Question

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

Answer 1

`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`