Question

The maximum possible number of real roots of equation $x^{5} - 6x^{2} - 4x + 5 = 0$ is

• A. 0
• B. 3
• C. 4
• D. 5

Answer 1

Answer: (B)

3

Answer 2

$f(x) = x^{5} - 6x^{2} - 4x + 5 = 0$

+ – – +

2 changes of sign ⇒ maximum two positive roots.

$f( - x) = - x^{5} - 6x^{2} + 4x + 5$

– – + +

1 changes of sign ⇒ maximum one negative roots.

⇒ total maximum possible number of real roots = 2 + 1 = 3.