Problem 11 - Precalculus Mathematics for Calculus

Question

Use Pascal's triangle to expand the expression. $$\left(x^{2} y-1\right)^{5}$$

Step-by-Step Solution

Answer

\(x^{10}y^5 - 5x^8y^4 + 10x^6y^3 - 10x^4y^2 + 5x^2y - 1\)

Step 1: Identify the general formula

To expand the expression \((x^2 y - 1)^5\) using Pascal's triangle, we first recognize that it follows the pattern \((a + b)^n\). Here, \(a = x^2 y\), \(b = -1\), and \(n = 5\).

Step 2: Write out Pascal's triangle

Pascal's triangle gives us the binomial coefficients for the expansion. For \(n = 5\), the coefficients are in the sixth row: 1, 5, 10, 10, 5, 1.

Step 3: Apply the binomial theorem

The binomial theorem states that \((a + b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k\). Use the coefficients from Step 2 to expand:\[((x^2 y) - 1)^5 = 1(x^2 y)^5(-1)^0 + 5(x^2 y)^4(-1)^1 + 10(x^2 y)^3(-1)^2 + 10(x^2 y)^2(-1)^3 + 5(x^2 y)^1(-1)^4 + 1(x^2 y)^0(-1)^5.\]

Step 4: Simplify each term

Calculate each term one by one:1. \(1(x^2 y)^5(-1)^0 = x^{10}y^5\).2. \(5(x^2 y)^4(-1)^1 = -5(x^2)^4y^4 = -5x^8y^4\).3. \(10(x^2 y)^3(-1)^2 = 10(x^2)^3y^3 = 10x^6y^3\).4. \(10(x^2 y)^2(-1)^3 = -10(x^2)^2y^2 = -10x^4y^2\).5. \(5(x^2 y)^1(-1)^4 = 5x^2y\).6. \(1(x^2 y)^0(-1)^5 = -1\).

Step 5: Write the expanded expression

Combine the simplified terms to get the complete expansion:\[x^{10}y^5 - 5x^8y^4 + 10x^6y^3 - 10x^4y^2 + 5x^2y - 1.\]

Key Concepts

Pascal's Triangle

Pascal's Triangle is a fascinating mathematical tool that provides us with a quick way to determine the coefficients needed for binomial expansions. If you've ever struggled with figuring out what numbers should multiply each term in a polynomial like \(a + b\)^n, then Pascal's Triangle is your best friend. Pascal's Triangle is structured like a triangular array of numbers. Each row corresponds to the coefficients of the expanded polynomial expression for a specific power of \(n\). For example, the row that corresponds to \(n = 5\) is 1, 5, 10, 10, 5, 1. This tells us that when we expand \(x^2y - 1\)^5, these numbers will be the multipliers (or coefficients) in the expanded form.
To construct Pascal's Triangle, start with a '1' at the top. Each subsequent number in the rows is the sum of the two numbers directly above it from the previous row. Reflect on how easily this approach saves you from complex calculations as it grows larger with each power of \(n\). This simplicity is why Pascal's Triangle remains an essential concept in algebra and polynomial expansion.

Polynomial Expansion

Expanding polynomials, especially ones with higher powers, can seem intimidating at first glance. However, with some straightforward strategies in hand, you can simplify the process significantly.
Key among these strategies is the use of the Binomial Theorem, which tells us how to expand expressions of the form \(a + b\)^n into a sum of terms involving coefficients, powers of \(a\), and powers of \(b\). With the polynomial \(x^2y - 1\)^5, each term in the expansion corresponds to a part of this sum. Applying the theorem, we multiply each term generated from Pascal's Triangle by the appropriate powers of \(x^2y\) and \(-1\), then find the sum. Using the coefficients from Pascal's Triangle, start from \(x^{10}y^{5}\), then subtract and add progressively less powerful terms like \(-5x^{8}y^{4}\) and \(10x^{6}y^{3}\), until reaching an altogether simple constant term, often marked by \(-1\) for this example.
All of these intricate details combine to form the full expansion, delivering a complete polynomial that retains all of the initial expression's characteristics while displaying them across various terms.

Algebraic Expressions

Algebraic expressions are collections of numbers, letters (which typically represent variables), and operation symbols, all woven together to describe a particular mathematical relationship or problem. When dealing with expressions where variables are raised to powers and combined with constants, like in the exercise \(x^2y - 1\)^5, it's crucial to understand how each part functions within an equation.
Here, \(x^2y\) represents a term with multiple variables and a combined degree, and \(-1\) as a lone constant that affects each term in the expansion by alternately adding and subtracting. Being adept in handling algebraic expressions involves recognizing opportunities to simplify and restructure them to reveal their intricacies, much like peeling back layers. Skilled use of tools like the Binomial Theorem and Pascal's Triangle help achieve this by breaking down complex expressions into manageable components.
Mastering these concepts empowers you with the ability to analyze and expand similar complex algebraic expressions with confidence and clarity.