y²+8y+k
Suppose it factors as
(y+a)(y+b) where a and b are positive integers.
Then FOIL-ing that out,
y²+by+ay+ab
Factor y out of the middle two terms:
y²+(b+a)y+ab
Compare that to
y²+8y+k
So the coefficient of y which is b+a=8, and the last term ab=k
b a b+a ab=k factorization
1 7 8 7=k y²+8y+7 = (y+1)(y+7)
2 6 8 12=k y²+8y+12 = (y+2)(y+6)
3 5 8 15=k y²+8y+15 = (y+3)(y+5)
4 4 8 16=k y²+8y+16 = (y+4)(y+4) = (y+4)²
Answer: all positive integral valuse for k are 7,12,15, and 16
Now you do the other one the same way.
Edwin