Asked by ABC
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).
Answers
Answered by
Anonymous
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
(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
Answered by
Anonymous
check it
(It works :)
(It works :)
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