Given:
[tex]\frac{2a^2-a-1}{a^2-1}\times\frac{4a^2+4a}{2a^2+7a+3}[/tex]Factorize the given expression.
[tex]\text{ Use -a=-2a+a.}[/tex][tex]\frac{2a^2-2a+a-1}{a^2-1}\times\frac{4a^2+4a}{2a^2+7a+3}[/tex]Take out the common terms.
[tex]\frac{2a(a-1)+(a-1)}{a^2-1}\times\frac{4a^2+4a}{2a^2+7a+3}[/tex][tex]\frac{(a-1)(2a+1)}{a^2-1}\times\frac{4a^2+4a}{2a^2+7a+3}[/tex][tex]Use\text{ }a^2-1=(a-1)(a+1).[/tex][tex]\frac{(a-1)(2a+1)}{(a-1)(a+1)}\times\frac{4a^2+4a}{2a^2+7a+3}[/tex][tex]\text{Take out the co}mmon\text{ term }4a^2+4a=4a\mleft(a+1\mright).[/tex][tex]\frac{(a-1)(2a+1)}{(a-1)(a+1)}\times\frac{4a(a+1)}{2a^2+7a+3}[/tex][tex]\text{Use 7a=6a+a in }2a^2+7a+3.[/tex][tex]\frac{(a-1)(2a+1)}{(a-1)(a+1)}\times\frac{4a(a+1)}{2a^2+6a+a+3}[/tex]Take out common term.
[tex]\frac{(a-1)(2a+1)}{(a-1)(a+1)}\times\frac{4a(a+1)}{2a(a^{}+3)+(a+3)}[/tex][tex]\frac{(a-1)(2a+1)}{(a-1)(a+1)}\times\frac{4a(a+1)}{(a^{}+3)(2a+1)}[/tex]The restrictions are
[tex]a\ne1,-1,-3\text{ and }\frac{-1}{2}\text{.}[/tex]Simplify the given expression
[tex]\frac{(a-1)(2a+1)}{(a-1)(a+1)}\times\frac{4a(a+1)}{(a^{}+3)(2a+1)}[/tex]Cancel out the common factors.
[tex]\frac{4a}{(a^{}+3)}[/tex]The answer is :
[tex]\frac{2a^2-2a+a-1}{a^2-1}\times\frac{4a^2+4a}{2a^2+7a+3},\text{ }a\ne1,-1,-3\text{ and }\frac{-1}{2}\text{.}[/tex][tex]\frac{4a}{a^{}+3},\text{ }a\ne-3.[/tex]
