Question

Find the quotient and the remainder of

(4x^3+10x^2+12x+1) / (2x^2+3x).

Answer 1

`4x^3=2x\cdot2x^2`, and
`2x(2x^2+3x)=4x^3+6x^2`. Subtract this from the numerator to get a remainder of

`(4x^3+10x^2+12x+1)-(4x^3+6x^2)=4x^2+12x+1`

`4x^2=2\cdot2x^2`, and
`2(2x^2+3x)=4x^2+6x`. Subtract this from the previous remainder to get a new remainder of

`(4x^2+12x+1)-(4x^2+6x)=6x+1`

`6x` is not divisible by
`2x^2`, so we're done, and we've found that

`(4x^3+10x^2+12x+1)/(2x^2+3x)=2x+(4x^2+12x+1)/(2x^2+3x)`

`(4x^3+10x^2+12x+1)/(2x^2+3x)=2x+2+(6x+1)/(2x^2+3x)`

so that the quotient is
`2x+2` with remainder
`6x+1`.