SOLUTION: When ax³ + bx² + cx - 4 is divided by (x+2), the remainder is double that obtained when the expression is divided by (x+1). Show that c can have any value and find b in terms of

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Question 1157973: When ax³ + bx² + cx - 4 is divided by (x+2), the remainder is double that obtained when the expression is divided by (x+1). Show that c can have any value and find b in terms of a.
Answer by KMST(5377) About Me  (Show Source):
You can put this solution on YOUR website!
P%28x%29=ax%5E3%2Bbx%5E2%2Bcx-4
If you divide P%28x%29 by x%2B1 , you obtain a quotient Q%28x%29 and remainder r that is a constant.
You may remember that in then P%28-1%29=r
If you did not remember, you would understand that if the reminder is r ,
it means that P%28x%29=%28x%2B1%29Q%28x%29%2Br and that
for x=-1 , P%28-1%29=%28-1%2B1%29Q%28-1%29%2Br=0%2AQ%28-1%29%2Br=0%2Br=r .
So a%28-1%29x%5E3%2Bb%28-1%29%5E2%2Bc%28-1%29-4=r-->highlight%28-a%2Bb-c-4=r%29 .
Similarly the remainder, when dividing P%28x%29 by x%2B2 is
a%28-2%29x%5E3%2Bb%28-2%29%5E2%2Bc%28-2%29-4=2r-->highlight%28-8a%2B4b-2c-4=2r%29 .
Then, -8a%2B4b-2c-4=2%28-a%2Bb-c-4%29%29-->-8a%2B4b-2c-4=-2a%2B2b-2c-8%29-->+4b-2b-2c%2B2c=8a-2a-8%2B4-->2b=6a-4-->highlight%28b=3a-2%29