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 .

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- Jan 13th 2009, 01:46 AMmathaddictfactors of a polynomial
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 .

- 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 :

$\displaystyle f(x)=(x-k)^2 Q(x)$

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 $\displaystyle f(k)=f'(k)=f^{(2)}(k)=\dots=f^{(n-1)}(k)=0$ - Jan 15th 2009, 06:34 AMmathaddictRe :
Thanks a lot Moo for telling me these extra things which are not in my book .