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    Home / Algebra Trig Review / Algebra / Simplifying Rational Expressions
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    Simplifying Rational Expressions

    Simplify each of the following rational expressions.Show All Solutions Hide All Solutions

    1. \(\displaystyle \frac{{2{x^2} - 8}}{{{x^2} - 4x + 4}}\)
      Show Solution

      There isn’t a lot to these problems. Just factor the numerator and denominator as much as possible then cancel all like terms. Recall that you can only cancel terms that multiply the whole numerator and whole denominator. In other words, you can’t just cancel the \({x^2}\) and you can’t cancel a 4 from the 8 in the numerator and the 4 in the denominator. Things just don’t work this way.

      If you need convincing of this consider the following number example.

      \[8.5 = \frac{{17}}{2} = \frac{{8 + 9}}{2} \ne \frac{{4 + 9}}{1} = 13\]

      Here is the answer to this problem.

      \[\frac{{2{x^2} - 8}}{{{x^2} - 4x + 4}} = \frac{{2\left( {{x^2} - 4} \right)}}{{{x^2} - 4x + 4}} = \frac{{2\left( {x - 2} \right)\left( {x + 2} \right)}}{{{{\left( {x - 2} \right)}^2}}} = \frac{{2\left( {x + 2} \right)}}{{x - 2}}\]
    2. \(\displaystyle \frac{{{x^2} - 5x - 6}}{{6x - {x^2}}}\)
      Show Solution

      In this one you’ve got to be a little careful. First factor the numerator and denominator.

      \[\frac{{{x^2} - 5x - 6}}{{6x - {x^2}}} = \frac{{\left( {x - 6} \right)\left( {x + 1} \right)}}{{x\left( {6 - x} \right)}}\]

      At first glance it doesn’t look like anything will cancel. However, remember that

      \[a - b = - \left( {b - a} \right)\]

      Using this on the term in the denominator gives the following.

      \[\frac{{{x^2} - 5x - 6}}{{6x - {x^2}}} = \frac{{\left( {x - 6} \right)\left( {x + 1} \right)}}{{x\left( {6 - x} \right)}} = \frac{{\left( {x - 6} \right)\left( {x + 1} \right)}}{{ - x\left( {x - 6} \right)}} = \frac{{x + 1}}{{ - x}} = - \frac{{x + 1}}{x}\]

      Also recall that

      \[\frac{a}{{ - b}} = \frac{{ - a}}{b} = - \frac{a}{b}\]

      so it doesn’t matter where you put the minus sign.

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