p(x)=x*(x^{n1}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)=x^{2}x+a: both roots are complex)
Instead is n is odd p(x)>0 for x>=0 and p(x)<0 when x*(x^{n1}1)>a and (x<0).
Then: at least one real root (r) exists.
If now we express a in the form of zz^{n }then the polynomial will have a root for x=z (z is a negative number)
p(x)=(xz)(ax^{n1}+bx^{n2}+...+w) [p(x)=(xz)*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) x^{5}x+240
240= 3(3)^{5}
p(x)=(x+3)*( x^{4}3x^{3}+9x^{2}27x+80)
Edited on March 23, 2019, 9:05 am

Posted by armando
on 20190322 10:14:29 