Let the quadratic polynomial be $f(x)=a(x-3)(x-r), \quad a\neq 0$ where $3$ is one root and $r$ is the other root.

We are given: $f(-1)+f(2)=0$

1. Compute $f(-1)$ and $f(2)$

Using $f(x)=a(x-3)(x-r)$,

$f(-1)=a(-1-3)(-1-r)=a(-4)(-1-r)=4a(1+r)$

and

$f(2)=a(2-3)(2-r)=a(-1)(2-r)=a(r-2)$

So, $f(-1)+f(2)=4a(1+r)+a(r-2)=0$

Since $a\neq 0$, divide by $a$: $4(1+r)+(r-2)=0$

2. Solve for $r$

$4+4r+r-2=0$ $2+5r=0$ $5r=-2$ $r=-\frac{2}{5}$

3. Identify the interval

Now, $-\frac{2}{5}=-0.4$ which lies in the interval $(-1,0)$

So the other root lies in Option C.

4. Check options systematically

• A: $(-3,-1)$ → $-\frac25 \notin (-3,-1)$
• B: $(1,3)$ → $-\frac25 \notin (1,3)$
• C: $(-1,0)$ → $-\frac25 \in (-1,0)$
• D: $(0,1)$ → $-\frac25 \notin (0,1)$

Therefore, the correct answer is C.