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If the remainder on division of x3 + 2x2 + k x + 3 by x – 3 is 21,find the quotient and the value of k . hence find the zeros of the cubic polynomial x3 + 2x2+ k x -18 ( 5 marks)

Let p(x) = x3 + 2x2 + kx + 3
Given that p(3) = 21
ie, 27 + 18 + 3k + 3 = 21
3k  = -27
k = -27 / 3 = -9

p(x) = x3 + 2x2 + kx - 18
       = x3 + 2x2 - 9x - 18
when x = -2, p(-2) = 0
ie, x + 2 is a factor, dividing p(x) by x + 2 ,we get,
x2 - x - 6 = (x - 3)(x + 2)
 Hence the zeros of x3 + 2x2 - 9x - 18 are -2 , -2 ,3


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