Friday 20 December 2013

POLYNOMIALS A1

When mx3+nx2-18x-45 is divided by x2-9 the remainder is zero. Calculate the values of m and n.

solution

x2-9 = 0

(x)2-(3)2=0

(x+3)(x-3) = 0

x+3= 0  OR  x-3 = 0

x= 3  OR  x= -3

When x=3,

 m(3)3+n(3)2-18(3)-45=0

27m + 9n -54 -45 = 0

27m + 9n -99 = 0

27m + 9n = 99  …………………..(1)

When x=-3,
 m(-3)3+n(-3)2-18(-3)-45=0

-27m + 9n + 54 - 45 = 0

-27m + 9n + 9 = 0

-27m + 9n = -9 ………………………………(2)

Solving 1 and 2 simultaneously,

  27m + 9n = 99  …………………..(1)
-27m + 9n = -9 ………………..……(2)

By using elimination method,
      27m + 9n = 99  
-  (-27m) + 9n = -9
    54m + 0    = 108

54m = 108
54         54

m = 2

from equation (1),

27(2) + 9n = 99  

54 + 9n = 99  

9n = 99 - 54

9n = 45

9n = 45
9       9

n=5


Hence m=2 and n=5.

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