Resolve each of the following quadratic trinomials into factors:

3 + 23y – 8y2 = - 8y2 + 23y + 3

Here, coefficient of y2 = -8, coefficient of y = 23 and constant term = 3


We shall now split up the coefficient of x i.e., 23 into two parts whose sum is 23 and product is -8 (3) = - 24


Clearly,


24 - 1 = 23 and,


24 (-1) = - 24


So, we write middle term 23y as 24y - y


Thus, we have


-8y2 + 23y + 3 = - 82 + 24y - y + 3


= -8y (y – 3) - 1 (y – 3)


= - (8y + 1) (y – 3)


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