91Ó°ÊÓ

Factoring quadratics is an essential skill when working with rational expressions. A quadratic expression is typically in the form of \(ax^2 + bx + c\). The goal of factoring is to express this quadratic as a product of two binomials.
To factor successfully, you need to find two numbers that multiply to \(a \times c\) and add up to \(b\).
In our example, the expression \(2p^2 + p - 1\) needed to be factored. First, we multiply the leading coefficient \(2\) by the constant \(-1\), giving us \(-2\). We seek two numbers that multiply to \(-2\) and add to \(1\). These numbers are \(2\) and \(-1\).
We then rewrite the middle term:

• \(2p^2 + 2p - p - 1\)

Group and factor by grouping:

• \[2p(p + 1) - 1(p + 1) = (2p - 1)(p + 1)\]

Now, the quadratic expression is factored, making it easier to simplify further in rational expressions.

Dividing fractions is a step in simplifying complex rational expressions. This involves understanding that dividing by a fraction is the same as multiplying by its reciprocal.
The reciprocal of a fraction \(\frac{a}{b}\) is \(\frac{b}{a}\).
In our problem, we had to divide the first fraction by the second. This means:

• \[\frac{6(2p^2 + p - 1)}{4(p + 1)^2} \div \frac{3(2p - 1)}{2(p + 5)}\]

Convert this division into a multiplication:

• \[ \frac{6(2p^2 + p - 1)}{4(p + 1)^2} \times \frac{2(p + 5)}{3(2p - 1)} \]

This step is crucial as it allows us to handle the problem using multiplication, making the process of canceling common factors straightforward later on.

Canceling common factors is the final critical step in simplifying rational expressions. It requires identifying and removing factors that appear in both the numerator and the denominator of a fraction.
This is possible only if the terms are factored completely.During the simplification process, we need to carefully look for common factors:

• In our expression: \[\frac{6 \cdot (2p-1)(p+1)}{4 \cdot (p+1)^2} \times \frac{2(p+5)}{3(2p-1)}\]

We notice that \((2p-1)\) and \((p+1)\) can be canceled:

• The \((2p-1)\) cancel each other since they appear in both the numerator and the denominator.
• One factor of \((p+1)\) in the numerator cancels with one factor in the denominator.

After canceling, the expression simplifies significantly:

• \[\frac{12(p + 5)}{12} = p + 5\]

The simplification results in a more manageable form, showing how effective it is to cancel out common factors correctly.