1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Polynomial factoring

  1. Feb 18, 2017 #1
    1. The problem statement, all variables and given/known data
    ##\frac{3x^2+x+C}{x+5}##
    find value of constant C such that the clause can be cancelled in some manner. What will be the cancelled form of the clause.
    2. Relevant equations
    -presumably C is a constant, and also an integer.
    -polynomial factorization will be attempted
    -cancelling of rational numbers keeps the value the same. In cancelling, you divide numerator and denominator by the same number (other number than 1 or 0).
    3. The attempt at a solution

    we know initially that we have to have something like
    ##\frac{(x+5)*(?? ~~??}{x+5}##
    this is because there will be no further factorization for the divisor. For the clause to be cancellable, I think the (x+5) must be one of the factors in the upstairs of the division line.

    and from an educated guess point of view, I think we have to get the 3x^2 term finished somehow. So, that 3x^2 is the product of something.
    ##\frac{(x+5)*(3x ~~??}{x+5}##

    if the other integer is an unknown one. maybe it can be represented by letter r
    (x+5)(3x+r) should equal what we originally had upstairs (3x^2 +x +C) and we know C will be an integer also
    At this point I was honestly a little bit stumped and I had to review some older Khan academy problems about factoring. I'm still alittle bit unsure why the so-called analytical method of finding the factorization works...


    I was fiddling around with the clause (x+5)*(3x+r)
    and if you multiply it out, it comes out such as
    3x^2 +xr+15x+5r

    Well... looking at that clause in that form it looks like xr +15x=1x must be a true equation.
    Because this would allow you to choose the r so that the factors are correct for the original clause's terms for x^1 terms

    So, I suppose firstly r= 1-15
    r= -14
    C= 5*(-14) = -70

    [(x+5)(3x-14)]/(x+5)
    =3x-14

    [(x+5)(3x-14)]= 3x^2 -14x +15x -70
     
  2. jcsd
  3. Feb 18, 2017 #2

    fresh_42

    Staff: Mentor

    Looks good, but what is your question?
     
  4. Feb 18, 2017 #3

    Ray Vickson

    User Avatar
    Science Advisor
    Homework Helper

    Your method works because when you expand ##q(x) = (x+5)(3x+r)## you get a second-degree polynomial in ##x## that must be the same as ##p(x) = 3x^2+x+C##. That means that the coefficients in the two polynomials must be the same; that is, if ##p(x) = q(x)## for ALL ##x##, then all the coefficients in ##p(x)## must be the same as all the coefficients in ##q(x)##. That gives you two conditions for the two unknown constants ##r## and ##C##.

    BTW: a much faster and much easier way is to use the following general theorem: "The remainder when a polynomial ##p(x)## is divided by ##(x-a)## is equal to ##p(a)##". So, in your case you have ##a = -5## and to have ##p(x) = 2x^2 +x+C## divisible by ##(x+5)## the remainder must equal zero. That means that you need ##0 = p(-5) = 3(-5)^2 + (-5) + C = 70 - C, ## so ##C = 70##.
     
  5. Feb 18, 2017 #4
    thank you for the tip sir... sadly I never bothered to learn much about polynomial division. I was once taught about that in high school but I never learned how to do it properly. It's like a dark area for my math knowledge. I do know how to divide regular numbers in the so-called American long division algorithm, though.
     
  6. Feb 18, 2017 #5

    Mark44

    Staff: Mentor

    https://en.wikipedia.org/wiki/Polynomial_long_division
     
  7. Feb 18, 2017 #6

    fresh_42

    Staff: Mentor

  8. Feb 18, 2017 #7

    Ray Vickson

    User Avatar
    Science Advisor
    Homework Helper

    The result is incredibly easy to obtain. If we divide ##p(x)## by ##(x-a)## we have some quotient ##q(x)## and some remainder ##r## (a number). Basically, that means that
    ##p(x) = q(x) \, (x-a) + r.##
    What do you get when you put ##x = a?##
     
  9. Feb 18, 2017 #8
    Well, you said originally that for the clause to cancellable... then the remainder will be zero when you do the cancellation thing.

    when you have the thing set-up that way. it looks like it will become
    p(a)=q(a) * 0 + r
    <=> p(a)=r

    so you were saying that essentially p(a) = 0, because r=0

    Im going to sleep now because it is 01:47 so it's getting pretty late.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Polynomial factoring
  1. Factoring Polynomials (Replies: 3)

  2. Polynomial Factors (Replies: 3)

  3. Factoring Polynomials (Replies: 2)

  4. Factoring Polynomial (Replies: 4)

  5. Factoring a polynomial (Replies: 4)

Loading...