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Limit factorization

  1. Jun 19, 2013 #1
    Hello, why I can't directly find lim x->3 (x^2+2x-15)/(x^2-5x+6) but I have to factorize them ?
    Is there any intuitive way to understand that ?

  2. jcsd
  3. Jun 19, 2013 #2
    You have a 0/0 expression and can use L'Hôpital's rule
  4. Jun 19, 2013 #3


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    Understand what? That you have to factor them? As Janhaa said, that's not true- there are other methods. Of course, you cannot just set x= 3 because that gives 0/0 which has no meaning.

    That you can factor them? That's due to a fundamental algebra rule: a polynomial is 0 for x= a if and only if it has x- a as a factor.

    Here, because setting x= 3 the numerator becomes [tex]3^2+ 2(3)- 15= 0[/tex] so we know that x- 3 is a factor. And since 3(a)= -15, requires that a= -5, we know that the other factor must be x- (-5)= x+ 5:
    [tex]x^2+ 2x- 15= (x- 3)(x+ 5)[/tex].

    At x= 3, the denominator becomes [tex]3^2- 5(3)+ 6= 0[/tex] so we know that x- 3 is a factor of this also. and since 3(a)= 6 requires that a= 2, we know the other factor must be x- 2:
    [tex]x^2-5x+ 6= (x- 2)(x- 3)[/tex].

    Then the fraction is [tex]\frac{x^2+ 2x- 15}{x^2- 5x+ 6}= \frac{(x- 3)(x+ 5)}{(x- 3)(x- 2)}[/tex]
    Now, as long as x is NOT 3[/tex], we can cancel those terms to get [tex]\frac{x+ 5}{x- 2}[/tex] and setting x= 3 in that we have [tex]\frac{3+ 5}{3- 2}= 8[/tex] which tells us that [tex]\lim_{x\to 3}\frac{x^2+ 2x- 15}{x^2- 5x+ 6}= 8[/tex].

    Notice I said "as long as x is NOT 3". We cannot divide by 0 so that reduction is NOT true for x= 3. AT x= 3, this function is not defined- it has no value. But there is a theorem (unfortunately, often overlooked in introductory courses) that says "if f(x)= g(x) in some neighborhood of a but NOT necessarily at x= a (what is called a "deleted neighborhood) then [tex]\lim_{x\to a}f(x)= \lim_{x\to a}g(x)[/tex]".
  5. Jun 19, 2013 #4
    Why not all trinomials (for example x^2+3x-15) can be written as (x+a)(x+b) ?
  6. Jun 19, 2013 #5


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    They can all the written like that - it's called the fundamental theorem of algebra. Unfortunately though sometimes a and b have to be complex numbers to do it, such as
    [tex] x^2 + 1 = (x+i)(x-i) [/tex]
  7. Jun 20, 2013 #6


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    Again, a polynomial, p(x), has factor x- a if and only if p(a)= 0. To find factors for [itex]x^2+ 3x- 15[/itex] we need to solve [itex]x^2+ 3x- 15= 0[/itex], for example, by the quadratic formula:
    [tex]x= \frac{-3\pm\sqrt{9- 4(-15)}}{2}= \frac{-3\pm\sqrt{9+ 60}}{2}= \frac{-3\pm\sqrt{69}}{2}[/tex]
    The two solutions are
    [tex]-\frac{3}{2}+ \frac{\sqrt{69}}{2}[/tex] and
    [tex]-\frac{3}{2}- \frac{\sqrt{69}}{2}[/tex]

    so that can be factored as
    [tex]x^2+ 3x- 15= \left(x+ \frac{3}{2}- \frac{\sqrt{69}}{2}\right)\left(x+ \frac{3}{2}+ \frac{\sqrt{69}}{2}\right)[/tex]

    Of course, when we talk about "factoring a polynomial", especially when we are trying to factor in order to solve a polynomial, we really mean "factoring with integer coefficients" which can only be done if the polynomial has rational roots: if a/b is a root then the polynomial has a factor of x- a/b= (bx- a)/b.
  8. Jun 21, 2013 #7
    I mean why you can't solve it directly but you have to factorize it since the original trinomial and its factorization are the same thing
  9. Jun 26, 2013 #8


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    Ignore the limit for the moment, and ask yourself if there is a good way to simplify the quotient of two polynomials without factorizing them.
  10. Jun 27, 2013 #9
    What you mean ?
  11. Jun 27, 2013 #10


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    The question is "what do you mean?" You keep asking why you must factor the polynomials after others tell you that you don't have to.

    But the question pasmith is asking is "Do you know how to simplify an algebraic fraction."
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