Problem 38 - A Graphical Approach to College Algebra

Question

Write the binomial expansion for each expression. $$\left(\frac{1}{k}-\sqrt{3} p\right)^{3}$$

Step-by-Step Solution

Answer

The expansion is \( \frac{1}{k^3} - \frac{3\sqrt{3}p}{k^2} + \frac{9p^2}{k} - 3\sqrt{3}p^3 \).

Step 1: Understand the Formula

First, recall the binomial theorem which states that for any positive integer \( n \), \((a + b)^n = \sum_{k=0}^n \binom{n}{k} a^{n-k} b^k\). Here, \( a = \frac{1}{k} \), \( b = -\sqrt{3}p \), and \( n = 3 \).

Step 2: Apply the Binomial Theorem

Use the binomial theorem to expand \( \left(\frac{1}{k} - \sqrt{3} p\right)^3 \). The terms are given by the formula: \[ \sum_{j=0}^{3} \binom{3}{j} \left(\frac{1}{k}\right)^{3-j} \left(-\sqrt{3}p\right)^j \].

Step 3: Calculate the Binomial Coefficients

Compute the binomial coefficients for \( n = 3 \), which are \( \binom{3}{0} = 1 \), \( \binom{3}{1} = 3 \), \( \binom{3}{2} = 3 \), and \( \binom{3}{3} = 1 \).

Step 4: Calculate Each Term in the Expansion

Calculate each term using the formula in Step 2: - For \( j = 0 \), the term is \( 1 \cdot \left(\frac{1}{k}\right)^3 \cdot (1) = \frac{1}{k^3} \).- For \( j = 1 \), the term is \( 3 \cdot \left(\frac{1}{k}\right)^2 \cdot (-\sqrt{3}p) = -\frac{3\sqrt{3}p}{k^2} \).- For \( j = 2 \), the term is \( 3 \cdot \left(\frac{1}{k}\right)^1 \cdot (\sqrt{3}p)^2 = \frac{9p^2}{k} \).- For \( j = 3 \), the term is \( 1 \cdot (1) \cdot (-\sqrt{3}p)^3 = -3\sqrt{3}p^3 \).

Step 5: Combine All Terms

Write down the complete expansion by combining all calculated terms with the correct sign: \[ \frac{1}{k^3} - \frac{3\sqrt{3}p}{k^2} + \frac{9p^2}{k} - 3\sqrt{3}p^3 \].

Key Concepts

Binomial Theorem

The Binomial Theorem is a fundamental principle in algebra that provides a quick way to expand expressions that are raised to a power. It applies to expressions of the form \((a + b)^n\), where \(a\) and \(b\) are any numbers or algebraic expressions, and \(n\) is a non-negative integer.

When you apply the binomial theorem, you rely on a formula that involves binomial coefficients. These coefficients are found in Pascal's Triangle and are represented by \(\binom{n}{k}\), meaning "\(n\) choose \(k\)". This gives the number of ways to choose \(k\) elements from a set of \(n\). In the expansion, you would multiply each term by these coefficients.

For example, if we want to expand \((a + b)^3\), the theorem tells us:

• The term for \(k = 0\) is \(\binom{3}{0} a^3 b^0 = a^3\).
• The term for \(k = 1\) is \(\binom{3}{1} a^2 b = 3a^2b\).
• The term for \(k = 2\) is \(\binom{3}{2} a b^2 = 3ab^2\).
• The term for \(k = 3\) is \(\binom{3}{3} b^3 = b^3\).

This results in \(a^3 + 3a^2b + 3ab^2 + b^3\).Using this theorem simplifies the process of expanding such expressions without tedious direct multiplication.

Polynomial Expansion

Polynomial expansion involves expressing a polynomial, such as \((x + y)^3\), in an extended form. The expansion breaks down the terms and provides a detailed representation of how each part contributes to the whole expression.

The process of polynomial expansion using the Binomial Theorem ensures that we correctly account for the coefficients and powers of each term separately. In an exercise like \((\frac{1}{k} - \sqrt{3} p)^3\), we treat it as expanding \(a + b\), where \(a = \frac{1}{k}\) and \(b = -\sqrt{3} p\).

Through expansion, we create a sum of terms where each term consists of a product of the binomial coefficient, a powered variant of \(a\), and a powered variant of \(b\).

• The first term comes from \((\frac{1}{k})^3\).
• The second term comes from multiplying \(3\times(\frac{1}{k})^2\) by \(-\sqrt{3}p\).
• And so on, until all terms are expanded and simplified.

Polynomial expansions are immensely helpful in simplifying calculations and in understanding the contributions of individual parts in an algebraic expression. It's particularly useful in calculus and when solving for roots and factors of polynomials.

Algebraic Expressions

Algebraic expressions form the basic building blocks of algebra and consist of variables, constants, and operations. They vary in complexity from a simple term like \(x\) to more complicated expressions like \(\left(\frac{1}{k} - \sqrt{3} p\right)^3\).Understanding the components of algebraic expressions is crucial for performing operations such as addition, subtraction, multiplication, division, and expansion.
Variables in expressions, like \(x\) or \(p\), represent numbers whose exact values are often unknown but can be solved through equations. Constants are numbers that remain unchanged, such as \(\frac{1}{k}\) where \(k\) is a constant.In many homework exercises, we manipulate these expressions using rules and theorems, such as the Binomial Theorem, to discover more about their properties or to simplify them. Understanding how to expand and simplify complex algebraic expressions paves the way for solving high-level mathematical problems efficiently.