Problem 28 - Elementary and Intermediate Algebra

Question

Perform the operations. Simplify, if possible. $$ \frac{9}{b^{2}-2 b+1}-\frac{2}{b-1} $$

Step-by-Step Solution

Answer

\(\frac{11 - 2b}{(b-1)^2}\)

Step 1: Identify the Denominators

The denominators in the two fractions are different. The first denominator is \(b^2 - 2b + 1\) and the second is \(b - 1\). Notice that the first denominator can be factored since it is a perfect square trinomial.

Step 2: Factor the Denominator

Factor the quadratic expression in the first denominator: \(b^2 - 2b + 1 = (b-1)^2\). Now, both denominators involve \(b-1\).

Step 3: Rewrite the First Fraction

Since \((b-1)^2 = (b-1)(b-1)\), rewrite the first fraction as: \(\frac{9}{(b-1)(b-1)}\).

Step 4: Find the Common Denominator

The common denominator of the fractions is \((b-1)^2\). The second fraction, \(\frac{2}{b-1}\), needs to be converted to this common denominator.

Step 5: Convert Second Fraction

To convert \(\frac{2}{b-1}\) to a fraction with denominator \((b-1)^2\), multiply the numerator and the denominator by \((b-1)\). This gives you \(\frac{2(b-1)}{(b-1)^2}\).

Step 6: Perform the Subtraction

With the common denominator, the expression becomes:\(\frac{9}{(b-1)^2} - \frac{2(b-1)}{(b-1)^2} \)Combine under a single denominator:\(\frac{9 - 2(b-1)}{(b-1)^2}\)

Step 7: Simplify the Numerator

Expand and simplify the numerator:\(9 - 2(b-1) = 9 - 2b + 2 = 11 - 2b\).

Step 8: Write the Simplified Expression

The simplified expression is:\(\frac{11 - 2b}{(b-1)^2}\). No further simplification is possible.

Key Concepts

Factoring Polynomials

Factoring polynomials is the process of breaking down a polynomial into a product of simpler terms or factors. It is an essential skill in algebra that helps in simplifying expressions and solving equations. For example, in the exercise given, the expression in the denominator, \(b^2 - 2b + 1\), is a perfect square trinomial. A perfect square trinomial takes the form \((a - b)^2 = a^2 - 2ab + b^2\).
For the trinomial \(b^2 - 2b + 1\), you can see that it matches the form of \((b - 1)^2\) because:

• \(a = b\)
• \(b = 1\)

Thus, \(b^2 - 2b + 1 = (b-1)^2\). Factoring allows the denominators to be expressed in a simplified form. This simplification is crucial as it sets the foundation for adding, subtracting, or simplifying algebraic fractions.

Common Denominator

Finding a common denominator is necessary when performing operations such as addition or subtraction on fractions. In algebraic fractions, the process is similar to arithmetic fractions.

In this exercise, the fractions \(\frac{9}{b^2 - 2b + 1}\) and \(\frac{2}{b-1}\) initially have different denominators. However, once the first denominator is factored to \((b-1)^2\), it shows that the \'commonality\' lies in \(b-1\) being a factor of both denominators.

To establish a common denominator for these fractions, consider the highest degree of \(b-1\) present, which is \((b-1)^2\). Converting \(\frac{2}{b-1}\) to have a denominator of \((b-1)^2\) involves multiplying both the numerator and denominator by \(b-1\):

\[\frac{2}{b-1} \cdot \frac{b-1}{b-1} = \frac{2(b-1)}{(b-1)^2}\]

This unification under a common denominator allows for the fractions to be written together, paving the way for subtraction.

Simplifying Expressions

Simplifying expressions involves reducing them to their simplest form. This includes combining like terms, performing arithmetic operations, and reducing fractions.

After creating a common denominator in the problem, the expression becomes:

• \(\frac{9}{(b-1)^2} - \frac{2(b-1)}{(b-1)^2}\)

With the common denominator already established, these can be combined into a single fraction:

\[\frac{9 - 2(b-1)}{(b-1)^2}\]

Next, simplify the numerator. Begin by expanding \(-2(b-1)\):

\(-2(b-1) = -2b + 2\)

Therefore, the simplified numerator becomes:

• \(9 - 2b + 2 = 11 - 2b\)

As a final step, since the numerator and denominator have no common factors other than 1, the simplified expression is:

\[\frac{11 - 2b}{(b-1)^2}\]

No further simplification is possible, so this is the reduced form of the original algebraic fractions.