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Polynomial Long Division

I get normal long division but this doesn't make sense. How can doing it by only dividing by the leading term work? The problem is $$\dfrac{3x^3 - 2x^2 + 4x - 3 } {x^2 + 3x + 3},$$ not $$\dfrac{3x^3 - 2x^2 + 4x - 3} {x^2}.$$

Glorfindel
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johnmath
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    You are dividing by the whole divisor, but you can think of it like you are choosing a term which will get rid of the largest degree in the dividend, cancel it out, and deal with the aftermath as you move on. So to do that, you only need to look at the largest degree in the divisor. – turkeyhundt Jan 06 '15 at 00:41
  • It works the same way as euclidean division works for integers: you divide the highest (group of) figure(s) of the dividend by the highest figure of the divisor. Yet in the end you get the quotient and the remainder. – Bernard Jan 06 '15 at 00:44
  • You are involving the other terms, in the subtraction steps. – Deepak Jan 06 '15 at 01:07
  • possible duplicate of Why can/do we multiply all terms of a divisor with polynomial long division? – Caleb Stanford Jan 06 '15 at 02:16
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    There is one important difference between polynomial long division and long division for numbers: Polynomial long division doesn't involve a carry. Consider $3x(x^2+ax+b)$ what contribution do the $ax+b$ make to the leading term (the $x^3$ term) of the answer? A: they don't have any effect ... the leading term of $3x^3+ax^2+bx$ is $3x^3$, which doesn't involve an $a$ nor a $b$. – John Joy Jan 06 '15 at 18:28

2 Answers2

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As said, "it works the same way as euclidean division works for integers" (quoting Bernard). So, if you take $$A=\dfrac{3x^3 - 2x^2 + 4x - 3 } {x^2 + 3x + 3}$$ The division of $3x^3$ by $x^2$ (the highest terms) gives $3x$. So $$A=3x+\Big(\dfrac{3x^3 - 2x^2 + 4x - 3 } {x^2 + 3x + 3}-3x\Big)=3x -\frac{11 x^2+5x+3}{x^2 + 3x + 3}=3x-B$$ Now, consider $$B=\frac{11 x^2+5x+3}{x^2 + 3x + 3}$$ The division of $11x^2$ by $x^2$ (the highest terms) gives $11$. So $$B=\frac{11 x^2+5x+3}{x^2 + 3x + 3}=11+\Big(\frac{11 x^2+5x+3}{x^2 + 3x + 3}-11\Big)=11-\frac{28 x+30}{x^2 + 3x + 3}=11-C$$ Now, the highest degree in the numerator of $C$ is smaller that the highest degree in its denominator so you stop and have the remainder. So, by the end $$A=\dfrac{3x^3 - 2x^2 + 4x - 3 }{x^2 + 3x + 3}=3x - B=3x-(11-C)=3x-\Big(11-\frac{28 x+30}{x^2 + 3x + 3}\Big)=3x-11+\frac{28 x+30}{x^2 + 3x + 3}$$

Gaurav Lodhi
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Claude Leibovici
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Some trick on how to do this quickly: $3x^3 - 2x^2 + 4x - 3 = 3x(x^2+3x+3) - 11x^2 - 5x - 3 = 3x(x^2+3x+3) - 11(x^2+3x+3) + 28x + 30 = (3x-11)(x^2+3x+3) + 28x+30$.

DeepSea
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