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TopNewestQ2. Let the fourth root be \(\alpha\)

Then the polynomial can be written as \(P(x) = (x-1)(x-2)(x-3)(x-\alpha)\)

Now \(P(0)= (0-1)(0-2)(0-3)(0-\alpha) = 6\alpha\)

AND \(P(4) = (4-1)(4-2)(4-3)(4-\alpha) = 24-6\alpha\)

Therefore \(P(0) + P(4) = 6\alpha+24-6\alpha = 24\) !Done

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Use the Table of Differences method to solve first one.

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The first one can be done by Lagrange's Interpolation method.

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thanks

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