(X^3+3x^2-x+4)/(x-1)
if
F(x)=X^3+3x^2-x+4
f(1)=1+3-1+4=7 the remainder is going to be 7 (I'll just use for checking purposes)
x^2+4x +3
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x-1 |X^3+3x^2-x+4
x^3- x^2
-------------------
4x^2 -x+4
4x^2-4x
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3x+4
3x-3
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7
\(\frac{(x^3+3x^2-x+4)}{(x-1)}=x^2+4x+3+\frac{7}{x-1}\)