Let p(x) be a fifth degree polynomial with real coefficients such that p(x)+1 is divisible by (x-1)3 and p(x)-1 is divisible by (x+1)3. Find the value of ∫p(x)dx from -10 to 10.
The trick here is to differentiate p(x) first!
From "p(x)+1 is divisible by (x-1)^3" we can differentiate and see that p'(x) is divisible be (x-1)^2.
Similarly, from "p(x)-1 is divisible by (x+1)^3" we can differentiate and see that p'(x) is divisible be (x+1)^2.
Given that p(x) is a fifth degree polynomial then p'(x) is a fourth degree polynomial; but we already have four roots of p'(x). So then p'(x) = k*(x-1)^2*(x+1)^2 for some constant k.
Rewrite p'(x) as k*(x^2-1)^2, then it is easier to see that p'(x) is in fact an even function. Then p(x) is an odd function plus some constant. Then p(1)+p(-1) equals two times that constant.
Go back to "p(x)+1 is divisible by (x-1)^3"; expressed differently this statement can be written as p(x) = (x-1)^3*(something) - 1. Then evaluating at x=1 we see that p(1) = -1.
Similarly, going back to "p(x)-1 is divisible by (x+1)^3" we can conclude that p(-1)=1.
Now p(1)+p(-1) = 2*constant = -1 + 1 = 0. Then the constant term of p(x) is zero. Thus p(x) is in fact an odd function.
Finally we are ready to integrate! p(x) is an odd function being integrated on a symmetric interval. This is a well known property that the definite integral will evaluate to zero.
Solution
