# Sum of the squares of remainders

1. Nov 1, 2012

### V0ODO0CH1LD

1. The problem statement, all variables and given/known data

The division of a polynomial f(x) by (x – 1)(x – 2) has remainder x + 1. If the remainder of the division of f(x) by (x – 1) and (x – 2) are, respectively, a and b. Then what is a^2 + b^2?

2. Relevant equations

I guess the remainder theorem could be useful here.

3. The attempt at a solution

f(x) = q(x)(x – 1) + a; where q(x) is the quotient of the division of f(x) by (x – 1).

f(x) = p(x)(x – 2) + b; where p(x) is the quotient of the division of f(x) by (x – 2).

q(x)(x – 1) + a = p(x)(x – 2) + b

I don't know how to move forward using the fact the remainder of the division of f(x) by (x – 1)(x – 2).

2. Nov 1, 2012

### haruspex

What happens if you consider f(1)?

3. Nov 1, 2012

### V0ODO0CH1LD

f(1) = a and f(2) = b; right? Which means that a = -p(x) + b and b = q(x) + a.

So a^2 + b^2 = (-p(x) + b)^2 + (q(x) + a)^2. I still don't see how to use the fact the remainder of the division of f(x) by (x – 1)(x – 2).

EDIT: Wait; a + p(x) = b and b = q(x) + a. So a + p(x) = q(x) + a and p(x) = q(x)!! Is that right?

4. Nov 1, 2012

### Fightfish

You have established that $f(1) = a$ and $f(2) = b$.
But we also know that
$f(x) = g(x)(x-1)(x-2) + (x+1)$
for some quotient $g(x)$.
Can you compute what $f(1)$ and $f(2)$ are?

5. Nov 1, 2012

### V0ODO0CH1LD

f(1) = (1 + 1) and f(2) = (1 + 2). So a = 2 and b = 3. And a^2 + b^2 = 13.

How do you guys see problems so clearly? Now that you pointed out everything became obvious, but before that I could never have seen that. Is that practice? I feel like I can grasp concepts really well, but when it comes to problem solving I can hardly interpret the question at first or second sight. How can I improve that?