Factor Theorem Question

Vivara

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!

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.

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!

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).