Problem 35 - Beginning and Intermediate Algebra

Question

Use the binomial theorem to expand each expression. $$(b+3)^{5}$$

Step-by-Step Solution

Answer

The expanded expression for \((b+3)^5\) using the binomial theorem is: \((b+3)^5 = b^5 + 15b^4 + 90b^3 + 270b^2 + 405b + 243\)

Step 1: Identify the expansion terms using the Binomial Theorem formula

Using the Binomial Theorem, we need to expand the expression \((b+3)^5\), so we will have the following form: \((b+3)^5 = \sum_{k=0}^{5} {5 \choose k} b^{5-k} (3)^k\) Now, we need to evaluate each term in the sum individually.

Step 2: Calculate the binomial coefficients

We will calculate the binomial coefficients \({5 \choose k}\) for each term in the sum: \({5 \choose 0} = \frac{5!}{0!(5-0)!} = 1\) \({5 \choose 1} = \frac{5!}{1!(5-1)!} = 5\) \({5 \choose 2} = \frac{5!}{2!(5-2)!} = 10\) \({5 \choose 3} = \frac{5!}{3!(5-3)!} = 10\) \({5 \choose 4} = \frac{5!}{4!(5-4)!} = 5\) \({5 \choose 5} = \frac{5!}{5!(5-5)!} = 1\)

Step 3: Insert the coefficients and calculate each term individually

Now we will insert the coefficients back into our expression and calculate the individual terms: \((b+3)^5 = \sum_{k=0}^{5} {5 \choose k} b^{5-k} (3)^k\) \(= {5 \choose 0} b^{5} (3)^0 + {5 \choose 1} b^{4} (3)^1 + {5 \choose 2} b^{3} (3)^2 + {5 \choose 3} b^{2} (3)^3 + {5 \choose 4} b^{1} (3)^4 + {5 \choose 5} b^{0} (3)^5\) \(= 1b^5(1) + 5b^4(3) + 10b^3(9) + 10b^2(27) + 5b^1(81) + 1b^0(243)\)

Step 4: Simplify the expanded expression

Now, let's simplify the expanded expression: \((b+3)^5 = b^5 + 15b^4 + 90b^3 + 270b^2 + 405b + 243\) So, the final expanded expression for \((b+3)^5\) is: \((b+3)^5 = b^5 + 15b^4 + 90b^3 + 270b^2 + 405b + 243\)

Key Concepts

Binomial Coefficients

Binomial coefficients are the heart of the binomial theorem. They are the numerical factors that arise in the expansion of binomial expressions. You often see them expressed as "n choose k," written mathematically as \( \binom{n}{k} \). The formula to find these coefficients is:
\[\binom{n}{k} = \frac{n!}{k!(n-k)!}\]where \(!\) represents the factorial of a number.

• Step-by-step calculation: For each specific \(k\), calculate the binomial coefficient using the formula. For example, when expanding \((b+3)^5\), we need coefficients like \(\binom{5}{0}\), \(\binom{5}{1}\), etc.
• Use in expansions: In our specific example, \(\binom{5}{2} = 10\), which means that it's the coefficient for the term involving \(b^3\) and \(3^2\).

These numbers help determine the "weight" or importance of each term in the polynomial expansion.

Algebraic Expressions

Algebraic expressions are combinations of constants, variables, and mathematical operations such as addition, subtraction, multiplication, and exponentiation. An expression can be as simple as a single number or more complex with various terms.

• Variables and constants: In the expression \((b+3)^5\), \(b\) is the variable while \(3\) is the constant. Variables represent unknown values that we can change.
• Operations within expressions: Expressions like \(b^4(3)^1\) indicate that we raise \(b\) to the fourth power and multiply by \(3\) raised to the first power.

When expanding, each term of the expression becomes more precise, splitting into various smaller components involving coefficients, powers of variables, and constants.

Polynomial Expansion

The polynomial expansion process involves rewriting a binomial power expression as a long sum of terms. This methodolog is at the core of understanding how simple binomial products translate into more complex polynomial expressions.

• Using the Binomial Theorem: The Binomial Theorem helps you expand an expression like \((b+3)^5\) by giving you a structured way to calculate each term's coefficient, the power of \(b\), and any constants involved.
• Sum of terms: The expansion results in a series of terms: \(b^5\), \(15b^4\), \(90b^3\), \(270b^2\), \(405b\), and \(243\). Each term depends on the binomial coefficients and the powers of the original terms \(b\) and \(3\).

Polynomial expansion through the binomial theorem thus allows one to take a compact expression and detail it out, showing each piece in its simplest form, contributing to an overall polynomial.