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# Please do it for me

1. 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 $$P(6)$$.

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

Note by Arko Roychoudhury
1 year, 10 months ago

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Q2. 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

- 1 year, 9 months ago

Use the Table of Differences method to solve first one.

- 1 year, 9 months ago

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

- 1 year, 10 months ago

thanks

- 1 year, 10 months ago

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