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)................... (Using factor/remainder theorem).

Let us assume the polynomial to be [math]f\left( x \right)[/math].When [math]f\left( x \right)[/math] is divided by [math]\left( {x - 1} \right)[/math], we get the remainder as [math]5[/math].Therefore, [math]f\left( 1 \right) = 5[/math]When [math]f\left( x \right)[/math] is divided by [math]\left( {x - 2} \right)[/math], we get the remainder as [math]7[/math].Therefore, [math]f\left( 2 \right) = 7[/math]Now, when same polynomial [math]f\left( x \right)[/math]is divided by [math]\left( {x - 1} \right)\left( {x - 2} \right)[/math], the remainder is given by:[math]f\left( x \right) = q\left( x \right).\left( {x - 1} \right)\left( {x - 2} \right) + r\left( x \right)[/math]When, [math]x =1[/math][math] \Rightarrow f\left( 1 \right) = 0.\left( {x - 2} \right)q\left( 1 \right) + r\left( 1 \right) = 5[/math]When, [math]x = 2[/math][math] \Rightarrow f\left( 2 \right) = 0.\left( {x - 1} \right)q\left( 2 \right) + r\left( 2 \right) = 7[/math]Solving above two equations, we get remainder [math]r\left( x \right) = 2x + 3[/math]