If a polynomial is divided by $(x-1)$ then remainder is 5 and if divided by $(x-2)$ the remainder is 7. What will be the remainder is the polynomial is divided by $(x-1)(x-2)$ ?
As the degree is unknown so we can't write the polynomial with arbitrary coefficients. So we have to assume the polynomial as $f(x)$. Now we can write …
$$f(x)=(x-1)g(x)+5$$
$$f(x)=(x-2)h(x)+7$$
Where $g(x)$ & $h(x)$ are some polynomial of x. then I subtract these two equations, but can't go further. Am I going correct? Should I need to use differentiation?
(If you are using some theorem please provide a link so that I can learn that)
Best Answer
By remainder theorem, “When f(x) is divided by (x−1), the remainder is 5” can be translated to:-
(1) … $f(1) = 5$
Similarly, we have:-
(2) … $f(2) = 7$
When $f(x)$ is divided by $(x-1)(x-2)$, then basically we have:-
(3) … $f(x) = (x-1)(x-2) \times $Quotient + Remainder
Since the degree of the remainder must be one lower than that of the divisor, [$= 2$ from $(x-1)(x-2)$], the remainder can have degree = 1 (or lower) only. The remainder should then take the form $ax + b$, which is a general expression of degree 1 (or lower if a = 0) in x. Therefore, (3) becomes:-
(4) …$f(x) = (x-1)(x-2)Q(x) + (ax + b)$
(1) and (2) can be used to find the values of $a$ and $b$ from (4).