"

14 Simplifying Rational Expressions

* A rational expression
is an expression that can be simplified  in the form \displaystyle{ \frac{P}{Q} } where P and  Q are polynomials.
* To simplify a rational expression, we can:
       – combine numerator and denominator into single quotients, then divide :       \displaystyle{ \frac{\frac{P}{Q}}{\frac{S}{T}}  = \frac{P}{Q} . \frac{T}{S}}     
       – reduce to the same denominator :          \displaystyle{ \frac{P}{Q}     +  \frac{S}{T}  =   \frac{P T + Q S }{ST} }
     
      – reduce to lower terms :          \displaystyle{ \frac{N.R}{D.R} = \frac{N}{D}  }

 

 

Watch a Video

View on YouTube
 

 

Show Examples

 

Exercise 1

 

Show/Hide Solution.

We have

x^2 -{6}x + {8} = (x-{2})(x-{4})

x^2 +{6}x -{16} = (x-{2})(x+{8}).

Thus, by simplifying the common factor (x-{2}), we obtain

\displaystyle{\frac{x^2 -{6}x + {8} }{ x^2 +{6}x -{16}} = \frac{(x-{2})(x-{4})}{(x-{2})(x+{8})} =\frac{x-{4}}{x+{8}}}.

 

 

Exercise 2

 

Show/Hide Solution.

We have

\displaystyle{\frac{ (x-{4})( {3}x+ {2} ) }{ x^3-{64} } } = \displaystyle{\frac{ (x-{4})( {3}x+ {2} ) }{(x-{4}) (x^2 +{4}x + {16}) } } = \displaystyle{\frac{ {3}x+ {2} }{x^2 +{4}x + {16} } }.

 

Exercise 3

 

Show/Hide Solution.

We have

{3} x^3 -{7}x^2 + {4} x = x({3} x^2 -{7}x + {4} )= x ({3}x-{4})(x-{1})

x^2 -{7}x + {6} = ( x -{1})(x-{6})

Thus, by simplifying the common factor (x- {1}), we obtain

\displaystyle{\frac{{3} x^3 -{7}x^2 + {4} x }{ x^2 -{7}x + {6}} = \frac{x({3}x-{4})(x-{1})}{(x-{1})(x-{6})} =\frac{{3}x^2-{4} x}{x-{6}}}.

License

Icon for the Creative Commons Attribution-NonCommercial-NoDerivatives 4.0 International License

Guide to Precalculus Review Copyright © 2025 by Samia CHALLAL is licensed under a Creative Commons Attribution-NonCommercial-NoDerivatives 4.0 International License, except where otherwise noted.