Question 29072
The given (y^2+7y-8) is an expression and not an equation!
We may factorise (y^2+7y-8) as follows
y^2+7y-8 
= y^2+(8y-y)-8   [splitting the middle term(the term in y) into two parts so that their sum is the middle term and their product is the product of the square term and the constant term]
[the product = -8y^2 = -(1X2X2X2)y^2= (8y)X(-y)]
=(y^2+8y)-y-8  (by additive asssociativity)
=y(y+8)-1(y+8)
=yp-p where p = (y+8)
=p(y-1)
=(y+8)(y-1)
Answer:(y^2+7y-8) = (y+8)(y-1)

To get the clue as to how the two parts should be 
you have to proceed as follows:
Multiply the square term and the constant term.
(y^2)X(-8) = -8y^2.
Observe that the product is negative. So the two parts should be 
such that one positive and the other negative.
Take into consideration all the numerical factors in the product 
and group them into two parts.
And the sum of these two parts should be the middle term which carries here a positive sign. So of the two parts the larger numerical number should be given the positive sign and the the other should be given the negative sign.