p(x)=x*(xn-1-1)+a and a integer and a>0
If n is even p(x)>0 for all values of x, and there are no posible roots and the polynomial do not admit real factors (ex: p(x)=x2-x+a: both roots are complex)
Instead is n is odd p(x)>0 for x>=0 and p(x)<0 when |x*(xn-1-1)|>a and (x<0).
Then: at least one real root (r) exists.
If now we express a in the form of z-zn then the polynomial will have a root for x=z (z is a negative number)
p(x)=(x-z)(axn-1+bxn-2+...+w) [p(x)=(x-z)*q(x)]
and the coefficients of q(x) will be the potencies of z.
So the answer should be for n=odd number.
Ex:
p(x) x5-x+240
240= -3-(-3)5
p(x)=(x+3)*( x4-3x3+9x2-27x+80)
Edited on March 23, 2019, 9:05 am
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Posted by armando
on 2019-03-22 10:14:29 |
posible solution