If $f\left(x\right)$  is divided by $x-3$  then the remainder is $\text{5}$ , and when it is divided by $\left(x-4\right)$  the remainder is $\text{'}0\text{'}$ . Then the remainder when it is divided by $\left(x-3\right)\left(x-4\right)$  is

By remainder theorem

$f\left(3\right)=5$ , and$f\left(4\right)=0$

$\therefore$ $f\left(x\right)÷\left(x-3\right)\left(x-4\right)$

$f\left(x\right)=\left(x-3\right)\left(x-4\right)q\left(x\right)+\left(ax+b\right)$

$f\left(3\right)=3a+b$

$5=3a+b\to \left(1\right)$

$f\left(4\right)=4a+b$

$0=4a+b$

$\Rightarrow b=-4a\to \left(2\right)$

From $\left(1\right)\&\left(2\right)$

$5=3a-4a$

$5=-a$

$a=-5$

From equation $\left(2\right)$

$b=\left(-4\right)\times \left(-5\right)$

$=20$

$\therefore$ The required remainder

$=ax+b$

$=-5x+20$