Multiple roots theorem proof (1 Viewer)

[WEMG]

For the proof for multiple roots theorem, what is the reason we cannot let Q(a)=0?
Thanks in advance.

 

[deterministic]

suppose we have a polynomial P(x) where "a" is a root of multiplicity n. so we can write:
P(x)=(x-a)^n*Q(x) where Q is another polynomial. Suppose Q(a) is now zero, then (x-a) must be a factor of Q(x) as a is a root of Q(x), so we can write Q(x) = (x-a)*D(x) where D(x) is another polynomial. So then P(x)=(x-a)^(n+1)*D(x). But this contradicts that "a" is a root of multiplicity n, so the assumption that Q(a)=0 is incorrect thus Q(a) cannot be zero.

 

[WEMG]

Thanks for clearing it up for me!
The Terry Lee textbook didn't mention anything about Q(a) =/= 0 but it was stated in the cambridge textbook but didnt explain why.
Thanks!