Problem 33 - Algebra and Trigonometry Real Mathematics, Real People

Question

Use the Binomial Theorem to expand and simplify the expression. \((4 y-3)^{3}\)

Step-by-Step Solution

Answer

The expanded and simplified form of the given binomial expression \((4 y-3)^{3}\) is \(64y^3 - 144y^2 + 108y - 27\).

Step 1: Identify the Values of a, b, and n

In the expression \( (4y - 3)^{3} \), \(4y\) is the first term (a), \(-3\) is the term to be subtracted (b), and 3 is the power to which the binomial is raised (n). So the values of a, b, and n are \(4y\), \(-3\), and \(3\) respectively.

Step 2: Apply the Binomial Theorem and Expand the Binomial

The binomial theorem states that \((a + b)^n = \Sigma_{k=0}^{n} {n \choose k} a^{n-k} b^{k}\). Let's apply this to \( (4y - 3)^{3} \). So, the expression expands to: \[{3 \choose 0}(4y)^3(-3)^0 + {3 \choose 1 }(4y)^2(-3)^1 + {3 \choose 2 }(4y)^1(-3)^2 + {3 \choose 3 }(4y)^0(-3)^3.\]

Step 3: Calculate the Terms

Each term involves the product of a binomial coefficient, a power of \(4y\), and a power of \(-3\). Let's calculate them:Term 1: \({3 \choose 0}(4y)^3(-3)^0 = 1*64y^3*1 = 64y^3\)Term 2: \({3 \choose 1}(4y)^2(-3)^1 = 3*16y^2*-3 = -144y^2\)Term 3: \({3 \choose 2}(4y)^1(-3)^2 = 3*4y*9 = 108y\)Term 4: \({3 \choose 3}(4y)^0(-3)^3 = 1*1*-27 = -27\)

Step 4: Consolidate the Terms

The complete expression becomes:\[64y^3 - 144y^2 + 108y - 27.\]

Key Concepts

Polynomial Expansion

When we talk about polynomial expansion, we often refer to the process of expressing a power of a binomial as a sum of terms involving coefficients, variables, and exponents. In simple terms, it's transforming an expression like \((a + b)^n\) into a series of terms. This transformation is facilitated by using the Binomial Theorem, which provides a structured way to expand such expressions.

For example, when expanding \((4y - 3)^3\), the polynomial form consists of terms like \(64y^3\), \(-144y^2\), \(108y\), and \(-27\).

Each term in the expanded expression can be seen as a combination of powers of \(4y\) and \(-3\), multiplied by a specific coefficient. This polynomial represents the expanded form of the original expression.

Combinatorics

Combinatorics is a fascinating branch of mathematics. It studies the counting, arrangement, and combination of objects. This concept comes into play with the Binomial Theorem. It helps determine the coefficients of terms in the expansion.

In the context of polynomial expansion, the formula for each term's coefficient is based on combinations, denoted as \({n \choose k}\). This represents the number of different ways to choose \(k\) elements from \(n\) elements, and it plays a crucial role in the expansion.

In our example, we used combinatorial numbers like \({3 \choose 0}\), \({3 \choose 1}\), etc., to establish the coefficients that go with each term in the expansion of \((4y - 3)^3\).

Combinatorial calculations are useful beyond polynomial expansion. They appear in probability, statistics, and numerous other areas of mathematics.

Binomial Coefficients

Binomial coefficients are the stars of the Binomial Theorem show. They are the coefficients of the variables' terms in a binomial expansion. The coefficients are denoted by \({n \choose k}\). This expression is read as "n choose k."

To calculate binomial coefficients, you usually use a formula involving factorials: \[{n \choose k} = \frac{n!}{k!(n-k)!}.\] This formula counts how many ways you can select \(k\) items from \(n\) items without regard to order.

In the expression \((4y - 3)^3\), binomial coefficients like \({3 \choose 0}\), \({3 \choose 1}\), \({3 \choose 2}\), and \({3 \choose 3}\) help us form the terms \(64y^3\), \(-144y^2\), \(108y\), and \(-27\). These coefficients determine how the terms are weighted in the final expanded polynomial expression.