A polynomial of degree 5 with leading coefficient 2 is such that \(P(1)=1,P(2)=4,P(3)=9,P(4)=16,P(5)=25\). Find the value of \(P(6) \).

Given that a polynomial \(P(x) \) with leading coefficient 1 of degree 4 with roots 1,2, and 3. Find the value of \(P(0) + P(4) \).

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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 – Anik Mandal · 11 months, 1 week ago

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Use the Table of Differences method to solve first one. – Anik Mandal · 11 months, 1 week ago

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The first one can be done by Lagrange's Interpolation method. – Akshat Sharda · 11 months, 2 weeks ago

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– Arko Roychoudhury · 11 months, 2 weeks ago

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