1. ## remainder theorem

When a polynomial f(x) is divided by x-1 and x-2, the remainders are 0 and -4 respectively. Find the remainder
when f(x) is divided by (x-1)(x-2).

The answer of this question is -4x+4

Can anyone show me how to work out the solution of this question? Thanks

2. ## Re: remainder theorem

We know:

$\frac{f(x)}{x-2}=Q_1(x)-\frac{4}{x-2}$

$\frac{f(x)}{x-1}=Q_2(x)$

Subtracting the latter from the former, we find:

$f(x)\left(\frac{1}{x-2}-\frac{1}{x-1} \right)=Q_1(x)-Q_2(x)-\frac{4}{x-2}$

Let $Q_3(x)=Q_1(x)-Q_2(x)$ and we may write:

$\frac{f(x)}{(x-1)(x-2)}=Q_3(x)+\frac{4(1-x)}{(x-1)(x-2)}$

Hence the remainder is $4(1-x)=4-4x$.

3. ## Re: remainder theorem

I don't understand why we need to subtract the later from the former and also why do we have to let Q3 =Q2-Q1? I hope to understand the logic and so can solve similar problem in my textbook . Thanks

4. ## Re: remainder theorem

Originally Posted by flower
I don't understand why we need to subtract the later from the former and also why do we have to let Q3 =Q2-Q1? I hope to understand the logic and so can solve similar problem in my textbook . Thanks
You don't have to let Q3 = Q2-Q1, it's just to make it easier to se that when we subtract the later from the former we actually get a new quotient with a new remainder which we want to write so that we have f(x)/((x-1)(x-2)) and r/((x-1)(x-2)). And in this case r = 4(1-x)

5. ## Re: remainder theorem

Thank you very much. I understand this question now.

However, I still can't solve the following two questions
1. f(x) is a polynomial. When 3x-2 divides f(x) , the remainder is K. when 2-3x divides f(x) , the remainder is

I try to set up this way following your method to last question,
3x-2/f(x) = Q1(x) + K/f(x)

so 2-3x/f(x) = -(3x-2) / f(x) + Q1(x) + (-k)/f(x)

so the answer show be -K which is different from the textbook answer.

2) Let f(x) be a polynomial. If f(x) is divisible by x-1, which of the following must be a factor of f(2x+1)?

I have no idea how to do this question at all.

6. ## Re: remainder theorem

1.) The remainder theorem tells us that since the two divisors have the same root, the function will have the same remainder when divided by them.

2.) We may write:

$f(x)=(x-1)Q(x)$ and so:

$f(2x+1)=((2x+1)-1)Q(2x+1)=2xQ(2x+1)$

Thus, $x$ is a factor of $f(2x+1)$.

7. ## Re: remainder theorem

Thank you very much. I understand them all now.