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Simplifying Rational Expressions

Factor numerator and denominator and cancel

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Simplifying Rational Expressions (7.4)

The area of a rectangle is \begin{align*}2x^4 - 2\end{align*}. The width of the rectangle is \begin{align*}x^2 + 1\end{align*}. What is the length of the rectangle?


Recall that a rational function is a function, \begin{align*}f(x)\end{align*}, such that \begin{align*}f(x)=\frac{p(x)}{q(x)}\end{align*}, where \begin{align*}p(x)\end{align*} and \begin{align*}q(x)\end{align*} are both polynomials. A rational expression, is just \begin{align*}\frac{p(x)}{q(x)}\end{align*}. Like any fraction, a rational expression can be simplified. To simplify a rational expression, you will need to factor the polynomials, determine if any factors are the same, and then cancel out any like factors.

Fraction: \begin{align*}\frac{9}{15} = \frac{\cancel{3} \cdot 3}{\cancel{3} \cdot 5} = \frac{3}{5}\end{align*}

Rational Expression: \begin{align*}\frac{x^2+6x+9}{x^2+8x+15} = \frac{\cancel{(x+3)}(x+3)}{\cancel{(x+3)}(x+5)} = \frac{x+3}{x+5}\end{align*}

With both fractions, we broke apart the numerator and denominator into the prime factorization. Then, we canceled the common factors.

Important Note: \begin{align*}\frac{x+3}{x+5}\end{align*} is completely factored. Do not cancel out the \begin{align*}x\end{align*}’s! \begin{align*}\frac{3x}{5x}\end{align*} reduces to \begin{align*}\frac{3}{5}\end{align*}, but \begin{align*}\frac{x+3}{x+5}\end{align*} does not because of the addition sign. To prove this, we will plug in a number for \begin{align*}x\end{align*} to and show that the fraction does not reduce to \begin{align*}\frac{3}{5}\end{align*}. If \begin{align*}x=2\end{align*}, then \begin{align*}\frac{2+3}{2+5} = \frac{5}{7} \ne \frac{3}{5}\end{align*}.

Example A

Simplify \begin{align*}\frac{2x^3}{4x^2-6x}\end{align*}.

Solution: The numerator factors to be \begin{align*}2x^3=2 \cdot x \cdot x \cdot x\end{align*} and the denominator is \begin{align*}4x^2-6x=2x(2x-3)\end{align*}.

\begin{align*}\frac{2x^3}{4x^2-6x} = \frac{\cancel{2} \cdot \cancel{x} \cdot x \cdot x}{\cancel{2} \cdot \cancel{x} \cdot(2x-3)} = \frac{x^2}{2x-3}\end{align*}

Example B

Simplify \begin{align*}\frac{6x^2-7x-3}{2x^3-3x^2}\end{align*}.

Solution: If you need to review factoring, see the Factoring When the Leading Coefficient Equals 1 concept and the Factoring When the Leading Coefficient Doesn't Equal 1 concept. Otherwise, factor the numerator and find the GCF of the denominator and cancel out the like terms.

\begin{align*}\frac{6x^2-7x-3}{2x^3-3x^2} = \frac{\cancel{(2x-3)}(3x+1)}{x^2\cancel{(2x-3)}} = \frac{3x+1}{x^2}\end{align*}

Example C

Simplify \begin{align*}\frac{x^2-6x+27}{2x^2-19x+9}\end{align*}.

Solution: Factor both the top and bottom and see if there are any common factors.

\begin{align*}\frac{x^2-6x+27}{2x^2-19x+9} = \frac{\cancel{(x-9)}(x+3)}{\cancel{(x-9)}(2x-1)} = \frac{x+3}{2x-1}\end{align*}

Special Note: Not every polynomial in a rational function will be factorable. Sometimes there are no common factors. When this happens, write “not factorable.”

Intro Problem Revisit

Recall that the the area of a rectangle is the length times the width. To find the length, we can therefore divide the area by the width. So we're looking for \begin{align*}\frac{2x^4 - 2}{x^2 + 1}\end{align*}.

If we factor the numerator and the denominator, we get:

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

Therefore, the length of the rectangle is \begin{align*}2(x^2 - 1) = 2x^2 - 2\end{align*}.

Guided Practice

If possible, simplify the following rational functions.

1. \begin{align*}\frac{3x^2-x}{3x^2}\end{align*}

2. \begin{align*}\frac{x^2+6x+8}{x^2+6x+9}\end{align*}

3. \begin{align*}\frac{2x^2+x-10}{6x^2+17x+5}\end{align*}

4. \begin{align*}\frac{x^3-4x}{x^5+4x^3-32x}\end{align*}


1. \begin{align*}\frac{3x^2-x}{3x^2} = \frac{\cancel{x}(3x-1)}{3 \cdot \cancel{x} \cdot x} = \frac{3x-1}{3x}\end{align*}

2. \begin{align*}\frac{x^2+6x+8}{x^2+6x+9} = \frac{(x+4)(x+2)}{(x+3)(x+3)}\end{align*} There are no common factors, so this is reduced.

3. \begin{align*}\frac{2x^2+x-10}{6x^2+17x+5} = \frac{\cancel{(2x+5)}(x-2)}{\cancel{(2x+5)}(3x+1)} = \frac{x-2}{3x+1}\end{align*}

4. In this problem, the denominator will factor like a quadratic once an \begin{align*}x\end{align*} is pulled out of each term.

\begin{align*}\frac{x^3-4x}{x^5+4x^3-32x} = \frac{x(x^2-4)}{x(x^4+4x^2-32)} = \frac{x(x-2)(x+2)}{x(x^2-4)(x^2+8)} = \frac{\cancel{x (x-2)(x+2)}}{\cancel{x (x-2)(x+2)}(x^2+8)} = \frac{1}{x^2+8}\end{align*}


Rational Expression
A fraction with polynomials in the numerator and denominator.


  1. Does \begin{align*}\frac{x-2}{x-6}\end{align*} simplify to \begin{align*}\frac{1}{3}\end{align*}? Explain why or why not.
  2. Does \begin{align*}\frac{5x}{10x}\end{align*} simplify to \begin{align*}\frac{1}{2}\end{align*}? Explain why or why not.
  3. In your own words, explain the difference between the previous two expressions and why one simplifies and one does not.

Simplify the following rational expressions.

  1. \begin{align*}\frac{4x^3}{2x^2+3x}\end{align*}
  2. \begin{align*}\frac{x^3+x^2-2x}{x^4+4x^3-5x^2}\end{align*}
  3. \begin{align*}\frac{2x^2-5x-3}{2x^2-7x-4}\end{align*}
  4. \begin{align*}\frac{5x^2+37x+14}{5x^3-33x^2-14x}\end{align*}
  5. \begin{align*}\frac{8x^2-60x-32}{-4x^2+26x+48}\end{align*}
  6. \begin{align*}\frac{6x^3-24x^2+30x-120}{9x^4+36x^2-45}\end{align*}
  7. \begin{align*}\frac{6x^2+5x-4}{6x^2-x-1}\end{align*}
  8. \begin{align*}\frac{x^4+8x}{x^4-2x^3+4x^2}\end{align*}
  9. \begin{align*}\frac{6x^4-3x^3-63x^2}{12x^2-84x}\end{align*}
  10. \begin{align*}\frac{x^5-3x^3-4x}{x^4+2x^3+x^2+2x}\end{align*}
  11. \begin{align*}\frac{-3x^2+25x-8}{x^3-8x^2+x-8}\end{align*}
  12. \begin{align*}\frac{-x^3+3x^2+13x-15}{-2x^3+7x^2+20x-25}\end{align*}

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