Please do it for me

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 $P(0) + P(4)$ .

#Algebra

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`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
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`\frac{2}{3}`	$\frac{2}{3}$
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`\sum_{i=1}^3`	$\sum_{i=1}^3$
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Comments

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

Anik Mandal - 5 years, 5 months ago

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

Akshat Sharda - 5 years, 5 months ago

thanks

arko roychoudhury - 5 years, 5 months ago

Use the Table of Differences method to solve first one.

Anik Mandal - 5 years, 5 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$