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

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Answer 1 · 2015-10-24T15:16:46

$2x^2-x+3$

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

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