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