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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 |