4-2 Skills Practice Powers Of Binomials Answer Key

Mastering Binomial Expansion: A Complete Guide to 4-2 Skills Practice Problems

Understanding the powers of binomials is a cornerstone of algebra that unlocks doors to more advanced mathematics, from calculus to discrete math. The 4-2 skills practice section in many algebra textbooks is designed to move you from basic recognition to fluent application of the Binomial Theorem. This guide provides a comprehensive walkthrough, serving as a detailed answer key and conceptual tutorial. We will not just list answers; we will build the reasoning behind each step, ensuring you can tackle any similar problem with confidence. The core principle involves expanding expressions like (a + b)^n systematically, and the key tool is the Binomial Theorem, which connects algebra to the elegant patterns found in Pascal's Triangle.

The Foundation: The Binomial Theorem Explained

Before diving into the practice problems, a solid grasp of the theorem is non-negotiable. The Binomial Theorem states that for any positive integer n:

(a + b)^n = C(n,0)a^n b^0 + C(n,1)a^(n-1) b^1 + C(n,2)a^(n-2) b^2 + ... + C(n,n)a^0 b^n

Where C(n, r) (read as "n choose r") is the binomial coefficient, calculated as n! / (r! (n-r)!). These coefficients are precisely the numbers in the nth row of Pascal's Triangle. The exponents on a decrease from n to 0, while the exponents on b increase from 0 to n. This pattern is your roadmap for every expansion.

Step-by-Step Solutions for 4-2 Skills Practice

Let's apply this theorem to typical problems found in a 4-2 skills practice set. We will solve them methodically.

Problem 1: Expanding (x + 2)^4

This is a classic starting point. Here, a = x, b = 2, and n = 4.

1. Identify the row of Pascal's Triangle for n=4: The 4th row (starting from row 0) is 1, 4, 6, 4, 1. These are your coefficients C(4,0) through C(4,4).
2. Set up the terms: The exponents on x start at 4 and decrease. The exponents on 2 start at 0 and increase.
   • Term 1: C(4,0) * x^4 * 2^0 = 1 * x^4 * 1 = x^4
   • Term 2: C(4,1) * x^3 * 2^1 = 4 * x^3 * 2 = 8x^3
   • Term 3: C(4,2) * x^2 * 2^2 = 6 * x^2 * 4 = 24x^2
   • Term 4: C(4,3) * x^1 * 2^3 = 4 * x * 8 = 32x
   • Term 5: C(4,4) * x^0 * 2^4 = 1 * 1 * 16 = 16
3. Combine: (x + 2)^4 = x^4 + 8x^3 + 24x^2 + 32x + 16

Problem 2: Expanding (3a - b)^3

Note the subtraction. Treat b as -b. Here, a = 3a, b = -b, n = 3.

1. Pascal's Row for n=3: 1, 3, 3, 1
2. Set up terms, carefully handling the negative sign:
   • Term 1: C(3,0) * (3a)^3 * (-b)^0 = 1 * 27a^3 * 1 = 27a^3
   • Term 2: C(3,1) * (3a)^2 * (-b)^1 = 3 * 9a^2 * (-b) = -27a^2b
   • Term 3: C(3,2) * (3a)^1 * (-b)^2 = 3 * 3a * (b^2) = 9ab^2 (Note: (-b)^2 = +b^2)
   • Term 4: C(3,3) * (3a)^0 * (-b)^3 = 1 * 1 * (-b^3) = -b^3
3. Combine: (3a - b)^3 = 27a^3 - 27a^2b + 9ab^2 - b^3 The alternating signs are a direct result of the odd/even powers of the negative term.

Problem 3: Finding a Specific Term – The 4th Term of (2x + 5)^6

Often, you're asked for a specific term without expanding everything. The r+1th term in the expansion of (a+b)^n is given by: T_(r+1) = C(n, r) * a^(n-r) * b^r We want the 4th term, so r+1 = 4, meaning r = 3.

• a = 2x, b = 5, n = 6, r = 3
• T_4 = C(6, 3) * (2x)^(6-3) * (5)^3
• C(6,3) = 20 (from Pascal's row: 1,6,15,20,15,6,1)
• (2x)^3 = 8x^3
• 5^3 = 125
• T_4 = 20 * 8x^3 * 125 = 20 * 1000x^3 = 20,000x^3

Problem 4: Expanding with Negative Exponents – (x^2 - 1/x)^5

This tests your ability to handle fractional or negative exponents