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

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Answer 1 · 2015-06-29T20:36:49

$(6x^4 – 3x^3 + 5x^2 +2x -6)/(3x^2-2) = 2x^2 –x +3$

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

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So,

$(6x^4 – 3x^3 + 5x^2 +2x -6)/(3x^2-2) = 2x^2 –x +3$

Check:

$(3x^2-2)( 2x^2 –x +3)= 6x^4 – 3x^3 + 9x^2 -4x^2 +2x -6 = 6x^4 – 3x^3 + 5x^2 + 2x -6$

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