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