# IB HL Maths - Polynomials

1. May 7, 2012

### thornluke

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.

Cheers.

Last edited: May 7, 2012
2. May 7, 2012

### Staff: Mentor

b can have any value and it will satisfy this equality, since +b appears on both sides.

3. May 7, 2012

### HallsofIvy

Staff Emeritus
This is wrong. $f(1)= 3(1)^2- a(1)+ b$ alright, but $f(-1)= 3(-1)^2- a(-1)+ b= 3+ a+ b$, 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 $f(x)= 3x^2+ b$ 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.

4. May 7, 2012

### thornluke

I just realised a mistake in the thread. It should be 3x5 not 3x2