Notes
Example Problem
Given: Rational Expression
\(\displaystyle \frac{x^2-9}{x^2+4x+3}\)
Step 1: Factor numerator and denominator
\(\displaystyle \frac{x^2-9}{x^2+4x+3}=\displaystyle \frac{(x+3)(x-3)}{(x+3)(x+1)}\)
Step 2: If possible, cancel common factors
\(\displaystyle \frac{(x+3)(x-3)}{(x+3)(x+1)}=\displaystyle \frac{(x-3)}{(x+1)}\)
Step 3: Add an \(x \ne\) for any roots that you cancel
\(\displaystyle \frac{(x-3)}{(x+1)} \,\,\,\, x\ne-3\)
Practice Problems
Simplify Each Rational Expression
\(\textbf{1)}\) \( \displaystyle\frac{x^3+2x^2}{x^2+3x+2} \)
The answer is \( \displaystyle\frac{x^2}{x+1}, \,\,\,\,\,\, x\ne-2 \)
\(\textbf{2)}\) \( \displaystyle\frac{x^4+2x^3-15x^2}{x^2-3x} \)
The answer is \( x(x+5), \,\,\, x\ne3, x\ne0 \)
\(\,\,\,\,\,\displaystyle\frac{x^4+2x^3-15x^2}{x^2-3x}\)
\(\,\,\,\,\,\displaystyle\frac{x^2\left(x+5\right)\left(x-3\right)}{x\left(x-3\right)}\)
\(\,\,\,\,\,\displaystyle\frac{x\left(x+5\right)}{1},\,\,\, x\ne3, x\ne0\)
\(\textbf{3)}\) \( \displaystyle\frac{x^2-6x-40}{2x^2+3x-20} \)
The answer is \( \displaystyle\frac{x-10}{2x-5}, \,\,\,\,\,\, x\ne -4 \)
\(\,\,\,\,\,\displaystyle\frac{x^2-6x-40}{2x^2+3x-20} \)
\(\,\,\,\,\,\displaystyle\frac{\left(x-10\right)\left(x+4\right)}{\left(2x-5\right)\left(x+4\right)} \)
\(\,\,\,\,\,\displaystyle\frac{\left(x-10\right)}{\left(2x-5\right)} \,\,\,x\ne -4 \)
\(\textbf{4)}\) \( \displaystyle\frac{25-x^2}{x^2+4x-5} \)
The answer is \( \displaystyle\frac{5-x}{x-1}, \,\,\,\,\,\, x\ne -5 \)
\(\,\,\,\,\,\, \displaystyle\frac{25-x^2}{x^2+4x-5} \)
\(\,\,\,\,\,\, \displaystyle\frac{(5+x)(5-x)}{(x+5)(x-1)} \)
\(\,\,\,\,\,\, \displaystyle\frac{(x+5)(5-x)}{(x+5)(x-1)} \)
\(\,\,\,\,\,\,\)The answer is \( \displaystyle\frac{5-x}{x-1}, \,\,\,\,\,\, x\ne -5 \)
\(\textbf{5)}\) \( \displaystyle\frac{x^3+5x^2+3x+15}{x^3+3x} \)
The answer is \( \displaystyle\frac{x+5}{x} \)
\(\,\,\,\,\,\displaystyle\frac{x^3+5x^2+3x+15}{x^3+3x}\)
\(\,\,\,\,\,\displaystyle\frac{x^2\left(x+5\right)+3\left(x+5\right)}{x\left(x^2+3\right)}\)
\(\,\,\,\,\,\displaystyle\frac{\left(x^2+3\right)\left(x+5\right)}{x\left(x^2+3\right)}\)
\(\,\,\,\,\,\displaystyle\frac{x+5}{x}\)
\(\textbf{6)}\) \( \displaystyle\frac{x^3-2x^2+x}{x^2-2x+1} \)
The answer is \( x, \,\,\,\,\,\, x\ne 1 \)
\(\,\,\,\,\,\displaystyle\frac{x^3-2x^2+x}{x^2-2x+1} \)
\(\,\,\,\,\,\displaystyle\frac{x\left(x^2-2x+1\right)^2}{x^2-2x+1} \)
\(\,\,\,\,\,x, \,\,\,\,\,\, x\ne 1 \)
\(\textbf{7)}\) \( \displaystyle\frac{x^2-1}{x-1} \)
The answer is \( x+1, \,\,\,\,\,\, x\ne1 \)
![Link to Youtube Video Solving Question Number 7](https://andymath.com/wp-content/uploads/2020/08/play-video.jpg")
\(\textbf{8)}\) \( \displaystyle\frac{x^3-1}{x^2-1} \)
The answer is \( \displaystyle\frac{x^2+x+1}{x+1} \,\,\,\,\,\, x\ne 1 \)
