Problem 3 - Algebra 1 Course in Mathematics for the IIT-JEE and Other Engineering Exams

Question

Expand \((2 x-3)^{6}\) by the binomial theorem.

Step-by-Step Solution

Answer

\((2x-3)^6 = 64x^6 - 576x^5 + 2160x^4 - 4320x^3 + 4860x^2 - 2916x + 729\).

Step 1: Understand the Binomial Theorem

The binomial theorem states that \((a + b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k\). In this case, \(a = 2x\), \(b = -3\), and \(n = 6\). We will use this theorem to expand \((2x-3)^6\).

Step 2: Calculate the Binomial Coefficients

The binomial coefficient \(\binom{n}{k}\) is calculated as \(\frac{n!}{k!(n-k)!}\). For \(n = 6\), calculate \(\binom{6}{0}, \binom{6}{1}, \binom{6}{2}, \ldots, \binom{6}{6}\). The values are 1, 6, 15, 20, 15, 6, and 1, respectively.

Step 3: Apply the Binomial Theorem

Substitute \(a = 2x\) and \(b = -3\) into the binomial formula for each term. Compute each term using the form \(\binom{6}{k} (2x)^{6-k} (-3)^k\) for \(k = 0\) to \(6\).

Step 4: Calculate Each Term

- When \(k = 0\), the term is \(1 \, (2x)^6 \, (-3)^0 = 64x^6\).- When \(k = 1\), the term is \(6 \, (2x)^5 \, (-3)^1 = -576x^5\).- When \(k = 2\), the term is \(15 \, (2x)^4 \, (-3)^2 = 2160x^4\).- When \(k = 3\), the term is \(20 \, (2x)^3 \, (-3)^3 = -4320x^3\).- When \(k = 4\), the term is \(15 \, (2x)^2 \, (-3)^4 = 4860x^2\).- When \(k = 5\), the term is \(6 \, (2x)^1 \, (-3)^5 = -2916x\).- When \(k = 6\), the term is \(1 \, (2x)^0 \, (-3)^6 = 729\).

Step 5: Combine the Terms

Add all the calculated terms: \(64x^6 - 576x^5 + 2160x^4 - 4320x^3 + 4860x^2 - 2916x + 729\). This sum is the expanded form of \((2x-3)^6\).

Key Concepts

Polynomial Expansion

Polynomial expansion involves expressing a power of a binomial as a sum of terms, making it more manageable to work with. It's like unfolding a compressed box to see everything that's inside. Using the Binomial Theorem is a common method for expanding binomials. This theorem helps us multiply a binomial by itself several times without doing each multiplication manually. For example, expanding \((2x - 3)^6\) directly could be cumbersome, but with the binomial theorem, we can systematically calculate each term.Polynomials are mathematical expressions that contain variables raised to powers, like \(x^2\) or \(x^3\). They can include constants and are added, subtracted, and multiplied together. Understanding polynomial expansion is crucial for solving this kind of algebraic problem because it allows us to convert a simple expression in its compact form into a detailed sum, which can be analyzed or solved further if necessary.

Algebraic Expressions

Algebraic expressions are the building blocks of algebra and are composed of variables, constants, and arithmetic operations like addition and multiplication. They can be as simple as \(2x\) or as complex as \(2x - 3)^6\). In the context of the binomial theorem, analyzing algebraic expressions allows us to substitute parts of them into the theorem's formula for expansion.Understanding how to manipulate algebraic expressions is key. This includes knowing how to distribute, combine like terms, and factorize. When you expand a binomial using the binomial theorem, each term generated is an algebraic expression in itself. By substituting values, like \(a = 2x\) and \(b = -3\), you can systematically break down the binomial and understand each individual part's contribution to the overall expression.

Factorials

Factorials are a mathematical operation denoted by an exclamation mark \(!\), and they are fundamental in calculating binomial coefficients. For a positive integer \(n\), the factorial \(n!\) is the product of all positive integers from 1 to \(n\). For example, \(6! = 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 720\).Factorials are used when determining terms in polynomial expansion via the binomial theorem. Specifically, the binomial coefficient \(\binom{n}{k}\) is calculated as \(\frac{n!}{k!(n-k)!}\). These coefficients are crucial because they dictate the weight of each term in the expansion. Understanding how to compute factorials makes it easier to implement the binomial theorem, allowing you to systematically identify and calculate each term when expanding a binomial.