How do find the quotient of (6x^3+5x^2+2x+1)/ (2x+3)?

1 Answer
Cesareo R.
May 12, 2016

6x^3+5x^2+2x+1 = (2x+3)(3x^2-2x+4)-11

Explanation:

Suppose that q(x) is the quocient and n(x) = 6x^3+5x^2+2x+1, d(x)=2x+3,r(x)=c_0. Then q(x)d(x)+r(x)=n(x) Putting q(x) = ax^2+bx+c (Here q(x) degree must be 3-1 because d(x) degree is 1. By the same reason r(x) degree is 0).
After multiplication of q(x)d(x)+r(x)-n(x)=0 we get:
(2a-6)x^3+(3a+2b-5)x^2+(3b+2c-2)x+3c+c_0-1=0. This equation must be identically null for all x. Solving for a,b,c,c_0 we obtain:
q(x)=3x^2-2x+4, r(x)=-11

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