# If a>0,b>0, the number of complex roots of the equation X^7-aX^4+bX^3 -8 =0

Need Explanation for this Polynomial euqation

Default Asked on July 7, 2018 in

Hello Suraj,

This polynomial equation can have a MAXIMUM of 6 complex roots, according to the Descartes’ Rule of signs.

Descartes’ Rule of signs says that,

“if the given polynomial is assumed to be f(x),

then the maximum number of positive real roots that the polynomial equation will have, will be equal to the number of sign changes between the                 consecutive terms of the polynomial f(x) [taken from left to right]. ”

Similarly,

“the maximum number of negative real roots will be equal to the number of sign changes in f(-x) [which is obtained by substituting (-x) in the place of x].”

In our case, the coefficients ‘a’ and ‘b’ are both positive numbers.

• f(x) = x^7 – ax^4+bx^3-8 has 3 sign changes. Hence, the given polynomial equation can have a maximum of 3 positive real roots.
• f(-x) = -x^7-ax^4-bx^3-8 which has ZERO sign changes. Hence, the given equation cannot have any negative real roots.

But since complex roots occur in pairs, the number of complex roots can be,

• 4 if there are 3 positive real roots.
• 6 if there is only 1 positive real root

However, I feel that some data is missing from the question and if made available, the exact number of Complex roots may be found out.

Request you to please re-confirm on the data provided in the question. Hope the above helps. Please post if you have any further queries.

Cheers,

Arvind BT

Expert Answered on August 2, 2018.