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

Step 1: Name the function.
Let $f(x)=2x^3-3x^2+6x+2$. We want to know how many real roots it has and where.
Step 2: Check the slope of $f$.
$f'(x)=6x^2-6x+6=6(x^2-x+1)$.
Step 3: See that the slope is always positive.
The discriminant of $x^2-x+1$ is $1-4=-3<0$, so $x^2-x+1>0$ for all $x$. Thus $f'(x)>0$ everywhere, meaning $f$ is always increasing. An always increasing cubic crosses zero exactly once, so there is only one real root.
Step 4: Hunt for the root by sign change.
$f(-1)=-2-3-6+2=-9<0$.
Step 5: Check the other end of the interval.
$f(0)=2>0$. The function moves from negative to positive between $-1$ and $0$.
Step 6: Conclude.
By the sign change, the single real root lies in $(-1,0)$. \[ \boxed{\text{only one real root in }(-1,0)} \]