By using the product rule, D((x-c)kg(x))=k(x-c)k-1g(x)+(x-c)kg'(x)=(x-c)k-1(kg(x)+(x-c)g'(x)), we see that if c is a repeated root of f(x) of order k, then it is also a repeated root of f'(x) of order k-1. This gives us a quick way to test if -1 is a repeated root without factoring: we need f(-1)=f'(-1)=0.
So let f(x)=x5-ax2-ax+1, so that f'(x)=5x4-2ax-a. f(-1)=-1-a+a+1=0. f'(-1)=5+2a-a=5+a. The only way to make this 0 is if a=-5.
Of course, we can approach this without calculus too. By grouping x5-ax2-ax+1=(x5+1)-a(x2+x)=(x+1)(x4-x3+x2-x+1)-ax(x+1)=(x+1)(x4-x3+x2-x+1-ax), and for -1 to be a repeated root, it must be a zero of the second factor. Plugging in -1 to that factor, we obtain 1+1+1+1+1+a=5+a. Setting this equal to 0, we get a=-5.
As an aside, one might ask "Other than -1, what other values can be double roots of f(x) for some choice of a?" By playing around with a graphing calculator (e.g., https://www.desmos.com/calculator/ijhxeoaipi), one can see that in addition to -1 being a double root when a=-5, there is a double root close to 0.9 when a is about 0.9 and another close to -0.5 when a is about -3.8. But how might one find them?
Suppose we had a value of a such that f(x) had a double root. Then f(x) and f'(x) would share a root, c. However, we can take the equations f(c)=0 and f'(c)=0 and solve them for a:
a=(c5+1)/(c2+c) and a=(5c4)/(2c+1)
Setting these two values of a equal to each other and simplifying, we get
While we cannot solve this equation exactly, we can get numerical solutions, c≈0.837, -0.5677, and plugging into either of the above equations for a would give the approximate value of a need to get these as double roots.