How do you long divide #(6x^4-13x^2+x+4)# by #(2x^2-3)#?

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Answer 1 · 2015-06-29T20:09:28

#(6x^4 - 13x^2 +x +4)/(2x^2-3) = 3x^2 – 2 + (x-2)/(2x^2-3)#

You use the process of long division, but you have to leave gaps for the missing powers of #x#.

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So,

#(6x^4 - 13x^2 +x +4)/(2x^2-3) = 3x^2 – 2 + (x-2)/(2x^2-3)#

Check:

#(2x^2-3)( 3x^2 – 2 + (x-2)/(2x^2-3))= (2x^2-3)(3x^2-2) + x - 2#

#= 6x^4 - 13x^2 +6 +x - 2 = 6x^4 - 13x^2 + x + 4#