I did not understand the solution provided by AB

According to the Fundamental Theorem of Algebra, a polynomial of degree $n$ has exactly $n$ roots (including real and non-real/complex roots).

• Degree of $f(x)$: 6
• Total Roots: 6
• Given Non-Real Roots: 4
• Remaining Real Roots: $6 - 4 = \mathbf{2}$

We now know there are exactly 2 real roots. These 2 roots must be either positive, negative, or zero.

Testing for Positive Roots ($x > 0$)

Look at the polynomial:

$f(x) = x^6 + 5x^5 + 11x^4 + 34x^2 + 20x + 24$

Notice the coefficients: 1, 5, 11, 34, 20, and 24.

• Every single coefficient is positive.
• If you plug in any positive number for $x$ (like $x = 1, 2, 0.5$, etc.), every term in the equation will be positive.
• A sum of positive numbers can never equal zero.

Conclusion: $f(x)$ will always be greater than zero for any $x > 0$. Therefore, there are 0 positive real roots.

Testing for Zero ($x = 0$)

If we plug in $x = 0$:

$f(0) = 0 + 0 + 0 + 0 + 0 + 24 = 24$

Since $f(0) \neq 0$, zero is not a root.

Solving the Mystery

We can now summarize the types of roots in a simple table:

Type of Root	Number of Roots	Reasoning
Non-Real (Complex)	4	Given in the question.
Positive Real	0	All coefficients are positive; sum can't be 0.
Zero	0	Constant term is 24, so $f(0) \neq 0$.
Negative Real	2	The only remaining spot for the 2 real roots.