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

Oct 24, 2015

$2 {x}^{2} - x + 3$

#### 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.
$\frac{6 {x}^{4}}{3 {x}^{2}}$ = $2 {x}^{2}$ = first expression

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

Subtract from original.

Remainder: $- 3 {x}^{3} + 9 {x}^{2} + 2 x - 6$

Divide first part of remainder by first part of divisor.
$\frac{- 3 {x}^{3}}{3 {x}^{2}}$ = $- 1 x$ = second expression

Multiply by the divisor
($- 1 x$) ( $3 {x}^{2} - 2$) = $- 3 {x}^{2} + 2 x$.

Subtract from remainder.

New remainder: $9 {x}^{2} - 6$

Divide first part of new remainder by first part of divisor.
$\frac{9 {x}^{2}}{3 {x}^{2}}$ = $3$ = last expression

Multiply by the divisor
($3$) ( $3 {x}^{2} - 2$) = $9 {x}^{2} - 6$.

Subtract from remainder = 0