Question

Given that the trinomial x^2 + bx + 8 is factorable as (x + p)(x + q), with p and q being integers, what are four possible values of b?

Answer 1

Answer:

Step-by-step explanation:

coefficient of x²=1

constant term=8

1×8=8

b=(p+q)

such that p×q=8

b=1+8,1×8=8

b=2+4,2×4=8

b=-1-8,-1×-8=8

b=-2-4,-2×-4=8

values of b are 9,6,-9,-6

so