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Question 180851: find the value of k so that 9x^3-2x^2+kx+6/(x+2) has a remainder of 8.
Answer by ptaylor(2198)   (Show Source):
You can put this solution on YOUR website! PERFORM ROUTINE LONG DIVISION AND YOU WILL END UP WITH A REMAINDER OF -74-2K AND THAT MUST EQUAL 8
Divide (x+2) into (9x^3-2x^2+kx+6)
First iteration, we get 9x^2
(9x^3-2x^2+kx+6)-(9x^3+18x^2)=-20x^2+kx+6
Divide (x+2) into (-20x^2+kx+6)
Second iteration, we get -20x
(-20x^2+kx+6)-(-20x^2-40x)=x(40+k)+6
Divide (x+2) into x(40+k)+6
Third iteration, we get x(40+k)+80+2k
(x(40+k))+6)-(x(40+k))+80+2k)=-74-2k --AND THIS IS THE REMAINDER
Now we are told that:
-74-2k=8 add 72 to each side
-2k=74+8=82
k=-41
Hope this helps---ptaylor
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