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!

Homework Help: IB HL Maths - Polynomials

  1. May 7, 2012 #1
    1. The problem statement, all variables and given/known data
    When 3x5 - ax + b is divided by x - 1 and x + 1 the remainders are equal. Given that a, b ε ℝ

    (a) the value of a;
    (b) the set of values of b.

    2. Relevant equations

    3. The attempt at a solution
    f(1) = f(-1)
    3 - a + b = -3 + a + b
    6 = 2a
    a = 3 ... [1]
    Substitute a -3 into 3 - a + b
    b = 0

    Can someone please explain to me why the set of values of b would be all real numbers?
    From my knowledge, I believe my way of understanding is unclear and may be incorrect.
    My interpretation: b not being affected by any coefficients and that b is simply a constant in a function.

    Last edited: May 7, 2012
  2. jcsd
  3. May 7, 2012 #2


    User Avatar

    Staff: Mentor

    b can have any value and it will satisfy this equality, since +b appears on both sides.
  4. May 7, 2012 #3


    User Avatar
    Science Advisor

    This is wrong. [itex]f(1)= 3(1)^2- a(1)+ b[/itex] alright, but [itex]f(-1)= 3(-1)^2- a(-1)+ b= 3+ a+ b[/itex], not -3+ a+ b.

    So this become a= 0.

    This makes no sense. Substititute a= -3 (or a= 0) into what equation? You don't know what 3- a+ b is supposed to be equal to. You seem to be assuming it is 0 and you have no right to do that.

    The requirement that f(-1)= f(1) gives a= 0 but gives NO condition on b.
    If [itex]f(x)= 3x^2+ b[/itex] then f(1)= 3+ b= f(-1). In fact, the condition that, for polynomial f, f(x)= f(-x) for all x, would tell us that there are no odd powers of x but would tell us nothing about the coefficients of the even powers of x.
  5. May 7, 2012 #4
    I just realised a mistake in the thread. It should be 3x5 not 3x2
    Sorry about that! :frown:

Share this great discussion with others via Reddit, Google+, Twitter, or Facebook