On R[X] f=X^4-aX^3-aX+1
So what does R[X] mean ?
Bob
]]>That's what I thought you meant and that's where my difficulty began.
If you go to this page of the MIF site
http://www.mathsisfun.com/data/function … x)&func2=2
you will find a very useful function grapher. It even allows you to put in an 'a' and adjust its value using the slider.
This shows no roots at all.
Bob
]]>I didn't post earlier because I do not understand what you mean by X1 X2 etc.
Bob
]]>By the way,any clues on 1st problem?
]]>After substitution into the general form you get.
Can you solve that or do you need a little more?
]]>Hi;
f(1)=9
f(2)=-5
f(3)=0
f(4)=35
Could you describe your solution step by step please? For f(1)=8 not 9 sorry.
]]>f(1)=9
f(2)=-5
f(3)=0
f(4)=35
f(1)=8
f(2)=-5
f(3)=0
f(4)=35
f=?
The thing is i know how to solve the problem but i cannot really solve the system.
f(3)=0 -> x1=3 solution ->f = (X-3)(aX^2+bX+c)==>
==>f=aX^3+(b-3a)X^2+(c-3b)X-3c
so f(3)=0 ==>-3c=0 =>c=0
f(1)=8 ==> a+b=-4
f(2)=-5 ==>-4a-2b=-5==>4a+2b=5
f(4)=35 ==> 16a+4b=35
Can anyone solve the system or at least tell me where i'm wrong ? I'm getting very mad here.
]]>On R[X] f=X^4-aX^3-aX+1
Proove that if |a|<1 then |x1|=|x2|=|x3|=|x4|=1;
Good luck.