Sum of the squares of remainders

  • #1
V0ODO0CH1LD
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0

Homework Statement



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?

Homework Equations



I guess the remainder theorem could be useful here.

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).
 
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  • #2
What happens if you consider f(1)?
 
  • #3
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
You have established that [itex]f(1) = a[/itex] and [itex]f(2) = b[/itex].
But we also know that
[itex]f(x) = g(x)(x-1)(x-2) + (x+1)[/itex]
for some quotient [itex]g(x)[/itex].
Can you compute what [itex]f(1)[/itex] and [itex]f(2)[/itex] are?
 
  • #5
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?
 

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