The polynomial has a factor in the form of (x-k)^2 , where k are natural numbers . Find the values of k and a .
It's f'(k), and yes, it means that you differentiate f, and then you take the value x=k.
Okay, let's say (x-k)² is a factor for f.
There exists a polynomial Q such that :
It is obvious that
Now if you differentiate, use the product rule and you'll have :
Is it clear that ?
In fact, it is true for higher powers : if f has a factor , then