Resolve each of the following quadratic trinomial into factor:

Solution:

Assuming $x=2 a-b$, we have :

$(2 a-b)^{2}+2(2 a-b)-8=x^{2}+2 x-8$

The given expression becomes $x^{2}+2 x-8 .                  \quad$ (Coefficient of $x^{2}=1$ and that of $x=2 ;$ constant term $=-8$ )

Now, we will split the coefficient of $x$ into two parts such that their sum is 2 and their product equals the product of the coefficient of $x^{2}$ and the cons $\tan t$ term, i.e., $1 \times(-8)=-8$.

Clearly,

$(-2)+4=2$

and

$(-2) \times 4=-8$

Replacing the middle term $2 x$ by $-2 x+4 x$, we get:

$x^{2}+2 x-8=x^{2}-2 x+4 x-8$

$=\left(x^{2}-2 x\right)+(4 x-8)$

$=x(x-2)+4(x-2)$

$=(x+4)(x-2)$

Relacing $x$ by $2 a-b$, we get:

$(x+4)(x-2)=(2 a-b+4)(2 a-b-2)$