Question

How do you divide `(6x^4-3x^3+5x^2+2x-6)/(3x^2-2)`?

Answer 1

`2x^2-x+3`

Explanation:

Set up a standard long-division problem.
`3x^2-2 ) 6x^4-3x^3+5x^2+2x-6`

Divide first part of dividend by first part of divisor.
`(6x^4)/(3x^2)` = `2x^2` = first expression

Multiply this by the divisor
(`2x^2`) ( `3x^2-2`) = `6x^4-4x^2`.

Subtract from original.

Remainder: `-3x^3+9x^2+2x-6`

Divide first part of remainder by first part of divisor.
`(-3x^3)/(3x^2)` = `-1x` = second expression

Multiply by the divisor
(`-1x`) ( `3x^2-2`) = `-3x^2+2x`.

Subtract from remainder.

New remainder: `9x^2 -6`

Divide first part of new remainder by first part of divisor.
`(9x^2)/(3x^2)` = `3` = last expression

Multiply by the divisor
(`3`) ( `3x^2-2`) = `9x^2-6`.

Subtract from remainder = 0