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.
