Given gives the same remainder when divided by

(x-1) and (x-2) . I found that a is -2 and the remainder is 2 .

When g(x) is divided by mx+n , it also gives the same remainder (2) , so find m and n .

I am not really sure about this part ..

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- Sep 17th 2009, 03:30 AMthereddevilsremainder theorem
Given gives the same remainder when divided by

(x-1) and (x-2) . I found that a is -2 and the remainder is 2 .

When g(x) is divided by mx+n , it also gives the same remainder (2) , so find m and n .

I am not really sure about this part .. - Sep 17th 2009, 04:17 AMHallsofIvy
How did you get a= -2? If a= -2, so that , then dividing by x- 1 gives with remainder 2 but dividing by x-2 gives with remainder -8, not 2.

Do you know the "remainder theorem": If polynomial p(x) is divided by x- a, then the remainder is p(a). With a= 1, the remainder is and with a= 2, the remainder is . To have the same remainder when divided by x-1 and x- 2, a must satisfy a+ 4= 8a+ 8. - Sep 17th 2009, 04:28 AMthereddevils
- Sep 17th 2009, 05:26 AMFinley
http://www.mathhelpforum.com/math-he...084cc13c-1.gif

I suppose you could let g(x) = 2 (to find the solutions for x, when the remainder is 2:

Factorising this would then reveal all of the possible solutions for x (including mx+n).