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

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Answer 1 · 2015-06-29T19:40:10

$(4x^3 - 6x^2 +5)/(x^2-4) = 4x – 6 + (16x-19)/(x^2-4)$

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

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So,

$(4x^3 - 6x^2 +5)/(x^2-4) = 4x – 6 + (16x-19)/(x^2-4)$

Check:

$(x^2-4)( 4x – 6 + (16x-19)/(x^2-4))= (x^2-4)(4x-6) + 16x - 19$

$= 4x^3 - 6x^2 –cancel(16x) +24 + cancel(16x) - 19 = 4x^3 - 6x^2 +5$

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