Resolve each of the following quadratic trinomials into factors:

Here, coefficient of x2 = 14, coefficient of x = 11y and constant term = - 15y2

We shall now split up the coefficient of middle term i.e., 11y into two parts whose sum is 11y and product is 14 (-15y2) = - 210y2


Clearly,


21y – 10y = 11y and,


(21y) (-10y) = - 210y2


So, we replace middle term 11xy = 21xy – 10xy


Thus, we have


14x2 + 11xy- 15y2 = 14x2 + 21xy - 10xy - 15y2


= 2x (7x – 5y) + 3y (7x – 5y)


= (2x + 3y) (7x - 5y)


14