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.

Let p(x)=x^3+2x^2+kx+3 
On dividing p(x) by x-3, the remainder is 21. Therefore,
 P(3)=21
 Substituting x=3 in p(x)
 P(3)=3^3 +2*(3)^2+k*3+3
  =27+18+3k+3
  =48+3k
 We know that, p(3)=21.
 So, 48+3k=21
  3k=21-48
  3k=-27
  k=-27/3 =-9
 now, p(x) =x^3+2x^2-9x-18 
 -2 is a factor of p(x) on inspection. Therefore, divide p(x) by x+2 to find the
zeroes of the polynomial.
 On dividing, we get the factors to be, (x^2-9)(x+2)
 (x^2-3^2)(x+2)
 Factorizing using the identity a^2-b^2=(a+b)(a-b) we get,
 (x+3)(x-3)(x+2)
 Therefore, the zeroes of the polynomials are -3,+3 and -2.

9/12 is the answer because 3/4 times 3 each is 9/12