Problem 9 - College Algebra

Question

In Exercises 9-30, use the Binomial Theorem to expand each binomial and express the result in simplified form. $$(x+2)^{3}$$

Step-by-Step Solution

Answer

The expanded form of \((x+2)^3\) is \(x^3 + 6x^2 + 12x + 8\).

Step 1: Write out each term of the Binomial Theorem for our expression

We plug in our \(a\), \(b\), and \(n\) into the Binomial Theorem and get \(C(3,0)*x^3*2^0 + C(3,1)*x^2*2^1 + C(3,2)*x^1*2^2 + C(3,3)*x^0*2^3\). Remember that \(C(n,k)\) (also denoted as \(nCk\)) stands for 'n choose k' and it can be calculated as \(n!/(k!(n-k)!)\) where \(!\) is the factorial function.

Step 2: Calculate the coefficients

Now we calculate each \(C(n,k)\) term. This gives us \(1*x^3*2^0 + 3*x^2*2^1 + 3*x^1*2^2 + 1*x^0*2^3\).

Step 3: Simplify each term

Simplify each term to get \(x^3 + 6x^2 + 12x + 8\).

Key Concepts

Binomial Expansion

Understanding the binomial expansion is essential for simplifying expressions where two terms are raised to a power. In our case, the binomial \(x+2\) raised to the third power is an embodiment of this concept. The binomial theorem provides a formula for expanding binomials of the form \( (a+b)^n \) into a sum involving terms of the form \( a^{n-k}b^{k} \) and binomial coefficients.

Let's apply the binomial theorem to the given exercise. The power of 3 indicates that there will be four terms in the expansion (since we count starting with zero). Specifically, for \( (x+2)^3 \) the expansion follows the pattern \( x^n, x^{n-1}2, x^{n-2}2^2, ..., 2^n \) with their respective binomial coefficients. In this pattern, \(n\) is the exponent and each term respectively decreases the power of \(x\) and increases the power of 2.

By understanding this pattern and the role of binomial coefficients, expanding any binomial expression becomes a more approachable task.

Factorial Function

Central to calculating binomial coefficients is the factorial function, denoted by \( ! \). The factorial of a non-negative integer \( n \) is the product of all positive integers less than or equal to \( n \). For instance, \( 3! = 3 \times 2 \times 1 = 6 \). It's also worth noting that \( 0! = 1 \)—a convention that aids in calculations involving the binomial theorem.

When we speak of the binomial theorem, the factorial function isn't just useful; it's indispensable for determining the binomial coefficients, which provide us the multipliers necessary for each term in the expansion. In our exercise, simplified factorial expressions play a part in calculating the coefficients \( C(3,0), C(3,1), C(3,2), \) and \( C(3,3) \) which indicate how many ways we can choose \( k \) elements from a larger set of \( n \) without regard to the order.

Binomial Coefficients

Now, let's delve into binomial coefficients. These are the numbers that appear as coefficients in the binomial theorem's expansion. They’re often referred to as 'n choose k' and are symbolized as \( C(n,k) \) or \( {n \choose k} \). A binomial coefficient represents the number of ways to pick \( k \) unordered outcomes from \( n \) possibilities, also known as combinations.

To calculate a binomial coefficient, we use the formula \( C(n,k) = \frac{n!}{k!(n-k)!} \), involving factorial functions. In our original exercise, the coefficients are \( C(3,0), C(3,1), C(3,2), \) and \( C(3,3) \) which, when calculated using the factorial function, give us 1, 3, 3, and 1, respectively. These coefficients dictate how many of each term we will have in our final expanded form, resulting in the simplified expression \( x^3 + 6x^2 + 12x + 8 \) for the original binomial \( (x+2)^3 \).

Mastery of binomial coefficients is not only crucial for the binomial theorem but also for understanding broader concepts in probability and combinations.