Practice Problems
State the domain of each rational expression.
\(\textbf{1)}\) \( \displaystyle\frac{x^3+2x^2}{x^2+3x+2} \)
The domain is \((-\infty,-2) \cup (-2,-1) \cup (-1,\infty)\)
\(x^2+3x+2 \ne 0\)
\((x+2)(x+1) \ne 0\)
\(x \ne -2, x \ne -1\)
The domain is \((-\infty,-2) \cup (-2,-1) \cup (-1,\infty)\)
![Link to Youtube Video Solving Question Number 1](https://andymath.com/wp-content/uploads/2020/08/play-video.jpg")
\(\textbf{2)}\) \( \displaystyle\frac{x^4+2x^3-15x^2}{x^2-3x} \)
The domain is \((-\infty,0) \cup (0,3) \cup (3,\infty)\)
\(x^2-3x \ne 0\)
\((x)(x-3) \ne 0\)
\(x \ne 0, x \ne 3\)
The domain is \((-\infty,0) \cup (0,3) \cup (3,\infty)\)
\(\textbf{3)}\) \( \displaystyle\frac{x^2-6x-40}{2x^2+3x-20} \)
The domain is \((-\infty,-4) \cup (-4,\frac{5}{2}) \cup (\frac{5}{2},\infty)\)
\(2x^2+3x-20 \ne 0\)
\((2x-5)(x+4) \ne 0\)
\(x \ne -4, x \ne \frac{5}{2}\)
The domain is \((-\infty,-4) \cup (-4,\frac{5}{2}) \cup (\frac{5}{2},\infty)\)
\(\textbf{4)}\) \( \displaystyle\frac{25-x^2}{x^2+4x-5} \)
The domain is \((-\infty,-5) \cup (-5,1) \cup (1,\infty)\)
\(x^2+4x-5 \ne 0\)
\((x+5)(x-1) \ne 0\)
\(x \ne -5, x \ne 1\)
The domain is \((-\infty,-5) \cup (-5,1) \cup (1,\infty)\)
\(\textbf{5)}\) \( \displaystyle\frac{x^3+5x^2+3x+15}{x^3+3x} \)
The domain is \((-\infty,0) \cup (0,\infty)\)
\(x^3+3x \ne 0\)
\((x)(x^2+3) \ne 0\)
\(x \ne 0\)
The domain is \((-\infty,0) \cup (0,\infty)\)
\(\textbf{6)}\) \( \displaystyle\frac{x^3-2x^2+x}{x^2-2x+1} \)
The domain is \((-\infty,1) \cup (1,\infty)\)
\(x^2-2x+1 \ne 0\)
\((x-1)(x-1) \ne 0\)
\(x \ne 1\)
The domain is \((-\infty,1) \cup (1,\infty)\)
\(\textbf{7)}\) \( \displaystyle\frac{x^2-1}{x-1} \)
The domain is \((-\infty,1) \cup (1,\infty)\)
\(x-1 \ne 0\)
\(x \ne 1\)
The domain is \((-\infty,1) \cup (1,\infty)\)
\(\textbf{8)}\) \( \displaystyle\frac{x^3-1}{x^2-1} \)
The domain is \((-\infty,-1) \cup (-1,1) \cup (1,\infty)\)
\(x^2-1 \ne 0\)
\((x+1)(x-1) \ne 0\)
\(x \ne -1, x \ne 1\)
The domain is \((-\infty,-1) \cup (-1,1) \cup (1,\infty)\)
See Related Pages\(\)
