Question

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.

Answer 1

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.