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Algebra Question: Simplifying Ration Questions

  1. Aug 11, 2008 #1
    Alright It's been a while since I did algebra and now I'm getting back into it in honors algebra 2 (shakes head) [highschool]. Anyways I got a question on factoring numerators.

    [tex]6x^{2} + 7x + 2[/tex]
    [tex]4x^{2} - 1[/tex]

    I don't know where to start, if someone can be a guidance this would be much appreciated I was trying to http://www.analyzemath.com/Rational_expressions/Simp_rat_expre.html" [Broken] but the walk through didn't really show me how they got to the next step.

    Thanks in advanced
    Last edited by a moderator: May 3, 2017
  2. jcsd
  3. Aug 11, 2008 #2


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    Try factoring. If you find a binomial which occurs in both the numerator and the denominator then you can eliminate as a factor of 1.

    For the numerator, start tries using:
    (2x + ?)(3x + ?), and if that does not help, try (6x + ?)(x + ?);
    but what you really will rely on is the presence of (2x+1) in numerator AND denominator. Otherwise, you will not be able to simplify the given expression.
  4. Aug 11, 2008 #3
    For the bottom, it's just the difference of squares factorization. a^2 - b^2 = (a-b)(a+b). The top is a bit trickier. But if you're new to factoring, maybe this will help.

    Set the numerator equal to (rx + s)(tx + v). Now multiply out this expression and equate its terms to the terms of the numerator. You should be able to find suitable r,s,t,v and then you'll have the factorization. Don't forget to multiply out to check.
  5. Aug 11, 2008 #4


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    Welcome to PF!

    Hi Lamonte ! Welcome to PF! :smile:

    Useful tip: This isn't real life, it's an exam question, so it probably has an easy answer.

    You know that they want you find a common factor, and you can probably see exactly what the two factors of the denominator are.

    So just try both of them in the numerator … if it's a sensible exam question, then one of them should work! :wink:
  6. Aug 12, 2008 #5


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    To add to tiny-tim's answer- even in "real life" where problems don't always have easy answers, since you are trying to simplify the fraction, the only reason you want to factor is to be able to cancel. You should be able to see quickly that 4x2- 1 factors as (2x-1)(2x+ 1). Then try each of those as a factor of 6x2+ 7x+ 2. If it is a factor, good, you can factor and cancel. If neither is a factor, also good. You know that cannot be simplified.
  7. Aug 12, 2008 #6
    As have been stated, the denominator should be easily factored using the following shortcuts (which you should have learned, but might have forgotten if it was a while ago):

    [tex](a-b)^2 = a^2 - 2ab + b^2[/tex]
    [tex](a+b)^2 = a^2 + 2ab + b^2[/tex]
    [tex](a+b)(a-b) = a^2 -b^2[/tex]

    These things come up over and over, you need to practice so that you spot them easily.

    As for the discriminant: First, I check to see if the expression is a complete square. If it is not, then I factor it using the quadratic equation. You can also factor by completing the square, but that is just more writing. It is good to know both ways though, since when you work with only variables it makes more sense to complete (as a way to derive the quadratic equation from Ax^2 + Bx + C, for example).

    I assume you remember what the terms I used mean, if you don't then you probably need to backtrack a bit to freshen your memory on earlier topics.

  8. Aug 12, 2008 #7
    Thanks for the help guys will give it a try and return with an answer
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