I need help to solve polynomial question

Question

Given that 2x^3 - x^2 + 6 = (3+2x)[(x+p)^2] + qx + r , where p, q and r are constant. By using long division, calculate the values of p, q and r.

Answer 1

2x^3 -x^2 + 6 = (3+2x)(x+p)^2 + qx +r

= (3+2x)(x^2 +2xp+p^2) + qx+ r

= 2x^3 + 4x^2p + 2xp^2 + 3x^2 +6xp + 2xp^2 +qx +r

= 2x^3 +(4p+3)x^2 + (2p^2 +6p +2p^2+q)x + r

set coefficients equal

-1 = 4p+3 for the x^2 terms

4p=-4

p =-1

2p^2 +6p+2p^2 +1 = 0

4p^2 +6p+q =-1

4-6 +q =-1

q =1

r=6

p=-1, q=1, r=6