SOLUTION: ax^3+bx^2+x-6 has factor (x+2) and (x-2) then in both cases remainder is 4 then find the value of a and b? thanks if anybody can solve it for me

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Question 799653: ax^3+bx^2+x-6 has factor (x+2) and (x-2) then in both cases remainder is 4 then find the value of a and b? thanks if anybody can solve it for me
Answer by josgarithmetic(39623) About Me  (Show Source):
You can put this solution on YOUR website!
Many details to the process, but you would want to do synthetic division using "divisors" of x+2, and of x-2, each separately. The roots correspondingly being checked are -2 and +2.

The dividend would be one of the factors of ax%5E3%2Bbx%5E2%2Bx%2B6, once a has been factored. This dividend is x%5E3%2B%28b%2Fa%29x%5E2%2Bx%2Fa%2B6%2Fa. During the division, the intermediary expressions become more complicated...

The root, -2 gives remainder %284%2B4b-8a%29%2Fa=4, equated according to the description of the problem; and the root +2 gives remainder %288%2B4b%2B8a%29%2Fa=4.
These seem to be two equations in the unknowns, a and b.

Finishing the solution for a and b from here should be no trouble, seems to appear straightforward.