Problem 36 - Beginning and Intermediate Algebra

Question

Use the binomial theorem to expand each expression. $$(t+9)^{3}$$

Step-by-Step Solution

Answer

Using the Binomial Theorem, we can expand the given expression \((t+9)^{3}\) as follows: Step 1: Identify n, a, and b: n = 3, a = t, b = 9 Step 2: Apply the Binomial Theorem formula: \((t + 9)^3 = \sum_{k=0}^{3} \binom{3}{k} t^{3-k} 9^k\) Step 3: Calculate each term of the sum: - k = 0: \(t^3\) - k = 1: \(27t^2\) - k = 2: \(243t\) - k = 3: \(729\) Step 4: Write the expanded form: \((t + 9)^3 = t^3 + 27t^2 + 243t + 729\)

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

In this expression, we have: - n = 3 (the exponent) - a = t (the first term inside the parentheses) - b = 9 (the second term inside the parentheses)

Step 2: Apply the Binomial Theorem formula

Now we will apply the Binomial Theorem formula: \((t + 9)^3 = \sum_{k=0}^{3} \binom{3}{k} t^{3-k} 9^k\)

Step 3: Calculate each term of the sum

Next, we'll calculate each term of the sum: - For k = 0: \(\binom{3}{0}t^{3-0}9^0 = 1 \cdot t^3 \cdot 1 = t^3\) - For k = 1: \(\binom{3}{1}t^{3-1}9^1 = 3 \cdot t^2 \cdot 9 = 27t^2\) - For k = 2: \(\binom{3}{2}t^{3-2}9^2 = 3 \cdot t \cdot 81 = 243t\) - For k = 3: \(\binom{3}{3}t^{3-3}9^3 = 1 \cdot 1 \cdot 729 = 729\)

Step 4: Write the expanded form

Combine the terms calculated in step 3 to get the expanded form: \((t + 9)^3 = t^3 + 27t^2 + 243t + 729\)

Key Concepts

Algebra Expansion

When we discuss algebra expansion, we are referring to the process of transforming expressions from their compact forms into sum forms using different algebraic methods. The expansion is like unwrapping a gift in math—it shows the component parts that make up an expression.

Using the binomial theorem for algebra expansion involves taking a binomial, like \((a + b)^n\), and rewriting it as a sum of terms. This process involves breaking down the power of a binomial into manageable parts, making it easier to understand and work with.

The beauty of algebra expansion is that it allows you to see the individual terms of the expression clearly. Once expanded, each term in the expression appears separately, often revealing patterns or terms that can simplify solving equations or inequalities.

A simple example is expanding \((t + 9)^3\). By expanding, we transform this compact notation into \(t^3 + 27t^2 + 243t + 729\). Every term of this expanded form contributes to a deeper understanding of the original expression's potential behavior and applications.

Polynomial Expression

A polynomial expression is a mathematical expression involving a sum of powers in one or more variables multiplied by coefficients. The essence of polynomial expressions is that they are constructed in a way that makes them highly versatile and applicable across different areas of mathematics.

Polynomials can range from simple ones like constants or linear expressions to more complicated ones like the cubic expression \((t + 9)^3\) we have expanded through the binomial theorem. It includes terms such as \(t^3\), \(27t^2\), \(243t\), and constant term \(729\), each showcasing a different degree.

• The degree of a polynomial is determined by the highest power of the variable.
• Polynomials are named based on their degree, like linear (degree 1), quadratic (degree 2), cubic (degree 3), and so on.
• Each term in a polynomial expression is structured as a product of a constant (the coefficient) and a variable raised to an exponent.

Understanding polynomials is key because they are central to many areas in algebra and calculus. They're used to model real-world situations, solve equations, and more, making them an essential concept in mathematics.

Combinatorial Coefficients

Combinatorial coefficients, often denoted as binomial coefficients, are central to expanding binomials using the binomial theorem. These coefficients are numeric factors that help determine the weight of each term when expanding expressions like \((t + 9)^3\).

We often represent combinatorial coefficients using the formula \(\binom{n}{k}\), which reads as 'n choose k'. This calculation helps find how many ways we can select k items from a total of n items without regard to order. They're found in every level of algebra dealing with expansions and much beyond in probability and statistics.

To compute these, we use the formula: \[\binom{n}{k} = \frac{n!}{k!(n-k)!}\]

For our example, some specific coefficients include:

• For \(k = 0\), \(\binom{3}{0} = 1\)
• For \(k = 1\), \(\binom{3}{1} = 3\)
• For \(k = 2\), \(\binom{3}{2} = 3\)
• For \(k = 3\), \(\binom{3}{3} = 1\)

These coefficients then appear as multipliers of each term in the expansion, affecting the sum's overall structure. They provide the mathematical balance needed to correctly expand and interpret the full polynomial.