Vtyjkyuk

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)

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) … $f(x) = (x-1)(x-2)$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).

HINT:

$f(x)$ can be written as $$(x-1)(x-2)g(x)+Ax+B=(x-1)(x-2)g(x)+C(x-1)+D(x-2)$$ where $g(x)$ is a finite polynomial.

$f(2)=7,Rightarrow,f = 7+(x!-!2)g. $ $ 5 = f(1) = 7-g(1),Rightarrow, g(1) = 2,Rightarrow, g = 2+(x!-!1)h$

Therefore $ f, =, 7+(x!-!2)underbrace(2+(x!-!1)h_large g), =, 2x+3 + (x!-!2)(x!-!1)h$

hint...Write $$f(x) =(x-1)(x-2)q(x)+ax+b$$ since the remainder on division by a quadratic will be of degree of at most one less. Use the known results to find $a$ and $b$.

$(ax + b)/(x-1)$, the quotient is a and the remainder is $a+b$
$(ax+b)/(x-2)$, the quotient is a and the remainder is $7$

so we have $a+b = 5$ and $2a+b = 7$, equating the two we have $a= 2$ and $b = 3$
so the remainder is $2x+3$ if said polynomial is divided by $(x-1)(x-2)$

Answer $2x+3$