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Homework Help: Factoring Involving the Difference of Two Squares

  1. Sep 18, 2017 #1
    1. The problem statement, all variables and given/known data
    Completely factor:

    $$5abc^4-80ab$$

    2. Relevant equations

    N/A

    3. The attempt at a solution

    $$5abc^4-80ab$$
    $$5ab\left(c^4-16\right)$$
    $$5ab\left(c^2+4\right)\left(c^2-4\right)$$

    The correct solution is ##5ab\left(c^2+4\right)\left(c+2\right)\left(c-2\right)##
    I can see why it factored out the ##\left(c^2-4\right)## further, but why isn't it required to do so with the ##\left(c^2+4\right)## as well?
     
  2. jcsd
  3. Sep 18, 2017 #2

    Mark44

    Staff: Mentor

    Because ##c^2 - 4## is a difference of squares while ##c^2 + 4## is the sum of two squares. The latter can't be factored into two linear binomials whose coefficients are real. It can, however, be factored into complex numbers, as ##(c + 2i)(c - 2i)##, but in ordinary factoring, you're only concerned with factors with real coefficients.
     
  4. Sep 18, 2017 #3
    Thanks for that quick response, Mark. But why can the difference between two squares be factored with real numbers however the sum between two squares cannot?
     
  5. Sep 20, 2017 #4

    Mark44

    Staff: Mentor

    Let's look at the equation ##x^2 - 1 = 0##. I don't think you have any problem with recognizing that the equation could be rewritten as (x - 1)(x + 1) = 0.

    Now consider the equation ##x^2 + 1 = 0##. Is is possible to rewrite this as (x + a)(x + b) = 0, assuming that both a and b are real numbers? If so, we must have a*b = 1 and a + b = 0 (the latter since the coefficient of x is 0).

    If a + b = 0, then a and b must be opposites (i.e., additive inverses), which would also include the possibility that a = b = 0.
    If ab = 1, then a and b need to be reciprocals (multiplicative inverses) of one another. This rules out the possibility that either one could be zero. It also says that both a and b have to be positive, or both have to be negative. Each of these latter conclusions rules out the possibility of a and b being opposites.

    The upshot is that there are no real numbers a and b so that (x + a)(x + b) = 0. If I relax the condition that a and b must be real, it turns out that there are solutions for a and b; namely, a = i and b = -i, where i is the imaginary number ##\sqrt{-1}##. Thus, allowing coefficients from the complex numbers, ##x^2 + 1## can be factored into ##(x + i)(x - i)##.
     
  6. Sep 24, 2017 #5
    Thanks a lot for that reply Mark. Been real busy with schoolwork so I got to this a little late, but it makes sense. I'll keep this in mind when I eventually get to imaginary numbers.
     
  7. Sep 24, 2017 #6
    I did a little fiddling and multiplied out ##\left(c+2\right)\left(c-2\right)## and came up with ##\left(c^2-4\right)##.
    In multiplying ##\left(c+2\right)\left(c+2\right)##, I got ##c^2+4c+4##.
    Is this another way of saying it as well?
     
  8. Sep 24, 2017 #7

    Mark44

    Staff: Mentor

    Well, sort of. The multiplication you're doing is the opposite operation of factoring. You found that (c + 2)(c - 2) = c2 - 4, but there aren't any real numbers a and b for which (x - a)(x - b) = x2 + k2.
     
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