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
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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]
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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