Problem 59 - College Algebra

Question

Explain how to use the Binomial Theorem to expand a binomial. Provide an example with your explanation.

Step-by-Step Solution

Answer

The Binomial Theorem provides an effective way to expand binomials. For example, using the theorem, \( (x+y)^3 = x^3 + 3x^2y + 3xy^2 + y^3 \). The coefficients in the expansion follow a pattern and can be easily identified using Pascal's Triangle.

Step 1: Understanding the Binomial Theorem

The Binomial theorem states that when expanding the expression \( (a+b)^n \), the equations forms a pattern governed by the formula \( (a+b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k}b^k \). In this formula, \( a \) and \( b \) are the terms of the binomial, \( n \) is the power to which the binomial is raised, \( k \) varies from 0 to \( n \), and \( \binom{n}{k} \) represents binomial coefficients.

Step 2: Finding Binomial Coefficients using Pascal's Triangle

The coefficients of a binomial expansion follow a pattern and can be found easily using Pascal's Triangle. To create Pascal's Triangle, start with a row of 1. Each ensuing row is gotten by adding the number above and to the right with the number above and to the left. If a number is not present, it is assumed to be 0.

Step 3: Expanding a Binomial Using the Binomial Theorem - Example

Given the binomial \( (x+y)^3 \), use the binomial theorem to expand it. First, identify \( n \), \( a \), and \( b \) in the binomial theorem equation, which are 3, x, y respectively. Now, find the coefficients using Pascal's triangle for row 3. The coefficients from Pascal triangle will be 1, 3, 3, 1. Insert these values into the equation. This results in \( (x+y)^3 = 1.x^3.y^0 + 3.x^2.y^1 + 3.x^1.y^2 + 1.x^0.y^3 \). Which simplifies to \( (x+y)^3 = x^3 + 3x^2y + 3xy^2 + y^3 \).

Step 4: Note Regarding Expanding Binomials

Remember that in the Binomial Theorem, the powers in the terms of the binomial start with the power equal to \( n \) for the first term (in the case, \( x \)) and decrease until reaching 0 while for the second term (in this case \( y \)), the powers start from 0 and increase until reaching \( n \). The sum of the powers in each term of the expansion is always equal to \( n \).