Question

The cubic equation $4x^3 + 3x^2 + 2x + 1 = 0$ has:

• A. 3 real roots
• B. A real root in the interval $[-2, -1]$
• C. No real roots
• D. 1 real root

Hint:

For any cubic equation, if the derivative has no real roots (discriminant $D < 0$), the cubic equation is guaranteed to have exactly one real root.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
By the Fundamental Theorem of Algebra, a cubic polynomial has exactly three roots (which can be real or complex).
To find the number of real roots, we can analyze the monotonicity of the function using its derivative.

Step 2: Detailed Explanation:
Let us define the function:
\[ f(x) = 4x^3 + 3x^2 + 2x + 1 \]
Since it is a polynomial of odd degree, it must have at least one real root.
Let us calculate the first derivative $f'(x)$:
\[ f'(x) = 12x^2 + 6x + 2 \]
To find if $f'(x)$ can be zero, calculate its discriminant $D$:
\[ D = 6^2 - 4 \cdot (12) \cdot (2) = 36 - 96 = -60 < 0 \]
Because the discriminant of the quadratic derivative is negative and the leading coefficient (12) is positive, $f'(x) > 0$ for all real $x$.
This means that $f(x)$ is strictly increasing on the entire real line.
A strictly monotonic function can cross the $x$-axis at most once.
Therefore, $f(x) = 0$ has exactly 1 real root (and 2 complex conjugate roots).
Let us check if the root is in the interval $[-2, -1]$:
- $f(-1) = 4(-1)^3 + 3(-1)^2 + 2(-1) + 1 = -4 + 3 - 2 + 1 = -2$
- $f(-2) = 4(-8) + 3(4) + 2(-2) + 1 = -32 + 12 - 4 + 1 = -23$
Since the function value does not change sign between $x = -2$ and $x = -1$ (both are negative), the root is not in $[-2, -1]$. (In fact, because $f(-1) = -2$ and $f(0) = 1$, the root lies in the interval $(-1, 0)$).
Thus, the equation has exactly 1 real root.

Step 3: Final Answer
The correct option is (D).