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Find the values of a, b and c

  1. Nov 21, 2004 #1
    I looked at this question and said WHAT? i dont know what its talking about can someone please help me out, I dont have a clue as to what to do, or where to start. :cry:

    For x^3-1=(x-1)(ax^2+bx+c) find the values of a, b, and c
     
  2. jcsd
  3. Nov 21, 2004 #2
    Try dividing [tex]x^3 -1 [/tex] by [tex]x - 1[/tex].
    Do you know how to do that?
     
  4. Nov 21, 2004 #3

    Hurkyl

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    Or he could multiply out the RHS and set both sides equal to each other.
    Or he could plug in values of x to generate a system of equations.
     
  5. Nov 21, 2004 #4
    Many polynomials can be factored into products of smaller polynomials.

    E.g -
    (x^2 - 1) = (x+1)(x-1)

    This is known as factorisation of polynomials.

    x^3-1 can be factored in a similar way.
    Now,
    x^3-1=(x-1)(ax^2+bx+c)

    What does this mean?
    This means that x^3 - 1 can be factored into two polynomials (x-1) and (ax^2+bx+c).

    You are supposed to find the coefficients of the second polynomial.

    Ways to do it?
    1> Factorise x^3-1 manually and see what u get?
    Suppose u get,
    x^3-1 = (x-1)(px^2+qx+r)
    then a = p,b = q and c = r.

    2> Now if x^3-1=(x-1)(ax^2+bx+c)
    then (x^3-1)/(x-1) = (ax^2+bx+c)
    That means u can get (ax^2+bx+c) by dividing x^3-1 by x-1.

    3> Multiply RHS. That is multiply (x-1) with (ax^2+bx+c) . Then compare the coefficients of this with x^3 - 1 so that u can determine a,b and c

    4> Note that,
    x^3-1=(x-1)(ax^2+bx+c)
    is true for all x
    So substitute x = 0 and u will note that u can get c
    sub in x = -1 and u will get an equation in terms of a and b
    sub in x = 2 and u will again get an equation in terms of a and b
    solve them simultaneously to find a and b.

    -- AI
     
  6. Nov 21, 2004 #5
    Ya, this question is rather simple.
    Three simple method:
    1. Long division. Don't forget to wite down 0x^2 and 0x
    2.Identity. Comparing coefficient. Expand the given function.
    Compare the term to x^3-1
    3.The most common one. It is learnt in Grade K10 I think.
    x^3-1=(x-1)(x^2+x+1)
     
  7. Nov 21, 2004 #6
    This is what I got:
    [tex]x^3-1=(x-1)(ax^2+bx+c)[/tex]

    [tex]ax^2+bx+c=\frac{x^3-1}{x-1}=x^2[/tex]

    [tex]ax^2+bx+c=x^2[/tex]

    [tex]a=\frac{x^2-bx-c}{x^2}[/tex]

    The [tex]x^2[/tex] get cancelled out and I got:

    [tex]a=bx-c[/tex]


    *NOTE: I might be wrong
     
    Last edited: Nov 21, 2004
  8. Nov 21, 2004 #7
    Yeah, Raza you are... wrong that is.
    [tex]a=\frac{x^2-bx-c}{x^2}[/tex] does not cancel out to get [tex]a=bx-c[/tex]
    Plus, you say that [tex]ax^2+bx+c=x^2[/tex]. If this is true for all (since a,b and c are constants) that would mean that either a,b and c are all = 0, (which the aren't) or that a = 1* and b and c = 0 (which, they don't)
     
    Last edited: Nov 21, 2004
  9. Nov 21, 2004 #8
    I think I've got the answer

    Hi, thanks for all ur help I think I've got it tell me if u think im wrong.
    the question was for X^3-1=(X-1)(ax^2+bx+c)
    what are the values of a, b and c

    well I brought x-1 to the LHS and divided x^3-1 by x-1 I used long division and i got an answer of x^2+x+1 with no remainder I made this equal to ax^2+bx+c and I think a=1 b=1 and c=1 ? I'm not sure
     
  10. Nov 21, 2004 #9
    Sounds right to me.
     
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