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If the remainder on division of x ³ +2x ² +kx+3 by x-3 is 21, find the quotient and the value of k. Hence, find the zeroes of the cubic polynomial x ³ +2x ² +kx-18.

$Given x^{3} + 2 x^{2} + k x + 3 It is divided by x - 3 \Rightarrow x - 3 = 0 \Rightarrow x = 3 From the given data we get f(3)=21 \Rightarrow (3)^{3} + 2 (3)^{2} + k (3) + 3 = 21 \Rightarrow 27 + 18 + 3 k + 3 = 21 \Rightarrow 48 + 3 k = 21 \Rightarrow 3 k = - 27 \Rightarrow k = - 9 (i i) Given x^{3} + 2 x^{2} + k x - 18 here k=-9 \Rightarrow x^{3} + 2 x^{2} - 9 x - 18 Now try {f(-2)=(-2)}^{3} + 2 (- 2)^{2} - 9 (- 2) - 18 \Rightarrow f (- 2) = - 8 + 8 + 18 - 18 = 0 So -2 is one zero of the polynomial So x+2 is one factor of the polynomial Let x^{3} + 2 x^{2} - 9 x - 18 = (x + 2) (a x^{2} + b x + c) \Rightarrow x^{3} + 2 x^{2} - 9 x - 18 = {a x}^{3} + x^{2} (b + 2 a) + x (c + 2 b) + 2 c On comparing coefficients we get a=1 and 2 c = - 18 \Rightarrow c = - 9 b + 2 a = 2 \Rightarrow b + 2 (1) = 2 \Rightarrow b = 0 So x^{3} + 2 x^{2} - 9 x - 18 = (x + 2) (x^{2} - 9) \Rightarrow x^{3} + 2 x^{2} - x - 18 = (x + 2) (x - 3) (x + 3) So the zeroes of the polynomial are -2,-3,3$

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