Problem 11 - College Algebra

Question

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

Step-by-Step Solution

Answer

The simplified form of the binomial \((3x+y)^3\) is \(27x^3 + 27x^2y + 9xy^2 + y^3\).

Step 1: Identify the components

Identify the components of the binomial. Here, our 'a' component is '3x', our 'b' component is 'y', and the power 'n' is 3.

Step 2: Application of the Binomial Theorem

Apply the Binomial Theorem which states that \((a+b)^n = \sum_{k=0}^{n} C(n, k) a^{n-k} b^k\), where \(C(n, k)\) are binomial coefficients. Therefore, we can write \((3x+y)^3 = C(3, 0) (3x)^3 y^0 + C(3, 1) (3x)^2 y^1 + C(3, 2) (3x)^1 y^2 + C(3, 3) (3x)^0 y^3 \)

Step 3: Calculate the Binomial Coefficients

Calculate the Binomial Coefficients. They are computed as \(C(n, k)=\frac{n!}{k!(n-k)!}\), where 'n!' denotes n-factorial, the product of all positive integers till n. Here, \(C(3, 0) = 1, C(3, 1) = 3, C(3, 2) = 3\), and \(C(3, 3) = 1\)

Step 4: Substitute the Binomial Coefficients and Simplify

Substitute the calculated binomial coefficients and simplify the result. This simplification would lead to \(1 * (3x)^3 * y^0 + 3 * (3x)^2 * y + 3 * (3x) * y^2 + 1 * y^3 = 27x^3 + 27x^2y + 9xy^2 + y^3\)

Key Concepts

Binomial Theorem

When we have an algebraic expression such as equal to the sum of its separate terms after being subtracted from each other.terms to get the final expanded form. It's one of the fundamental concepts in algebra that facilitates handling complex polynomial power expansions.

Imagine you're trying to open a combination lock that requires three numbers in the correct order. Think of the Binomial Theorem as a guide that tells you all the possible combinations so you eventually hit upon the correct one without trying each one by hand. Similarly, when you expand binomials, the theorem ensures you capture all the necessary terms.

Binomial coefficients

The binomial coefficients, commonly presented as \(C(n, k)\), are the specific numbers that feature as multipliers in the Binomial Theorem. Each coefficient corresponds to an individual term in the expansion and represents the number of ways to choose \(k\) items from \(n\) options, without considering the order — akin to choosing toppings for your pizza, where the order in which you add them doesn't matter.

To calculate these coefficients, we use factorial notation in a fractional form as \(C(n, k) = \frac{n!}{k!(n-k)!}\). With this, it's simple to figure out that in the given problem \(C(3, 1)\) equals 3 because there are three different ways to pick one item out of three. Understanding these coefficients gives you the insight to anticipate the number of terms and their values when expanding binomials.

Algebraic expression simplification

Simplifying algebraic expressions is a bit like cleaning your room: you want to organize similar items together to make it look tidy and manageable. When you apply the Binomial Theorem and expand the terms, you're essentially left with an untidy room full of mathematical objects. Simplifying means combining like terms — those with the same variables raised to the same power.

For example, in the expression \(27x^3 + 27x^2y + 9xy^2 + y^3\), there's no further simplification required because there are no like terms to combine. It's already neatly sorted. The key is to always look for terms you can merge to make the expression as straightforward as possible, which makes it easier to work with for further mathematical operations.

Factorial notation

The factorial notation is symbolized by an exclamation mark (!) and refers to the product of all positive integers up to a given number. For instance, the term \(5!\) (read as '5 factorial') is shorthand for \(5 \times 4 \times 3 \times 2 \times 1\). This notation is indispensable when working with binomial coefficients because it gives you a compact way to express the calculation of the coefficients.

As you saw in the problem, calculating \(C(3, 2)\) requires understanding factorial notation since you'll need to compute \(\frac{3!}{2!(3-2)!}\) to arrive at the coefficient value. Embracing this notation is crucial when working with combinatorics, probabilities, and, as you've seen, binomial expansions.