When a polynomial f(x) is divided by (x - 1), the remainder is 5 and when it is, divided by (x - 2), the remainder is 7. Find the remainder when it is divided by (x - 1)(x - 2).

By remainder theorem, on dividing f(x) by (x - a), the remainder left is f(a).

Given, when f(x) is divided by (x - 1), remainder = 5

∴ f(1) = 5

Given, when f(x) is divided by (x - 2), remainder = 7

∴ f(2) = 7

Suppose on dividing f(x) by (x - 1)(x - 2),

Quotient = q(x)

Remainder = ax + b

So, f(x) = (x - 1)(x - 2)q(x) + ax + b

Putting x = 1, we get:

$\Rightarrow f(1) = (1 - 1)(1 - 2)q(1) + a(1) + b = 5 \\[0.5em] \Rightarrow 0 + a + b = 5 \\[0.5em] \Rightarrow a + b = 5 \\[0.5em] a = 5 - b \text{ \space (Equation 1)}$

Putting x = 2, we get:

$\Rightarrow f(2) = (2 - 1)(2 - 2)q(2) + a(2) + b = 7 \\[0.5em] \Rightarrow 0 + 2a + b = 7 \\[0.5em] \Rightarrow 2a + b = 7$

Putting value of a from equation 1,

$\Rightarrow 2(5 - b) + b = 7 \\[0.5em] \Rightarrow 10 - 2b + b = 7 \\[0.5em] \Rightarrow b = 10 - 7 \\[0.5em] \Rightarrow b = 3 \\[0.5em] \text{and } a = 5 - b = 5 - 3 = 2.$

Remainder = ax + b = 2x + 3.

∴ The remainder when polynomial is divided by (x - 1)(x - 2) is 2x + 3.