Let $$f(x)$$ be a quadratic polynomial such that $$f(-1) + f(2) = 0$$. If one of the roots of $$f(x) = 0$$ is 3, then its other root lies in:

We are told that $$f(x)$$ is a quadratic polynomial, that one of its roots is $$3$$, and that the values of the polynomial at $$x=-1$$ and $$x=2$$ satisfy $$f(-1)+f(2)=0$$. Our aim is to find the interval in which the other root lies.

Because $$f(x)$$ is quadratic and one root is already known, we can write it in the standard “root-factor” form. A quadratic having real roots $$\alpha$$ and $$\beta$$ can always be expressed as

$$f(x)=k\,(x-\alpha)(x-\beta),$$

where $$k\neq 0$$ is a non-zero constant. Here, one root is given as $$3$$, so we set $$\alpha = 3$$ and denote the unknown second root by $$\beta$$. Thus,

$$f(x)=k\,(x-3)(x-\beta).$$

Next, we evaluate $$f(x)$$ at the two specified points.

First at $$x=-1$$:

$$\begin{aligned} f(-1) &= k\,((-1)-3)\bigl((-1)-\beta\bigr) \\ &= k\,(-4)\,(-1-\beta) \\ &= k\,\bigl(4(1+\beta)\bigr) \\ &= 4k(1+\beta). \end{aligned}$$

Now at $$x=2$$:

$$\begin{aligned} f(2) &= k\,(2-3)(2-\beta) \\ &= k\,(-1)\,(2-\beta) \\ &= -k(2-\beta). \end{aligned}$$

According to the condition in the problem, the sum of these two values is zero:

$$f(-1)+f(2)=0.$$ Substituting the expressions we just computed, we obtain

$$4k(1+\beta)\;+\;\bigl[-k(2-\beta)\bigr]=0.$$

We can factor out the common non-zero constant $$k$$ (since $$k\neq 0$$ for a non-trivial quadratic):

$$k\Bigl[\,4(1+\beta)-(2-\beta)\Bigr]=0.$$

Because $$k\neq 0,$$ the bracketed expression must itself be zero:

$$4(1+\beta)-(2-\beta)=0.$$

Expanding and combining like terms, we get

$$\begin{aligned} 4+4\beta-2+\beta &= 0 \\ (4-2)+ (4\beta+\beta) &= 0 \\ 2 + 5\beta &= 0. \end{aligned}$$

Solving for $$\beta,$$

$$5\beta = -2 \quad\Longrightarrow\quad \beta = -\dfrac{2}{5} = -0.4.$$

Thus the second root is $$\beta=-0.4$$. We now compare this value with the intervals given in the options:

• Option A: $$(-1,0)$$ — contains $$-0.4$$ ✅
• Option B: $$(1,3)$$ — does not contain $$-0.4$$ ❌
• Option C: $$(-3,-1)$$ — does not contain $$-0.4$$ ❌
• Option D: $$(0,1)$$ — does not contain $$-0.4$$ ❌

Only Option A encloses the value $$-0.4$$.

Hence, the correct answer is Option A.