Let f(x) = x^6 - 2x^5 + x^3 + x^2 - x - 1 and | vedclass

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(B) Given $f(x) = x^6 - 2x^5 + x^3 + x^2 - x - 1$ and $g(x) = x^4 - x^3 - x^2 - 1$.By performing polynomial division of $f(x)$ by $g(x)$,we get:$f(x)

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(B) Given $f(x) = x^6 - 2x^5 + x^3 + x^2 - x - 1$ and $g(x) = x^4 - x^3 - x^2 - 1$.By performing polynomial division of $f(x)$ by $g(x)$,we get:$f(x) = (x^2 - x)g(x) + (2x^2 - 2x - 1)$.Since $a, b, c, d$ are roots of $g(x) = 0$,we have $g(a) = g(b) = g(c) = g(d) = 0$.Thus,$f(a) = 2a^2 - 2a - 1$,$f(b) = 2b^2 - 2b - 1$,$f(c) = 2c^2 - 2c - 1$,and $f(d) = 2d^2 - 2d - 1$.Summing these,we get $\sum f(a) = 2\sum a^2 - 2\sum a - 4$.From $g(x) = x^4 - x^3 - x^2 - 1 = 0$,by Vieta's formulas,$\sum a = 1$ and $\sum ab = -1$.We know $\sum a^2 = (\sum a)^2 - 2\sum ab = (1)^2 - 2(-1) = 1 + 2 = 3$.Substituting these values,$\sum f(a) = 2(3) - 2(1) - 4 = 6 - 2 - 4 = 0$.

Let $f(x) = x^6 - 2x^5 + x^3 + x^2 - x - 1$ and $g(x) = x^4 - x^3 - x^2 - 1$ be two polynomials. Let $a, b, c,$ and $d$ be the roots of $g(x) = 0$. Then,the value of $f(a) + f(b) + f(c) + f(d)$ is

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