right]\)

Step 2: Calculate the terms

Now, we will calculate each term in the sums as follows: For the first sum, \(\binom{4}{0} 1^{4} (\sqrt{3})^{0} = 1\) \(\binom{4}{1} 1^{3} (\sqrt{3})^{1} = 4\sqrt{3}\) \(\binom{4}{2} 1^{2} (\sqrt{3})^{2} = 6\times 3 = 18\) \(\binom{4}{3} 1^{1} (\sqrt{3})^{3} = 4\times 3\sqrt{3} = 12\sqrt{3}\) \(\binom{4}{4} 1^{0} (\sqrt{3})^{4} = 81\) For the second sum, \(\binom{4}{0} 1^{4} (-\sqrt{3})^{0} = 1\) \(\binom{4}{1} 1^{3} (-\sqrt{3})^{1} = -4\sqrt{3}\) \(\binom{4}{2} 1^{2} (-\sqrt{3})^{2} = 6\times 3 = 18\) \(\binom{4}{3} 1^{1} (-\sqrt{3})^{3} = -4\times 3\sqrt{3} = -12\sqrt{3}\) \(\binom{4}{4} 1^{0} (-\sqrt{3})^{4} = 81\)

Step 3: Combine the like terms

Combine the terms for each sum and total expression as follows: \((1+\sqrt{3})^4 = 1 + 4\sqrt{3} + 18 + 12\sqrt{3} + 81 = 100 + 16\sqrt{3}\) \((1-\sqrt{3})^4 = 1 - 4\sqrt{3} + 18 - 12\sqrt{3} + 81 = 100 - 16\sqrt{3}\) Adding the two expressions: \([(1+\sqrt{3})^{4}+(1-\sqrt{3})^{4}]\) = \((100 + 16\sqrt{3}) + (100 - 16\sqrt{3}) = 200\) The expanded and simplified expression is \(200\).

Key Concepts

Polynomial Expansion

When dealing with expressions raised to a power, such as \((1 + \sqrt{3})^4\) and \((1 - \sqrt{3})^4\), we can use the Binomial Theorem to expand them efficiently. The theorem provides a formula that helps us write each power expression as a sum of terms involving binomial coefficients. These coefficients come from Pascal's Triangle and simplify the process of expansion.

• In our problem, we expand each binomial using: \((a + b)^{n} = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^{k}\).
• Here, \(a = 1\), \(b = \sqrt{3}\) or \(-\sqrt{3}\), and \(n = 4\).
• This helps us calculate each term like \(1^{4-k} (\sqrt{3})^{k}\) and then sum them up for the full expansion.

Every expansion results in a polynomial where each term is clearly defined by these coefficients and powers. This method is not only systematic but ensures every necessary component of the polynomial is accurately captured.
Understanding how to apply the Binomial Theorem can greatly reduce the complexity of polynomial expansion tasks, allowing for precise calculations.

Simplifying Expressions

After expanding polynomials, simplifying expressions involves combining like terms.This can drastically reduce the complexity of the expression. In the exercise, once the polynomial terms are fully expanded, we group and combine similar terms.

• The expression \((1 + \sqrt{3})^4\) results in terms like \(4\sqrt{3}\) and \(12\sqrt{3}\).
• Similarly, \((1 - \sqrt{3})^4\) yields terms such as \(-4\sqrt{3}\) and \(-12\sqrt{3}\).
• Each of these terms is organized by collecting coefficients of similar powers of \(\sqrt{3}\).

Once all like terms are combined, the expression is simplified. In the example, this results in two simpler polynomials: \(100 + 16\sqrt{3}\) and \(100 - 16\sqrt{3}\).
The final simplification step includes checking for any remaining operations or simplifications. Combining these two expanded forms \((100 + 16\sqrt{3}) + (100 - 16\sqrt{3}) = 200\) yields the cleaner solution.

Algebraic Manipulation

Algebraic manipulation is the process of using algebraic techniques to simplify, rearrange, or solve expressions.It often includes operations like expanding, factoring, and combining like terms. In this exercise, manipulation starts with the expansion by the Binomial Theorem.

• We substitute easy values (like \(a = 1\) and \(b = \sqrt{3}\)) into the binomial formula.
• Once expanded, we reorder terms to group like terms together, thus simplifying our calculations.
• Next, we consider the symmetry in these expressions, knowing \((1 + \sqrt{3})\) and \((1 - \sqrt{3})\) expansions lead to exact opposite terms with respect to \(\sqrt{3}\).

Upon completion, manipulation facilitates the understanding of how expressions interact and combine. In this exercise, seeing how the negativity of terms cancels out highlights the simplification towards \(200\).
Mastering algebraic manipulation allows for flexible problem-solving and adaptability in tackling various algebraic challenges.