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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?
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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?
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Mathematics
3 years
2021-08-02T01:05:09+00:00
2021-08-02T01:05:09+00:00 1 Answers
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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