Factorise $x^{2}-7 x+12$ by using the factor theorem.

Let $p(x)=x^{2}-7 x+12$

Now, if $p(x)=(x-a)(x-b),$ we know that the constant term will be $a b$.

So, $a b=12$

So, to look for the factors of $p(x),$ we look at the factors of $12 .$

The factors of $12$ are $±1,±2,±3,±4 ,$$±6$ and $±12$

Now, $p(3)=(3)^{2}-7(3)+12$

$=9-21+12$

$=21-21$

$\therefore p(3)=0$

Now, $p(4)=(4)^{2}-7(4)+12$

$=16-28+12$

$=28-28$

$\therefore p(4)=0$

$\therefore(x-3)$ and $(x-4)$ are factors of $p(x)$.

Therefore, $x^{2}-7 x+12=(x-3)(x-4)$