Problem 21 - Contemporary Precalculus

Question

Expand and (where possible) simplify the expression. $$(x+2)^{4}$$

Step-by-Step Solution

Answer

Answer: The expanded and simplified form of the expression (x+2)^4 is x^4 + 8x^3 + 24x^2 + 32x + 16.

Step 1: Binomial theorem formula

The binomial theorem formula is given by: $$(a+b)^n = \sum_{k=0}^{n} (\comb{n}{k}a^{n-k}b^k)$$ where \(n \comb{k}\) is the combination of picking k items from a set of n items, i.e.: $$\comb{n}{k} = \frac{n!}{k!(n-k)!}$$ In our case, \(a=x, b=2, n=4\)

Step 2: Calculate the binomial coefficients

We need to calculate the binomial coefficients for \(k=0\) to \(k=4\). $$\comb{4}{0} = \frac{4!}{0!(4-0)!} = 1$$ $$\comb{4}{1} = \frac{4!}{1!(4-1)!} = 4$$ $$\comb{4}{2} = \frac{4!}{2!(4-2)!} = 6$$ $$\comb{4}{3} = \frac{4!}{3!(4-3)!} = 4$$ $$\comb{4}{4} = \frac{4!}{4!(4-4)!} = 1$$

Step 3: Expand each term using the binomial theorem formula

Now expand each term using the binomial theorem formula: $$(x+2)^4 = \sum_{k=0}^{4} (\comb{4}{k}x^{4-k}2^k)$$ $$= \comb{4}{0}x^{4-0}2^0 + \comb{4}{1}x^{4-1}2^1 + \comb{4}{2}x^{4-2}2^2 + \comb{4}{3}x^{4-3}2^3 + \comb{4}{4}x^{4-4}2^4$$ $$= x^4 + 4x^32^1 + 6x^22^2 + 4x^12^3 + 2^4$$

Step 4: Simplify the expression

Combine and simplify the terms: $$= x^4 + 4x^32 + 6x^24 + 4x8 + 16$$ $$= x^4 + 8x^3 + 24x^2 + 32x + 16$$ The expanded and simplified expression for \((x+2)^4\) is: $$x^4 + 8x^3 + 24x^2 + 32x + 16$$

Key Concepts

Binomial Coefficients

When dealing with the expansion of binomials, binomial coefficients are the numbers that appear as multipliers of the terms in the expanded polynomial. They play a critical role in determining the weight each term carries in the expression.Binomial coefficients can be found using the combination formula, which for any non-negative integer values of n and k is denoted as \[\begin{equation}\comb{n}{k} = \frac{n!}{k!(n-k)!}\end{equation}\]In this notation, '!' represents the factorial operation, which means to multiply a series of descending natural numbers. For instance, \[\begin{equation}4! = 4 \times 3 \times 2 \times 1\end{equation}\]These coefficients correspond to the entries in Pascal's triangle. For example, the coefficients for expanding a binomial to the fourth power (such as in the exercise \[\begin{equation}(x+2)^{4}\end{equation}\]are the numbers from the fifth row of Pascal's triangle, which are 1, 4, 6, 4, and 1, as calculated in step 2 of the solution.

Binomial Theorem Formula

The binomial theorem provides a quick way to expand binomials raised to any given power without manually multiplying the binomial by itself multiple times. The general form of the binomial theorem for the expansion of the form \[\begin{equation}(a+b)^n\end{equation}\]is given by \[\begin{equation}(a+b)^n = \sum_{k=0}^{n} (\comb{n}{k}a^{n-k}b^k)\end{equation}\]The symbol \[\begin{equation}\sum_{k=0}^{n}\end{equation}\]is the summation sign, indicating that the expression following it is to be added together for each value of k from 0 to n. The binomial coefficient \[\begin{equation}\comb{n}{k}\end{equation}\]multiplies each term, determining its coefficient in the expanded form. This formula reveals the powerful connection between algebra and combinatorics, as each coefficient corresponds to the number of combinations for selecting k elements from a set of n.

Combinations

Combinations refer to a way of selecting items from a collection, such that the order of selection does not matter. In mathematics, they are a fundamental part of combinatorics, which is the study of counting and arrangement possibilities. The number of ways to choose k items from a set of n distinct items is given by the combination formula:\[\begin{equation}\comb{n}{k} = \frac{n!}{k!(n-k)!}\end{equation}\]This concept is essential when expanding binomials using the binomial theorem since the coefficients in the expansion correspond to the number of combinations of k elements chosen from n.For example, suppose we have a set of four elements and want to know how many ways we can select two elements. Using the combination formula, we find that \[\begin{equation}\comb{4}{2} = \frac{4!}{2!(4-2)!} = 6\end{equation}\]Hence, there are six different combinations, which is also reflected in the binomial expansion of \[\begin{equation}(x+2)^{4}\end{equation}\]as the coefficient of the third term \[\begin{equation}x^{4-2}2^{2}\end{equation}\]Enhancing understanding of combinations can improve a student's ability to work with the binomial theorem and assess probabilities in various situations.