Problem 38 - Algebra and Trigonometry with Analytic Geometry

Question

Express as a polynomial. $$ (x+3 y)^{3} $$

Step-by-Step Solution

Answer

The polynomial is \(x^3 + 9x^2y + 27xy^2 + 27y^3\).

Step 1: Understand the Problem

The problem asks us to express the expression \((x+3y)^3\) as a polynomial. This involves expanding the cube of a binomial.

Step 2: Recall the Binomial Theorem

To expand \((x+3y)^3\), we use the binomial theorem which states: \((a+b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k\). Here, \(a = x\), \(b = 3y\), and \(n = 3\).

Step 3: Calculate the coefficients

Calculate the binomial coefficients for \(n=3\):- \(\binom{3}{0} = 1\)- \(\binom{3}{1} = 3\)- \(\binom{3}{2} = 3\)- \(\binom{3}{3} = 1\)

Step 4: Apply the Binomial Theorem

Expand using the coefficients:\((x+3y)^3 = \binom{3}{0} x^3 (3y)^0 + \binom{3}{1} x^2 (3y)^1 + \binom{3}{2} x^1 (3y)^2 + \binom{3}{3} x^0 (3y)^3\).This simplifies to:\(x^3 + 3x^2(3y) + 3x(3y)^2 + (3y)^3\).

Step 5: Simplify the terms

Calculate each term:- \(x^3\) remains as is.- \(3x^2(3y) = 9x^2y\)- \(3x(3y)^2 = 27xy^2\)- \((3y)^3 = 27y^3\)

Step 6: Write the final polynomial

Combine the terms to express as a polynomial:\(x^3 + 9x^2y + 27xy^2 + 27y^3\).

Key Concepts

Binomial Theorem

The Binomial Theorem is a powerful tool in algebra that allows us to expand expressions of the form \((a + b)^n\). This theorem provides a systematic method for expanding any binomial raised to a positive integer power, \(n\). According to the theorem, each term in the expansion takes on the form \( \binom{n}{k} a^{n-k} b^k \), where \( \binom{n}{k} \) is the binomial coefficient, \(a\) and \(b\) are the terms being combined, and \(k\) is the specific term in the expansion.

This is particularly useful because it helps us break down complex polynomial expressions into simpler parts that can be more easily understood or computed. In our example, to expand \((x + 3y)^3\), we applied the theorem with \(a = x\), \(b = 3y\), and \(n = 3\). Using the binomial theorem not only simplifies the process but also ensures that nothing is overlooked during the expansion.

Binomial Coefficient

The binomial coefficients are the numbers that appear in the expansion of a binomial raised to a power. These coefficients can be calculated using the formula \( \binom{n}{k} = \frac{n!}{k!(n-k)!} \), where \(n!\) represents the factorial of \(n\).

In simpler terms, they are the coefficients for each term in the expansion of the polynomial. For \((x + 3y)^3\), we calculated the coefficients:

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

These numbers, known as the elements of Pascal's Triangle, help us organize and compute each corresponding term of the expanded polynomial efficiently.

The symmetry and pattern of these coefficients also provide valuable insights into algebraic relationships among binomial expressions.

Polynomial Simplification

Polynomial simplification involves breaking down expressions with multiple terms into a more concise and tractable form. After using the binomial theorem to expand a binomial expression, the next step is to simplify, combine, and rearrange like terms into a clearer polynomial form.

For our example of \((x + 3y)^3\), we initially expand the expression to \(x^3 + 3x^2(3y) + 3x(3y)^2 + (3y)^3\). Each of these terms can be simplified:

• \(x^3\) remains \(x^3\).
• \(3x^2(3y) = 9x^2y\).
• \(3x(3y)^2 = 27xy^2\).
• \((3y)^3 = 27y^3\).

Combining these simplifies the expression into \(x^3 + 9x^2y + 27xy^2 + 27y^3\).

This process of simplification is crucial in algebra as it makes the expression easier to work with or solve in subsequent operations.