The polynomial has a factor in the form of (x-k)^2 , where k are natural numbers . Find the values of k and a .

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- Jan 13th 2009, 01:46 AMmathaddictfactors of a polynomial
The polynomial has a factor in the form of (x-k)^2 , where k are natural numbers . Find the values of k and a .

- Jan 13th 2009, 02:33 AMMoo
- Jan 14th 2009, 11:27 PMmathaddict
- Jan 14th 2009, 11:38 PMMoo
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 - Jan 15th 2009, 06:34 AMmathaddictRe :
Thanks a lot Moo for telling me these extra things which are not in my book .