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Factor Theorem Question

  • 11-01-2013 9:02pm
    #1
    Vivara
    Registered Users, Registered Users 2 Posts: 333 ✭✭


    Would anyone know how to do the following question and offer an explanation as to how to do it?

    Show, using the factor theorem or otherwise, that (x – p – q – r) is a factor of
    (x – 4) (x + 2) (x + 1) – (q + r) (r + p) (p + q)

    If you could show how to do it both using the factor theorem and "otherwise", it would be most helpful.

    Thanks!


Comments

  • #2
    equivariant
    Registered Users, Registered Users 2 Posts: 966 ✭✭✭equivariant


    I don't believe that you have stated the question correctly. Have you included all the information given in the question?

    As you have stated it, the proposition is false. For example, if x=0 and p=q=r=1 then
    (x-p-q-r) = -3 and (x-4)(x+2)(x+1)-(q+r)(r+p)(p+q) = -16. Now -3 does not divide into -16, so your statement about factors must be false.

    Please check that you have stated the entire problem *exactly* as it was given to you.


  • #3
    Vivara
    Registered Users, Registered Users 2 Posts: 333 ✭✭Vivara


    equivariant wrote: »
    I don't believe that you have stated the question correctly. Have you included all the information given in the question?

    As you have stated it, the proposition is false. For example, if x=0 and p=q=r=1 then
    (x-p-q-r) = -3 and (x-4)(x+2)(x+1)-(q+r)(r+p)(p+q) = -16. Now -3 does not divide into -16, so your statement about factors must be false.

    Please check that you have stated the entire problem *exactly* as it was given to you.

    Oh my God, I feel so silly! You're right, and I was reading the wrong part of the question. I'm sorry.

    The question asked to prove it was a factor of (x – p) (x + q) (x – r) – (q + r) (r + p) (p + q)

    Which is quite simple. If you substitute in x for p + q + r it cancels. But is this proving it using the factor theorem?

    The question then asks: Hence, or otherwise prove that
    (x-4)(x+2)(x+1) + 18 = 0

    What I did is multiply it out and get x^3 – x^2 – 10x + 10 = 0, I divided in (x – 1) since (x – a) is a factor and got x^2 – 10.

    Thus, x = 1 and x = +root(10) and x = –root(10)

    But still, this is how I would normally solve it and I don't think I've used the first part as suggested by the "hence, or otherwise", have I?

    Thanks again for your help!


  • #4
    MathsManiac
    Registered Users, Registered Users 2 Posts: 1,595 ✭✭✭MathsManiac


    Yes, your method IS using the factor theorem, as you have used the fact that x = [an expression] is a root to show that x - [the expression] is a factor.

    For the second part, your method is an "otherwise". The intention of the "hence" was probably that you should try to think of values for p, q, and r that would make the expression in part (i) look like the expression in part (ii).


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