Find the remainder when the polynomial x^18 + x^13 + x^7 + x^4 + x is divided by x^3 - x .
+128408 x^15 + x^13 + x^11 + x^10 + x^9 + x^8 + x^7 +x^6 + x^5 + 2x^4 + x^3 + 2x^2 + 2x + 2
x^3 - x [ x^18 + x^13 + x^7 + x^4 ]
x^18 - x^16
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x^16 + x^13 + x^7 + x^4
x^16 - x^14
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x^14 + x^13 + x^7 + x ^4
x^14 -x^12
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x^13 + x^12 + x^7 + x^4
x^13 - x^11
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x^12 + x^11 + x^7 + x^4
x^12 - x^10
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x^11 + x^10 + x^7 + x^4
x^11 - x^9
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x^10 + x^9 + x ^7 + x^4
x^10 - x^8
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x^9 + x^8 + x^7 + x^4
x^9 -x^7
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x^8 + 2x^7 + x^4
x^8 -x^6
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2x^7 + x^6 + x^4
2x^7 - 2x^5
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x^6 + 2x^5 + x^4
x^6 -x^4
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2x^5 + 2x^4
2x^5 - 2x^3
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2x^4 + 2x^3
2x^4 - 2x^2
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2x^3 + 2x^2
2x^3 - 2x
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2x^2 + 2x = the remainder
Suppose that
\(\displaystyle \frac{x^{18}+x^{13}+x^{7}+x^{4}+x}{x(x^{2}-1)}=Q(x)+\frac{R(x)}{x(x^{2}-1)}\)
Q(x) will be of degree 15, ( but we don't need to know that anyway), and R(x) will be of degree 2 at most, 1 less than the degree of the denominator.
Note that we could cancel an x, top and bottom, in the original fraction, so that we might expect the remainder term to be of the form R*(x)/(x^2 - 1), where R*(x) is of degree 1, or a constant.
So, in the original equation, let
\(\displaystyle R(x) = ax^{2}+bx + c,\)
and multiply throughout by
\(\displaystyle x(x^{2}-1),\\ \text{then } \quad x^{18}+x^{13}+x^{7}+x^{4}+x=Q(x).x(x^{2}-1)+ax^{2}+bx+c.\)
Now let x = 0, 1, -1 and we have the equations
0 = c,
5 = a + b,
-1 = a - b.
so a = 2, b = 3, and c = 0.
That means that the remainder term is \(\displaystyle \frac{2x^{2}+3x}{x(x^{2}-1)},\text{which simplifies to }\frac{2x+3}{x^{2}-1}.\)
