Question 20701
numerator, {{{9 - x^2}}} can be thought of as {{{3^2-x^2}}}, which is difference of 2 squares. This factorises into (3-x)(3+x)


denominator, {{{10x^2 - 28x - 6}}} can be simplified into {{{2(5x^2-14x-3)}}}. This can then be factorised, by thinking about it a little as 2(5x+1)(x-3)


So we have, {{{((3-x)(3+x))/(2(5x+1)(x-3))}}}


This can be simplified further by realising that (3-x) is really just -(x-3).. this sort of thing is very useful to remember in manipulation of algebra terms. 


So, {{{(-(x-3)(3+x))/(2(5x+1)(x-3))}}} which then simplifies to {{{(-(3+x))/(2(5x+1))}}}


cheers
Jon.