We have to find the nature of the roots of the polynomial "2x^5+x^3+5x+1".
Assume that "f(x)=2x^5+x^3+5x+1". (1)
By the Descartes rule of signs, number of positive real real roots is 0 (because the number of changes of signs is 0).
Let's substitute -x to the (1):
"f(-x)=2(-x)^5+(-x)^3+5(-x)+1"
"f(-x)=-2x^5-x^3-5x+1" (2)
We see that in (2) the sign changes only once (from negative (-5x) to positive (1)).
Thus, by the Descartes rule of signs, number of negative real roots is 1 (because the number of changes of signs is 1).
By the condition, the degree of "f(x)" is 5. Thus, by the fundamental theorem of algebra, the total number of roots is 5. So, remaining roots "=5-1=4." These roots are imaginary.
Answer: 1 real negative and 4 imaginary roots.
Comments
Leave a comment