Question

1. simplify the rational expression \\(\frac{2x}{x^2 - 4} - \frac{1}{x^2 - 3x + 2} + \frac{x + 1}{x^2 + x - 2}\\). identify any non - permissible values.

Explanation:

Step 1: Factor the denominators

First, factor each quadratic (or cubic) denominator:

• For \(x^{2}-4\), use the difference of squares: \(x^{2}-4=(x - 2)(x + 2)\)
• For \(x^{2}-3x + 2\), factor the quadratic: \(x^{2}-3x + 2=(x - 1)(x - 2)\)
• For \(x^{2}+x - 2\), factor the quadratic: \(x^{2}+x - 2=(x + 2)(x - 1)\)

So the expression becomes:
\(\frac{2x}{(x - 2)(x + 2)}-\frac{1}{(x - 1)(x - 2)}+\frac{x + 1}{(x + 2)(x - 1)}\)

Step 2: Find the least common denominator (LCD)

The denominators are \((x - 2)(x + 2)\), \((x - 1)(x - 2)\), and \((x + 2)(x - 1)\). The LCD is the product of the unique factors, each raised to the highest power they appear. So the LCD is \((x - 2)(x + 2)(x - 1)\)

Step 3: Rewrite each fraction with the LCD

• For \(\frac{2x}{(x - 2)(x + 2)}\), multiply numerator and denominator by \((x - 1)\): \(\frac{2x(x - 1)}{(x - 2)(x + 2)(x - 1)}\)
• For \(\frac{1}{(x - 1)(x - 2)}\), multiply numerator and denominator by \((x + 2)\): \(\frac{1(x + 2)}{(x - 1)(x - 2)(x + 2)}\)
• For \(\frac{x + 1}{(x + 2)(x - 1)}\), multiply numerator and denominator by \((x - 2)\): \(\frac{(x + 1)(x - 2)}{(x + 2)(x - 1)(x - 2)}\)

Now the expression is:
\(\frac{2x(x - 1)}{(x - 2)(x + 2)(x - 1)}-\frac{x + 2}{(x - 1)(x - 2)(x + 2)}+\frac{(x + 1)(x - 2)}{(x + 2)(x - 1)(x - 2)}\)

Step 4: Combine the fractions

Since the denominators are the same, we can combine the numerators:
\[

$$\begin{align*} &\frac{2x(x - 1)-(x + 2)+(x + 1)(x - 2)}{(x - 2)(x + 2)(x - 1)}\\ &=\frac{2x^{2}-2x - x - 2+(x^{2}-2x+x - 2)}{(x - 2)(x + 2)(x - 1)}\\ &=\frac{2x^{2}-3x - 2+x^{2}-x - 2}{(x - 2)(x + 2)(x - 1)}\\ &=\frac{3x^{2}-4x - 4}{(x - 2)(x + 2)(x - 1)} \end{align*}$$

\]

Step 5: Factor the numerator

Factor \(3x^{2}-4x - 4\):
\(3x^{2}-4x - 4=(3x + 2)(x - 2)\)

So the expression becomes:
\(\frac{(3x + 2)(x - 2)}{(x - 2)(x + 2)(x - 1)}\)

Step 6: Cancel common factors

Cancel the common factor of \((x - 2)\) (note that \(x
eq2\) to avoid division by zero):
\(\frac{3x + 2}{(x + 2)(x - 1)}\)

Step 7: Identify non - permissible values

Non - permissible values are the values that make the original denominators zero.

• For \(x^{2}-4=(x - 2)(x + 2)=0\), \(x = 2\) or \(x=-2\)
• For \(x^{2}-3x + 2=(x - 1)(x - 2)=0\), \(x = 1\) or \(x = 2\)
• For \(x^{2}+x - 2=(x + 2)(x - 1)=0\), \(x=-2\) or \(x = 1\)

So the non - permissible values are \(x=-2\), \(x = 1\), and \(x = 2\)

Answer:

The simplified form of the rational expression is \(\frac{3x + 2}{(x + 2)(x - 1)}\) (with \(x
eq - 2,1,2\)) and the non - permissible values are \(x=-2\), \(x = 1\), and \(x = 2\)