# factors of a polynomial

• Jan 13th 2009, 01:46 AM
factors 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 AM
Moo
Hello,
Quote:

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 .

\$\displaystyle f(x)=2x^4-ax^3+19x^2-20x+12\$
If it has such a factor, then \$\displaystyle f(k)=0\$ and \$\displaystyle f'(k)=0\$

This should give you a system of 2 equations, and you can solve for a and k !
• Jan 14th 2009, 11:27 PM
Quote:

Originally Posted by Moo
Hello,

\$\displaystyle f(x)=2x^4-ax^3+19x^2-20x+12\$
If it has such a factor, then \$\displaystyle f(k)=0\$ and \$\displaystyle f'(k)=0\$

This should give you a system of 2 equations, and you can solve for a and k !

Thanks Moo , just wondering \$\displaystyle f(k')\$ , does it mean differentiate and why is it equals 0 .
• Jan 14th 2009, 11:38 PM
Moo
Quote:

Thanks Moo , just wondering \$\displaystyle f(k')\$ , does it mean differentiate and why is it equals 0 .

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 AM