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.