Solving a Polynomial Division Problem: Remainder = 5 & 7

[misogynisticfeminist]

Given that a polymial p(x) is

[tex] p(x)= (x-1)(x-2) q(x) + 2x+3 [/tex]

where q(x) is also a polynomial

Find the remainder when p(x) is divided by (x-1)(x+2) where the remainder divided by (x-1) and (x+2) is both 5 and 7 respectively. I don't know even where to start ! so please help, thanks alot.

 

[TenaliRaman]

Consider this,
P(x) = (x-1)(x+2)Q(x) + ax + b
Find P(1) and P(-2) (You know the remainders , since u know them , try to find a and b).

-- AI

 

[HallsofIvy]

Check your problem again. I suspect that p(x)= (x-1)(x+2)q(x)+ 2x+ 3 (or, conversely, you want to divide by (x-1)(x-2)). That way, the remainders are 5 and 7 as you say, no matter what q(x) is. The "quotient" when divided by (x-1)(x+2) is q(x) and the remainder is just 2x+ 3.

 

[misogynisticfeminist]

Halls, actually its really meant to be (x-2) but yes, the answer you gave is right too, but one thing still baffles me. If say,

[tex] \frac {(x-1)(x-2) q(x) + 2x+3}{(x-1)(x-2)}[/tex]

I cannot factor out (x-1)(x-2) so that they can cancel out in the fraction and then I get 2x+3.

And I understand why when you divide (x-1) and (x-2) individually q(x) can be ignored because it is multiplied by zero. But why is q(x) ignored when it is divided by the product of (x-1) and (x-2)?

 

[TenaliRaman]

One way to think abt it,
Dividend = Divisor * Quotient + Remainder.

Another way to think abt it,
Let Q'(x) = (x-1)(x-2)Q(x)
P(x) = Q('x) + 2x + 3
Q'(x) will give remainder 0 when divided by (x-1)(x-2) thereby P(x) will give remainder 2x+3.

-- AI