Problem 47 - Precalculus: Mathematics for Calculus

Question

Simplify using the Binomial Theorem. $$\frac{(x+h)^{3}-x^{3}}{h}$$

Step-by-Step Solution

Answer

The simplified expression is \(3x^2 + 3xh + h^2\).

Step 1: Recall the Binomial Theorem

The Binomial Theorem allows us to expand expressions of the form \((a+b)^n\). The theorem states that \((a+b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k\). Here, \(a\) is \(x\), \(b\) is \(h\), and \(n\) is 3.

Step 2: Apply the Binomial Theorem to \((x+h)^3\)

Expand \((x+h)^3\) using the Binomial Theorem: \((x+h)^3 = \binom{3}{0}x^3h^0 + \binom{3}{1}x^2h^1 + \binom{3}{2}x^1h^2 + \binom{3}{3}x^0h^3\). Calculate the coefficients: \((x+h)^3 = x^3 + 3x^2h + 3xh^2 + h^3\).

Step 3: Substitute the Expansion into the Original Expression

Substitute the expanded form of \((x+h)^3\) into the given expression: \(\frac{(x+h)^{3}-x^{3}}{h} = \frac{(x^3 + 3x^2h + 3xh^2 + h^3) - x^3}{h}\).

Step 4: Simplify the Numerator

Combine like terms in the numerator:The \(x^3\) terms cancel out, leaving: \(3x^2h + 3xh^2 + h^3\).

Step 5: Factor and Simplify

Factor \(h\) from the numerator: \(\frac{h(3x^2 + 3xh + h^2)}{h}\).Cancel out \(h\) from the numerator and the denominator to get the simplified expression: \(3x^2 + 3xh + h^2\).

Key Concepts

Polynomial Expansion

The Binomial Theorem is a crucial tool in algebra that allows us to expand expressions like \((a+b)^n\). This theorem helps us break down complex expressions into a sum of terms involving powers of \(a\) and \(b\). For example, if we have \((x+h)^3\), using the Binomial Theorem, we can express it as:

• \(x^3\) which is \(\binom{3}{0}x^3h^0\)
• \(3x^2h\) which is \(\binom{3}{1}x^2h^1\)
• \(3xh^2\) which is \(\binom{3}{2}x^1h^2\)
• \(h^3\) which is \(\binom{3}{3}x^0h^3\)

These terms reflect the various combinations of \(x\) and \(h\) raised to specific powers, weighted by the binomial coefficients \(\binom{n}{k}\), corresponding to the entries in Pascal’s Triangle. This process is fundamental in transforming expressions into more manageable forms.

Algebraic Simplification

Algebraic simplification is all about making expressions easier to work with by reducing the number of terms. In our specific problem, once we've expanded \((x+h)^3\), the next step is to simplify. After substituting our expansion into the initial expression, the numerator becomes \((x^3 + 3x^2h + 3xh^2 + h^3) - x^3\). Here, the \(x^3\) terms cancel each other out, which is a common occurrence when simplifying algebraic expressions.
The result is a simpler expression: \(3x^2h + 3xh^2 + h^3\). The next step involves factoring: we see that each term in the expression includes the factor \(h\). By factoring \(h\) out, the numerator becomes \(h(3x^2 + 3xh + h^2)\). This step is crucial as it allows us to cancel the \(h\) in the numerator with the \(h\) in the denominator, leading to the reduced form \(3x^2 + 3xh + h^2\).
Simplification, therefore, involves strategic steps of canceling, combining like terms, and factoring wherever possible.

Calculus Preparation

The simplified expression \(3x^2 + 3xh + h^2\) is not just an algebraic result but also a stepping stone toward calculus concepts, particularly limits and derivatives. In calculus, one frequently encounters expressions like \(\frac{(x+h)^n - x^n}{h}\) while dealing with difference quotients, which are foundational for deriving the derivative of a function.
The process of simplifying such expressions is crucial when we consider the limit as \(h\) approaches zero. In our expression's context, recognizing that \(h\) can be factored out and then eliminated informs us about the function's derivative at a specific point.

• Helps understand rates of change
• Prepares for the derivative concept using limit formulations
• Crucial for learning how small changes in variables affect the overall function

Understanding these simplifications enhance our preparation for calculus, providing insight into instantaneous rates of change and the slope of tangent lines at a point on a curve.