How do you factor #4y² +11y-3#?

1 Answer
Aug 9, 2015

#(4y-1)(y+3)#

Explanation:

Note 3 is a prime number so only has factors 3 and 1. 4 can have factors 2 and 2 or 4 and 1, therefore the factorisation will likely be of the form:

#(2y+-3)(2y+-1)#

or

#(4y+-(3,1))(y+-(3,1))#

In the second case by inspection we see that:

#4y^2+11y-3=(4y-1)(y+3)#