How do you factor 36y^2+12y-3536y2+12y35?

1 Answer
Oct 28, 2017

36y^2+12y-35=(6y-5)(6y+7)36y2+12y35=(6y5)(6y+7)

Explanation:

To factorize 36y^2+12y-3536y2+12y35, as the sign of constant term and coefficient of y^2y2 are of opposite sign (i.e. their product is negative),

we split the product of these two i.e. 35xx36=126035×36=1260 in two parts so that the difference of the two parts is equal to coefficient of middle term i.e. yy

The only two parts whose product is 12601260 and difference is 1212 are 3030 and 4242 and hence we split the middle term as follows:

36y^2+42y-30y-3536y2+42y30y35

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

= (6y-5)(6y+7)(6y5)(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 11 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).