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?

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?

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Thanh Thu 3 years 2021-08-02T01:05:09+00:00 1 Answers 17 views 0

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    2021-08-02T01:06:46+00:00

    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

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