The number a is called a double zero (or a zero of multiplicity 2) of the polynomial P if
P(x) = (x -a)^2 q(x) and q(a) not = 0.
Prove that if a is a double zero of P, then a is a zero of both P and P', and P"(a) not = 0.
Recall that a Zero of a polynomial function P is one number a is subset to R such that P(a) = 0. You may assume without proof that q is a polynomial function.
I do not understand the question at all, can anyone explain it and give some hints to prove it?
You are told that a is a double zero and told exactly what that means. Using that information, you are supposed to show that for ANY polynomial and ANY point satisfying this double zero property, the function evaluated at the point is 0, as is the derivative, but the second derivative is not.
Where did those "p" and "q" terms come from? In your original statement, P and q were functions, not constants.
If , P'= 2(x-a)q(x)+ (x-a)^2q'(x). Notice that each term contains a factor of "(x-a)" and so P'(a)= 0. Now differentiate again, using the product rule on each term.