The quotientire and reminder are x-2 and -2x+4 respectively when the polynomial x^3-3x^2+x+2 is divided by g(x) find g (x).

well, the highest power is 2 to get x-2 out of this

(1x^3-3x^2+1x+2)/(ax^2+bx+c)
=
[ x - 2 + (-2x+4)/(ax^2+bx+c)]
so multiply
[ x - 2 + (-2x+4)/(ax^2+bx+c)]
by (ax^2+bx+c)
= -2x+4
-2 (ax^2+bx+c)
+x (ax^2+bx+c)

= 4-2c
-2x-2bx+cx
-2ax^2+bx^2
+ax^3
which has to be x^3-3x^2+x+2
so
a = 1

-3 = -2+b
so
b = -1

1 = -2 -2(-1)+c
1 = c
so try
g(x) = x^2-x+1

check it

(It works :)

To find the polynomial g(x) given the quotient and remainder when divided by g(x), we can use long division.

The dividend is x^3 - 3x^2 + x + 2, the quotient is x - 2, and the remainder is -2x + 4.

To perform long division, we divide the dividend by the divisor (g(x)), as follows:

x - 2
___________________
g(x) | x^3 - 3x^2 + x + 2
- (x^3 - 2x^2)
________________
- x^2 + x
- (- x^2 + 2x)
________________
- x + 2
- (- x + 2)
________________
0

Looking at the remainder, which is 0, we can see that g(x) evenly divides the polynomial x^3 - 3x^2 + x + 2.

Hence, g(x) = x - 2.