Discussion
Solution of \(f(x) = x^{4} + 2x^{3} - 4x^{2} + 3x - 1 = 0\) is
Answer: C
The root is 0.658.
Test each option in \(f(x) = x^{4} + 2x^{3} - 4x^{2} + 3x - 1\). At \(x = 0.658\): \(x^{2} = 0.4330\), \(x^{3} = 0.2849\), \(x^{4} = 0.1875\), so
\[ \begin{aligned} f &= 0.1875 + 0.5698 - 1.7319 \\ &\quad + 1.974 - 1 \approx -0.0006 \end{aligned} \]
which is zero to the accuracy given; the root lies at about 0.6585, since \(f(0.66) \approx +0.002\). At \(x = 0.646\), \(f \approx -0.018\), thirty times further from zero, so it is not the root. At \(x = 0.333\), \(f \approx -0.36\), and at \(x = 1\), \(f = 1 + 2 - 4 + 3 - 1 = 1\), so neither of those is a solution either.
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