SOLUTION: when the polynomial 5x^3 + Mx + N is divided by x^2 + x + 1 the remainder is 0. what is M + N ? a -3 b 5 c -5 d 15 e none of the above

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Question 252530: when the polynomial 5x^3 + Mx + N is divided by x^2 + x + 1 the remainder is 0.
what is M + N ?
a -3 b 5 c -5 d 15 e none of the above

Found 2 solutions by drk, Edwin McCravy:
Answer by drk(1908)   (Show Source): You can put this solution on YOUR website!
Be careful setting this up. The best way to go here is polynomial division:
/
Dividing carefully gets a quotient of 5X - 5 with a left over part of Mx + N + 5, but the remainder = 0, so,
.
Now set Mx = 0 and solve for M, M = 0
Set (N + 5) = 0 and N = -5.
M + N = 0 + -5 = -5, [C]

Answer by Edwin McCravy(20059)   (Show Source): You can put this solution on YOUR website!


                                 5x -     5
           --------------------------------  
x² + x + 1 ) 5x³ + 0x² +         Mx +     N
             5x³ + 5x² +         5x
             ----------------------
                  -5x² +     (M-5)x +     N
                  -5x² -         5x -     5
                  -------------------------
                         [(M-5)+5]x + (N+5)

Remainder = [(M-5)+5]x + (N+5) = 0
              [M-5+5]x + (N+5) = 0
                    Mx +  N+5  = 0

The only this can be identically 0 for all values of x
is for the coefficient of x to be 0.

                      So M=0, so
                  
                         N+5 = 0
                           N = -5

Therefore M + N = 0 + (-5) = -5  choice c

Edwin

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