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Let p(x) be a monic polynomial of degree 4 such that p(1) = 1, p(2) = 2, p(3) = 3, and p(4) = 4.  Find p(5).

 Jan 5, 2021
 #1
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We  have    x^4 + bx^3  + cx^2   + dx + e

 

We  have  the following system......

 

1 + b + c + d + e   =  1    ⇒   b + c + d + e   = 0

2^4  + 8b + 4c + 2d + e  =  2   ⇒   8b + 4c + 2d + e  =  -14

3^4  + 27b + 9c + 3d + e  =  3  ⇒   27b + 9c + 3d + e  =  -78

4^4  + 64b + 16c + 4d + e  =  4 ⇒   64b + 16c + 4d + e  =  -252

 

 

b + c + d + e   = 0

 8b + 4c + 2d + e  =  -14

 27b + 9c + 3d + e  =  -78

 64b + 16c + 4d + e  =  -252

 

This isn't  difficult to solve, but it is tedious....I've  used some tecnology  to find the  values  of b,c,d,and e

 

b = -10   c   = 35    d   = -49   e  = 24

 

So....the polynomial is

 

x^4  -10x^3 + 35x^2 -49x  + 24

 

f(5)  =    5^4  - 10(5)^3  + 35(5)^2  - 49(5) + 24    =   29

 

 

cool cool cool

 Jan 6, 2021
edited by CPhill  Jan 6, 2021
 #2
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Let

 \(\displaystyle p(x)=x^{4}+ax^{3}+bx^{2}+cx+d, \\ \text{and} \\ q(x)=p(x)-x^{4}.\)

 

q(x) is a cubic, so its third difference column will be constant.

 

\(\displaystyle q(1)=p(1)-1^{4}=1-1=0, \\q(2)=p(2)-2^{4}=2-16=-14, \\ q(3)=p(3)-3^{4}=3-81=-78, \\ q(4)=p(4)-4^{4}=4-256=-252.\)

 

The third difference turns out to be -60, and running that down the diagonal leads to q(5) = -596.

 

From that, 

 

\(\displaystyle -596=p(5)-5^{4}, \\ p(5)=625-596=29.\)

 Jan 6, 2021

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