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).
+128407 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
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,
\(\displaystyle -596=p(5)-5^{4}, \\ p(5)=625-596=29.\)
