How do you factor #36y^2+12y-35#?

To factorize #36y^2+12y-35#, as the sign of constant term and coefficient of #y^2# are of opposite sign (i.e. their product is negative),

we split the product of these two i.e. #35xx36=1260# in two parts so that the difference of the two parts is equal to coefficient of middle term i.e. #y#

The only two parts whose product is #1260# and difference is #12# are #30# and #42# and hence we split the middle term as follows:

#36y^2+42y-30y-35#

= #6y(6y+7)-5(6y+7)#

= #(6y-5)(6y+7)#

Note - (1) If signs are same select two numbers whose product is product of constant term and coefficient of highest power and sum is equal to middle term.

(2) In case one finds difficult to get two factors whose sum or difference is equal to middle term, try out starting from #1# and build pairs #(1,1260),(2,630),(3,420),(4,315),(5,252),(6,210),(7,180),(9,140),(10,126),(12,105),(14,90),(15,84),(18,70),(20,63),(21,60),(28,45),(30,42),..#.

Here you find a difference (or sum in case you need that) at a very late stage and hence takes time. But if product is large and sum/difference is much smaller, one can try starting with a factor, which is close to square root of the product. Here product was #1260# and its square root is #~=35#, hence, one could start from #(35,36)# and then comes the pair #30,42)#.