Verified Answer :

  • The denominator is factored as \(x(1-y)\), leading to the form \(\frac{16x^2 + 21x + 54}{x(1-y)}\).
  • Restrictions for the expression are \(x eq 0\) and \(y eq 1\).
  • Dividing the terms by \(x\) results in the simplified form \(\frac{16x + 21 + \frac{54}{x}}{1-y}\).
  • The final expression is \(\boxed{\frac{16x + 21 + \frac{54}{x}}{1-y}}\).

Explanation

  1. Factorization and Domain Restrictions

The expression provided is \(\frac{16x^2 + 21x + 54}{x - yx}\). To understand its structure and behavior, we first examine the denominator, \(x - yx\). This expression can be factored by taking out the common factor \(x\), resulting in \(x(1 - y)\). Thus, the full expression is \(\frac{16x^2 + 21x + 54}{x(1 - y)}\). For this expression to be defined, the denominator must be non-zero, which requires \(x eq 0\) and \(y eq 1\).

  1. Simplifying the Expression

To simplify the expression, we can distribute the \(x\) in the denominator to each term in the numerator: \[\frac{16x^2 + 21x + 54}{x(1 - y)} = \frac{1}{\left(1 - y\right)} \cdot \left(\frac{16x^2}{x} + \frac{21x}{x} + \frac{54}{x}\right).\] Simplifying the terms inside the parentheses gives \(\left(16x + 21 + \frac{54}{x}\right)\). Therefore, the expression can be written as \(\frac{16x + 21 + \frac{54}{x}}{1 - y}\). This form clearly shows how the expression behaves as \(x\) becomes very large or very small, as well as its dependence on the variable \(y\).

Examples

Consider a scenario where you are calculating the average cost per unit produced, where the total cost is represented by a quadratic function \(16x^2 + 21x + 54\) and the number of units is given by \(x - yx\). This algebraic form helps managers analyze how the unit cost fluctuates based on production volume (\(x\)) and a scaling efficiency factor (\(y\)). By simplifying the expression, the manager can quickly determine how changing the production scale or the efficiency factor will impact the final unit cost.