Factorise $y^2 - 5y + 6$ by using the Factor Theorem.

(N/A) Let $p(y) = y^2 - 5y + 6$.
According to the Factor Theorem,if $(y - a)$ is a factor of $p(y)$,then $p(a) = 0$.
We look for factors of the constant term $6$,which are $\pm 1, \pm 2, \pm 3, \pm 6$.
Testing $y = 2$: $p(2) = (2)^2 - 5(2) + 6 = 4 - 10 + 6 = 0$.
Since $p(2) = 0$,$(y - 2)$ is a factor of $p(y)$.
Testing $y = 3$: $p(3) = (3)^2 - 5(3) + 6 = 9 - 15 + 6 = 0$.
Since $p(3) = 0$,$(y - 3)$ is a factor of $p(y)$.
Therefore,the factorisation of $y^2 - 5y + 6$ is $(y - 2)(y - 3)$.