(B) Given the polynomial equation $P(x) = x^4-2x^3+2x-1=0$. Since $1$ is a root of order $3$,$(x-1)^3$ must be a factor of $P(x)$. Expanding $(x-1)^3

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(B) Given the polynomial equation $P(x) = x^4-2x^3+2x-1=0$. Since $1$ is a root of order $3$,$(x-1)^3$ must be a factor of $P(x)$. Expanding $(x-1)^3 = x^3-3x^2+3x-1$. Dividing $x^4-2x^3+2x-1$ by $(x-1)^3$: $x^4-2x^3+2x-1 = (x-1)^3(x+1)$. Setting the factors to zero,we get $(x-1)^3 = 0$ and $x+1 = 0$. Thus,the roots are $1, 1, 1$ and $-1$. The other root is $-1$.

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If $1$ is a multiple root of order $3$ for the equation $x^4-2x^3+2x-1=0$,then the other root is

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$0$
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$-1$
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$1$
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$2$

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