Resolve each of the following quadratic trinomial into factor:

Solution:

The given expression is $\mathrm{a}^{2}-5 \mathrm{a}+6$.

Assuming $\mathrm{a}=\mathrm{x}-2 \mathrm{y}$, we have :

$(\mathrm{x}-2 \mathrm{y})^{2}-5(\mathrm{x}-2 \mathrm{y})+6=\mathrm{a}^{2}-5 \mathrm{a}+6 \quad$ (Coefficient of $\mathrm{a}^{2}=1$, coefficient of $\mathrm{a}=-5$ and constant term $\left.=6\right)$

Now, we will split the coefficient of a into two parts such that their sum is $-5$ and their product equals the product of the coefficient of $\mathrm{a}^{2}$ and the constant term, i.e., $1 \times 6=6$.

Clearly,

$(-2)+(-3)=-5$

and

$(-2) \times(-3)=6$

Replacing the middle term $-5 \mathrm{a}$ by $-2 \mathrm{a}-3 \mathrm{a}$, we have :

$a^{2}-5 a+6=a^{2}-2 a-3 a+6$

$=\left(a^{2}-2 a\right)-(3 a-6)$

$=a(a-2)-3(a-2)$

$=(a-3)(a-2)$

Replacing a by $(\mathrm{x}-2 \mathrm{y})$, we get:

$(a-3)(a-2)=(x-2 y-3)(x-2 y-2)$