1. ## remainder theorem

Given $g(x)=ax^3-x^2+7x-2$ 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 .

2. Originally Posted by thereddevils
Given $g(x)=ax^3-x^2+7x-2$ 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 .

How did you get a= -2? If a= -2, so that $g(x)= -2x^3- x^2+ 7x- 2$, then dividing by x- 1 gives $-2x^2- 3x+ 4$ with remainder 2 but dividing by x-2 gives $-2x^2- 5x- 3$ 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 $a(1^3)- 1^2+ 7(1)- 2= a-1+ 7 -2= a+ 4$ and with a= 2, the remainder is $a(2^3)- 2^2+ 7(2)- 2= 8a- 4+ 14- 2= 8a+ 8$. To have the same remainder when divided by x-1 and x- 2, a must satisfy a+ 4= 8a+ 8.

3. Originally Posted by HallsofIvy
How did you get a= -2? If a= -2, so that $g(x)= -2x^3- x^2+ 7x- 2$, then dividing by x- 1 gives $-2x^2- 3x+ 4$ with remainder 2 but dividing by x-2 gives $-2x^2- 5x- 3$ 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 $a(1^3)- 1^2+ 7(1)- 2= a-1+ 7 -2= a+ 4$ and with a= 2, the remainder is $a(2^3)- 2^2+ 7(2)- 2= 8a- 4+ 14- 2= 8a+ 8$. To have the same remainder when divided by x-1 and x- 2, a must satisfy a+ 4= 8a+ 8.
sorry , i think there is a typo in the equation g(x) but i don hv the question with me now . I know about part 1 and my answers are correct .

So can u just guide me through part 2 , how would i solve for m and n ??

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

$2=-2x^3-x^2+7x-2$

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