Question

The maximum possible number of real roots of the equation ${x^5} - 6{x^4} - 4x + 5 = 0$ is
A. 0
B. 3
C. 4
D. 5

Answer 1

Hint:Here in the question it is required to find out the maximum number of ‘possible’ real roots not the ‘exact’ number of possible real roots. So, it is necessary to know that the question is not asked based on the maxima-minima(derivatives) concept but on the Descartes rule of signs.

Formula used:Descartes Rule of signs:
1. The number of possible positive real zeroes is equal to the number of changes in the sign of the coefficients of $f(x)$ or less by an even integer.
2.The number of possible negative real zeroes is equal to the number of changes in the sign of the coefficients of $f( - x)$ or less by an even integer.

Complete Step by step solution:
Let us assume the given equation in the question as $f(x)$ for convenience.
That is $
  f(x) = {x^5} - 6{x^4} - 4x + 5 = 0 \\
    \\
 $
Then, according to Descartes’s rule of signs, in the above equation, the number of significant sign changes are 2. That is + to – and then – to +. Therefore the number of maximum possible positive real roots are 2.
For the number of the maximum number of possible negative roots of $f(x)$,
$
  f( - x) = {( - x)^5} - 6{( - x)^4} - 4( - x) + 5 \\
    \\
 $
That is \[f( - x) = - {x^5} - 6{x^2} - 4x + 5\]
Then again, according to Descartes’s rule of signs, in the above equation, the number of significant sign changes is 1. That is – to +. Therefore the number of maximum possible negative real roots is 1.
Therefore, the total number of maximum possible real roots are 3(2+1).

Hence the correct option is B.

Note: Here, it is important to note that Descartes’s rule of signs only applies to polynomial equations or expressions. Also, Descartes’s rules only speak about the maximum number of real roots not the exact number of real roots of the equation. And it says that the number of real roots may decrease by an even integer because the complex roots or zeros come in pairs as conjugate complex numbers.