How do you divide $\frac{6 x^{4} - 3 x^{3} + 5 x^{2} + 2 x - 6}{3 x^{2} - 2}$ $(6x^4-3x^3+5x^2+2x-6)/(3x^2-2)$ ?

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Answer 1 · 2015-10-24T15:16:46

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

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

Divide first part of dividend by first part of divisor.
$\frac{6 x^{4}}{3 x^{2}}$ $(6x^4)/(3x^2)$ = $2 x^{2}$ $2x^2$ = first expression

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

Subtract from original.

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

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

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

Subtract from remainder.

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

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

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

Subtract from remainder = 0

How do you divide 6x4−3x3+5x2+2x−63x2−2(6x^4-3x^3+5x^2+2x-6)/(3x^2-2)?