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Task:

Do polynomial division for

$$\frac{x^5+1}{x^4 + x^3 + x^2}$$

The question is really about finding all primitive functions via partial fractions, but I know how to do all the other steps without a problem once I have it in a form that I can use partial fraction reasoning on.

Attempted solution:

Because the exponent is higher for the numerator, we can use polynomial division:

 x - 1 + 1/ x^2 - 1/ x^3
 -----------------------------------
 x^5 + 1              | x^4 +x^3 + x^2
                      _____________
-(x^5+x^4+x^3)
----------------------
 -x^4 - x^3 + 1
-(-x^4-x^3 - x^2)
----------------------
 x^2 + 1
 -(x^2 + x + 1)
-----------------------
 -x
-(-x - 1 - 1/x)
-----------------------
1 + 1/x

This just seems to go on and on and on without reaching any meaningful end. It also seems possible to factor out $x^2$ from the denominator and maybe something from the numerator, but I cannot find a way to make them cancel out in a meaningful way. Perhaps forcibly factor out $x^2$ from the numerator?

Any advice on how to proceed with this question?

José Carlos Santos
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MathInferno
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  • I get $$x-1+\frac{x^2+1}{x^4+x^3+x^2}.$$ – Angina Seng Oct 20 '19 at 15:08
  • You usually stop dividing when the degree of the intermediate result (the polynomial below the dashed line) is less than the degree of the divisor. Whatever you have calculated so far becomes the "integer part", the last intermediate result becomes the numerator of the fractional part, and the original divisor becomes the denominator of the fractional part of the result. So this does not go on and on. You should have stopped at $x^2+1.$ – Reinhard Meier Oct 20 '19 at 15:18

2 Answers2

1

Note that$$x^5+1=(x^4+x^3+x^2)\times x-x^4-x^3+1$$and that$$-x^4-x^3+1=(x^4+x^3+x^2)\times(-1)+x^2+1.$$Therefore,$$x^5+1=(x^4+x^3+x^2)\times(x-1)+x^2+1$$and so$$\frac{x^5+1}{x^4+x^3+x^2}=x-1+\frac{x^2+1}{x^4+x^3+x^2}.$$

José Carlos Santos
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1

$\frac{x^5+1}{x^4 + x^3 + x^2}$

Using long division (we stop when we get to power less than 4):

enter image description here

NoChance
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