Question

Determine the quotient q(x) and remainder r(x) when f(x)=18x^4+27x^3+39x^2+22x+11

Is divided by g(x)=3x^2+3x+4

Answer 1

`18x^4=3x^2\cdot6x^2`, and

`6x^2(3x^2+3x+4)=18x^4+18x^3+24x^2`

Subtracting this from
`f(x)` gives a remainder of

`(18x^4+27x^3+39x^2+22x+11)-(18x^4+18x^3+24x^2)=9x^3+15x^2+22x+11`

`9x^3=3x^2\cdot3x`, and

`3x(3x^2+3x+4)=9x^3+9x^2+12x`

Subtracting this from the previous remainder gives a new remainder of

`(9x^3+15x^2+22x+11)-(9x^3+9x^2+12x)=6x^2+10x+11`

`6x^2=3x^2\cdot2`, and

`2(3x^2+3x+4)=6x^2+6x+8`

Subtracting this from the previous remainder gives a new remainder of

`(6x^2+10x+11)-(6x^2+6x+8)=4x+3`

`4x` is not divisible by
`3x^2`, so we're done. We ended up with

`(f(x))/(g(x))=\underbrace{6x^2+3x+2}_(q(x))+\underbrace{(4x+3)/(3x^2+3x+4)}_(r(x))`