Problem 16 - Algebra Form and Function

Question

Is the expression a polynomial in the given variable? $$ \left(a+\frac{1}{x}\right)^{2}-\left(a-\frac{1}{x}\right)^{2}, \text { in } a $$

Step-by-Step Solution

Answer

Expression: $$ \left(a+\frac{1}{x}\right)^{2}-\left(a-\frac{1}{x}\right)^{2} $$ Answer: Yes, the given expression is a polynomial in the variable 'a'.

Step 1: Rewrite the expression

We have the expression: $$ \left(a+\frac{1}{x}\right)^{2}-\left(a-\frac{1}{x}\right)^{2} $$ First, we'll expand both squares and then simplify the expression.

Step 2: Expand the squares

Apply the binomial theorem to expand each squared term: $$ \left(a^{2}+2a\frac{1}{x}+\frac{1}{x^{2}}\right)-\left(a^{2}-2a\frac{1}{x}+\frac{1}{x^{2}}\right) $$

Step 3: Simplify the expression

Now subtract the expressions in parentheses: $$ a^{2}+2a\frac{1}{x}+\frac{1}{x^{2}}-a^{2}+2a\frac{1}{x}-\frac{1}{x^{2}} $$ Combine like terms: $$ 2a\frac{1}{x}+2a\frac{1}{x}=4a\frac{1}{x} $$

Step 4: Determine if the expression is a polynomial in 'a'

The simplified expression is $$ 4a\frac{1}{x} $$ A polynomial in 'a' should have the form $$ c_{n}a^{n}+c_{n-1}a^{n-1}+\ldots+c_{1}a+c_{0} $$ where the coefficients 'c' are constants (they don't depend on the variable 'a') and 'n' is a non-negative integer. Our simplified expression can be written as: $$ a\left(4\frac{1}{x}\right) $$ which has the form of a polynomial in 'a', as all the coefficients are constants (they don't depend on the variable 'a') and 'n' (the power of 'a') is a non-negative integer (1 in this case). So, the given expression is a polynomial in the variable 'a'.

Key Concepts

Understanding Binomial Expansion

Binomial expansion is a powerful algebraic tool used to expand expressions raised to a power. It relies on the binomial theorem, which provides a formula to expand expressions like \((a + b)^n\). In our exercise, we use this concept to expand the squares: \(\left(a+\frac{1}{x}\right)^{2}\) and \(\left(a-\frac{1}{x}\right)^{2}\).

Here's how it works:

• Identify \(a\) and \(b\). In the expressions \((a + b)^2\), \(a\) is the variable and \(b\) is \(\frac{1}{x}\).
• Apply the formula: \((a+b)^2 = a^2 + 2ab + b^2\).
• For \(\left(a+\frac{1}{x}\right)^{2}\), it becomes \(a^2 + 2a\left(\frac{1}{x}\right) + \left(\frac{1}{x}\right)^2\).

By breaking down each term, you can see how they form the expanded polynomial. Binomial expansion makes it easier to simplify complex expressions like the one in our exercise.

Simplifying Expressions

Simplification is all about reducing expressions to their simplest form while retaining their original value. In our problem, after expanding the squares, we need to simplify the expression:

\[(a^2 + 2a\frac{1}{x} + \frac{1}{x^2}) - (a^2 - 2a\frac{1}{x} + \frac{1}{x^2})\]

Follow these steps to simplify:

• Distribute any negative signs across the terms in parentheses.
• Combine like terms: Look for terms involving the same powers or variables, like \(2a\frac{1}{x}\).
• Cancel terms where possible. For instance, \(a^2\) and \(-a^2\) cancel out.
• Identify and sum terms: Combine \(2a\frac{1}{x}\) and \(2a\frac{1}{x}\) to get \(4a\frac{1}{x}\).

The simplified result is a much cleaner expression, \(4a\frac{1}{x}\), which helps in determining its polynomial nature.

Recognizing Polynomial Form

Polynomials are algebraic expressions made up of terms where variables are raised to non-negative integer powers. The general form of a polynomial in variable \(a\) is:

\[c_na^n + c_{n-1}a^{n-1} + \ldots + c_1a + c_0\]

To check if an expression is a polynomial in a given variable like \(a\):

• Ensure that the variable \(a\) is raised only to non-negative integer powers.
• Check that the coefficients are constants, not containing the variable \(a\).

In the expression \(4a\frac{1}{x}\), \(a\) is to the power of 1, and \(\frac{1}{x}\) is a constant with respect to \(a\). Thus, the expression aligns with the polynomial form definition for variable \(a\). Understanding what qualifies as polynomial form helps you simplify and analyze algebraic expressions efficiently.