Question

The cubic equation \[ 2x^3-3x^2+6x+2=0 \]

• A. has \(3\) distinct real roots
• B. has only one real root in the interval \((-1,0)\)
• C. has two distinct real roots
• D. has only one real root in the interval \((0,1)\)

Hint:

If a polynomial satisfies \[ f'(x)>0 \] for all real \(x\), then the function is strictly increasing and can cross the \(x\)-axis at most once.

Answer 1

The Correct Option is B

Concept: To determine the nature and location of roots of a polynomial equation, we commonly use:

• Intermediate Value Theorem

• Sign analysis

• Derivative test for monotonicity
A cubic polynomial can have either:

• one real root and two complex roots, or

• three real roots.
The derivative helps determine whether the function is always increasing or decreasing.

Step 1: Define the polynomial function. Let \[ f(x)=2x^3-3x^2+6x+2 \] We first check the sign of the function at convenient points.

Step 2: Check the interval \((-1,0)\). Evaluate at \(x=-1\): \[ f(-1)=2(-1)^3-3(-1)^2+6(-1)+2 \] \[ =-2-3-6+2 \] \[ =-9 \] Thus, \[ f(-1)<0 \] Now evaluate at \(x=0\): \[ f(0)=2 \] Thus, \[ f(0)>0 \] Since the function changes sign between \(-1\) and \(0\), by the Intermediate Value Theorem there exists at least one real root in \((-1,0)\).

Step 3: Check whether more than one real root exists. Differentiate: \[ f'(x)=6x^2-6x+6 \] Factor: \[ f'(x)=6(x^2-x+1) \] Now examine: \[ x^2-x+1 \] Its discriminant is: \[ (-1)^2-4(1)(1)=1-4=-3<0 \] Therefore, \[ x^2-x+1>0 \] for all real \(x\). Hence, \[ f'(x)>0 \] for every real number. This means the function is strictly increasing everywhere. Therefore, the equation can have only one real root. Since we already found one real root in \((-1,0)\), that root is unique.

Step 4: Conclude the correct option. Hence, the equation has: \[ \boxed{\text{only one real root in }(-1,0)} \]