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

(N/A) Let $p(x) = x^{2}-7x+12$.
According to the factor theorem,if $(x-a)$ is a factor of $p(x)$,then $p(a) = 0$.
We look for factors of the constant term $12$. The possible factors are $\pm 1, \pm 2, \pm 3, \pm 4, \pm 6, \pm 12$.
Testing $x = 3$:
$p(3) = (3)^{2} - 7(3) + 12 = 9 - 21 + 12 = 0$.
Since $p(3) = 0$,$(x-3)$ is a factor of $p(x)$.
Testing $x = 4$:
$p(4) = (4)^{2} - 7(4) + 12 = 16 - 28 + 12 = 0$.
Since $p(4) = 0$,$(x-4)$ is a factor of $p(x)$.
Therefore,the factors of $x^{2}-7x+12$ are $(x-3)(x-4)$.