Answers: 1
![P(\frac{1}{2})=[\frac{1}{2}]^{3}-[\frac{1}{2}]^{2}+2*[\frac{1}{2}]-1\\\\=\frac{1}{8}-\frac{1}{4}+1-1\\\\=\frac{1}{8}-\frac{1*2}{4*2}+0\\\\=\frac{1}{8}-\frac{2}{8}\\\\=\frac{1-2}{8}\\\\=\frac{-1}{8}\neq 0](/tpl/images/0611/0754/530f3.png)
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Check 2 x-1 is a factor of p[x]=x^3-x^2+2x-1...
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