Let f (x) be a quadratic polynomial such that f(-2) + f(3) = 0. If one of the roots of f (x) = 0 is -1, then the sum of the roots of f (x) = 0 is equal to:

Key Ideas to Remember

1. Standard shape of a quadratic:
   Any quadratic polynomial can be written as $f(x)=ax^2+bx+c$ or, if you know its roots $\alpha$ and $\beta$, it can be factored as $f(x)=a\,(x-\alpha)(x-\beta)$ Here, $a\neq0$ is just a common scale‐factor.

2. Relation between roots and coefficients:
   For $f(x)=ax^2+bx+c$ with roots $\alpha,\beta$: $\alpha+\beta=-\frac{b}{a}\;\;\;\text{and}\;\;\;\alpha\beta=\frac{c}{a}$

3. Using a given root:
   If one root is $-1$, then $x+1$ is a factor. So we can rewrite the quadratic as $f(x)=a\,(x+1)(x-\beta)$ where $\beta$ is the unknown second root.

4. Extra condition $f(-2)+f(3)=0$ gives another equation:
   Plug $x=-2$ and $x=3$ into the factored form, add, and set the sum to zero. Because $a\neq0$, the bracketed expression itself must be zero, letting us solve for $\beta$.

Once $\beta$ is found, use $\alpha+\beta$ to get the desired answer.

What’s happening here?

Imagine a magic box (the quadratic polynomial) that eats a number $x$, does some secret squaring-and-adding, and spits out a new number $f(x)$.
We are told three simple facts:

1. If we feed in $-2$ and also $3$, and then add their outputs, we get zero.
2. One special input, $-1$, makes the box spit out exactly zero (so $-1$ is a root).
3. The question asks: If one root is 5 , what is the sum of both roots?

Think of the roots as two hidden keys. We already know one key ($-1$). We use the given clue (the sum $f(-2)+f(3)=0$) to crack the other key. Once both keys are known, we just add them to give the final answer.

Step-by-Step Solution

1. Write the quadratic with a known root
   Since one root is $-1$, factor: $f(x) = a\,(x+1)(x-\beta)$ where $a\neq0$ and $\beta$ is the second root.

2. Use the condition $f(-2)+f(3)=0$
   • Evaluate at $x=-2$: $f(-2)=a\,(-2+1)(-2-\beta)=a(-1)(-2-\beta)=a(2+\beta)$ • Evaluate at $x=3$: $f(3)=a\,(3+1)(3-\beta)=a(4)(3-\beta)=4a(3-\beta)$ • Add and set to zero: $f(-2)+f(3)=a(2+\beta)+4a(3-\beta)=a[(2+\beta)+12-4\beta]$ $\quad\;=a\,(14-3\beta)=0$ Because $a\neq0$, we must have $14-3\beta=0 \;\Longrightarrow\; \beta=\frac{14}{3}$

3. Find the sum of the roots
   Roots are $-1$ and $\frac{14}{3}$. Therefore $\alpha+\beta = -1 + \frac{14}{3}=\frac{-3+14}{3}=\frac{11}{3}$

4. Final Answer
   $\boxed{\dfrac{11}{3}}$