How do you factor the trinomial #y^2-7y-30#?

#y^2-7y-30 = y^2-10y+3y-30 = y(y-10)+3(y-10) #

# = (y-10)(y+3) = (y+3)(y-10)#[Ans]

For a simpler trinomial like this quadratic, in the form #y^2 + ay + b#, one can just look for a pair of numbers which multiply to get #b# and sum to #a#. In this case, they must multiply to get #-30# and add to #-7#, so the numbers are #3# and #-10# because #3*-10 = -30# and #3-10 = -7#.
If you can't immediately see that then you can use another method like completing the square:
#y^2 - 7y -30 = 0#
#(y - \frac{7}{2})^2 - (\frac{7}{2})^2 -30 = 0#
#(y - \frac{7}{2})^2 = \frac{169}{4}#
#y-\frac{7}{2} = +-sqrt\frac{169}{4} = +-\frac{13}{2}#
#y = \frac{7 +- 13}{2}#
#y = \frac{20}{2} or \frac{-6}{2} = 10 or -3#
So if our roots are 10 and -3, the factorisation of it is #(y-10)(y-(-3)) = (y-10)(y+3)#.
I hope I've explained that well enough.

To factor this, you have to find 2 numbers such that #a+b=-7# and #ab=-30# The 2 numbers, #a# and #b# are #-10# and #3#.

Since #y^2# can be expressed as #(y)(y)#, we get the following :

#y^2-7y-30=(y-10)(y+3)#

Using the FOIL method, we can expand this to check

#y^2+3y-10y-30#