Problem 88 - Beginning and Intermediate Algebra

Question

Find the following special products. $$\left(b-\frac{1}{5}\right)\left(b+\frac{1}{5}\right)$$

Step-by-Step Solution

Answer

The simplified product of the given expression is \(\boxed{b^2 - \frac{1}{25}}\).

Step 1: Identify the values of p and q

In this case, our p is "b" and our q is "1/5".

Step 2: Apply the difference of squares formula

Using the difference of squares formula, we can now find the product: \((p - q)(p + q) = p^2 - q^2\) Substitute the values of p and q: \((b - \frac{1}{5})(b + \frac{1}{5}) = b^2 - \left(\frac{1}{5}\right)^2\)

Step 3: Simplify the expression

Now, we need to simplify the expression by squaring q (1/5) and subtracting it from the square of p (b): \(b^2 - \left(\frac{1}{5}\right)^2 = b^2 - \frac{1}{25}\) The final product is: \(\boxed{b^2 - \frac{1}{25}}\)

Key Concepts

Difference of Squares Formula

The difference of squares formula is a useful algebraic identity for transforming the product of two binomials into a simpler expression. It is typically used when you have two identical terms with opposite signs, multiplied together, which is:

• \((p - q)(p + q) = p^2 - q^2\)

This formula recognizes that the middle terms (the products of the inner and outer parts of the binomial) cancel each other out. Thus, only the square of the first term subtracted by the square of the second term remains.
In our original exercise, we used the difference of squares formula to find the product of \((b - \frac{1}{5})(b + \frac{1}{5})\). First, identify \(p\) and \(q\) as \(b\) and \(\frac{1}{5}\), respectively. Then substitute them into the formula to simplify the expression. This results in \(b^2 - \left(\frac{1}{5}\right)^2\).
Understanding this concept will make recognizing when and how to apply the difference of squares formula much easier in your algebra problems, leading to efficient solutions.

Polynomial Simplification

Simplifying polynomials is a critical skill in algebra that allows you to manage complex expressions and equations with greater ease. It involves reducing the expression to its simplest form by performing operations such as addition, subtraction, multiplication, and factoring.

• Combine like terms, where you add or subtract coefficients of the same degree.
• Apply known algebraic identities to facilitate the transformation, like the difference of squares.

In the exercise, once the term \((b - \frac{1}{5})(b + \frac{1}{5})\) was expressed using the difference of squares formula, it needed to be simplified further. This meant calculating \(\left(\frac{1}{5}\right)^2\), which is \(\frac{1}{25}\). Subtracting this result from \(b^2\) gave us the cleaner expression, \(b^2 - \frac{1}{25}\).
Mastering polynomial simplification is crucial, as it lays the groundwork for more advanced topics in mathematics and helps ensure accuracy in problem-solving.

Binomial Products

Binomial products involve the multiplication of two expressions, each consisting of two terms. Understanding how to efficiently handle these products is key to simplifying algebraic expressions and solving equations.
These products can often be tackled using specific identities or formulas, such as:

• The square of a binomial: \((a + b)^2 = a^2 + 2ab + b^2\)
• The sum and difference of the same terms, leading to the difference of squares: \((a - b)(a + b) = a^2 - b^2\)

In our exercise, the binomial product was handled using the difference of squares identity because it fit the form \((b - \frac{1}{5})(b + \frac{1}{5})\). This allowed us to instantly simplify the expression to a difference of squares, \(b^2 - \left(\frac{1}{5}\right)^2\), without needing to expand and combine terms manually.
Grasping binomial products and patterns helps solve a variety of algebraic problems more efficiently and creates a foundation for understanding more complex algebraic structures.