Problem 52 - College Algebra

Question

The area of Sandy's yard is \(25 x^{2}-625 \mathrm{ft}^{2}\) . A patch of sod has an area of \(x^{2}-10 x+25 \mathrm{ft}^{2}\) . Divide the two areas and simplify to find how many pieces of sod Sandy needs to cover her yard.

Step-by-Step Solution

Answer

Sandy needs \(25(x+5)\) pieces of sod to cover her yard.

Step 1: Understand the problem

We are given two expressions representing areas: one for the yard and one for a piece of sod. We want to find out how many pieces of sod are needed to completely cover the yard. This involves dividing the yard area by the sod area.

Step 2: Write down the expressions

The area of the yard is given by the expression \(25x^2 - 625\) square feet. The area of one piece of sod is \(x^2 - 10x + 25\) square feet. We will divide these two expressions.

Step 3: Factor the expressions

Factor the quadratic expressions in the problem. The area of the yard, \(25x^2 - 625\), can be factored as \((5x + 25)(5x - 25)\) or \(25(x - 5)(x + 5)\). The expression for the sod, \(x^2 - 10x + 25\), can be factored as \((x - 5)^2\).

Step 4: Set up the division

Now divide the factored form of the yard area by the factored form of the sod area: \[\frac{25(x - 5)(x + 5)}{(x - 5)^2}\]

Step 5: Simplify the division

Simplify the division by canceling the common factor \(x - 5\):\[\frac{25(x + 5)}{x - 5}\]This reduces to \(25(x + 5)\).

Step 6: Find the number of sod pieces needed

To fully cover the yard, Sandy will need \(25(x + 5)\) pieces of sod. Substitute specific values for \(x\) if necessary to find the exact number of sod pieces.

Key Concepts

Quadratic Expressions

Quadratic expressions are polynomial expressions where the highest exponent of the variable is 2. Commonly, these expressions are structured in the format \(ax^2 + bx + c\) where \(a\), \(b\), and \(c\) are constants, and \(x\) is the variable.
A quadratic expression might look intimidating at first. However, it’s essential to become comfortable with them as they form the foundation for solving more complex algebraic problems. By factoring them, we can find roots and solve equations efficiently.

• Take \(x^2 - 10x + 25\) for example. Here, the \(x^2\) suggests it’s a quadratic expression.
• The expression \(25x^2 - 625\) is also quadratic and can often be simplified through factoring.

Notably, when dealing with quadratic expressions, it's helpful to look for patterns or special forms, such as perfect squares or the difference of squares, to facilitate factoring. Recognizing expressions like \((x-5)^2\) or \((x-5)(x+5)\) can significantly streamline solving or simplifying processes.

Simplifying Algebraic Fractions

Simplifying algebraic fractions involves reducing the fraction to its simplest form by canceling out common factors in the numerator and the denominator. This requires both an understanding of factoring polynomials and the structure of fractions.
When you're simplifying, your main focus should be:

• To ensure both parts of the algebraic fraction are fully factored.
• To cancel any common factors unearthed during the factoring process.

In the exercise, the yard's area is represented by the fraction \(\frac{25(x - 5)(x + 5)}{(x - 5)^2}\).

• The common factor here is \(x - 5\). Canceling it leaves you with \(25(x+5)\), a much simpler expression.

This simplification isn’t just mathematical housekeeping; it’s crucial for getting to the correct and simplest result when you calculate or solve equations.

Polynomial Division

Polynomial division, especially when encountered as part of a wider problem like this, can seem challenging. But, it can be straightforward if approached systematically.
There are several methods to divide polynomials, among which factorization is often the simplest. Here’s a quick breakdown of what polynomial division might involve when simplified through factoring:

• Factor both the numerator and the denominator.
• Cancel common components to simplify the division.

In our exercise, this method comes to the rescue. By rewriting the areas of the yard and the sod in their factored forms, you get \(\frac{25(x - 5)(x + 5)}{(x - 5)^2}\).

• Once factored, you easily cancel \(x-5\) from both the numerator and the denominator, turning the problem into a straightforward simplification yielding \(25(x + 5)\).

This approach not only provides the solution to how many pieces of sod are necessary, but also teaches the identification and cancellation of common factors in polynomial expressions.