The polynomial $\displaystyle 2x^4-ax^3+19x^2-20x+12=0$ 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 :
$\displaystyle f(x)=(x-k)^2 Q(x)$
It is obvious that $\displaystyle f(k)=0$
Now if you differentiate, use the product rule and you'll have :
$\displaystyle f'(k)=(x-k)^2 Q'(x)+2(x-k)Q(x)=(x-k)[(x-k)Q'(x)+2Q(x)]$
Is it clear that $\displaystyle f'(k)=0$ ?
In fact, it is true for higher powers : if f has a factor $\displaystyle (x-k)^n$, then $\displaystyle f(k)=f'(k)=f^{(2)}(k)=\dots=f^{(n-1)}(k)=0$