Question 134835
Looks like you are dealing with:


{{{((2x^2-3x-9)/(2x^2-18))((x^2+x-6)/(2x^2-x-6))}}}


And I take it that you get the idea that you need to factor each of the polynomials so that you can eliminate common factors in the numerators and denominators.


To factor {{{2x^2-18}}}, first factor out a 2:  {{{2(x^2-9)}}}


{{{x^2-9}}} is the difference of two squares, and factors like {{{a^2-b^2=(a+b)(a-b)}}}, or in this case: {{{(x+3)(x-3)}}}


Therefore the entire factorization is: {{{2x^2-18=2(x+3)(x-3)}}}


I think that is what you were looking for.  Write back if you need more help.


Good luck,
John