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Answer 1 · 2017-10-16T13:36:01

#(y+7/2-sqrt37/2)(y+7/2+sqrt37/2)#

We will find the zeros of the trynomial by the quadratic formula
(the zeros of the generic trynomial

#ax^2+bx+c#

are
#x=(-b+-sqrt(b^2-4ac))/(2a)# and the trynomial will be
#a(x-x_1)(x-x_2)#

Then our zeros are:

#y_(1,2)=(-7+-sqrt(7^2-4*3))/2=(-7+-sqrt37)/2=-7/2+-sqrt37/2#

and we can factor the given trynomial as:

#(y+7/2-sqrt37/2)(y+7/2+sqrt37/2)#