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Consider an unknown polynomial which when divided by $(x- 3)$ and $(x - 4)$ leaves remainders $2$ and $1$, respectively. Let $R(x)$ be the remainder when this polynomial is divided by $(x - 3) (x- 4)$.

If equation $R(x) = x^2 + mx + 1$ has two distinct real roots, then exhaustive values of $m$ are :

By the polynomial remainder theorem, R(3) equals 2 and R(4) equals 1. Assuming a linear remainder R(x) = ax + b, we solve the system 3a + b = 2 and 4a + b = 1 to get a = -1 and b = 5, yielding R(x) = -x + 5. The question states R(x) = x^2 + mx + 1, so substituting our linear expression gives the quadratic equation x^2 + (m+1)x - 4 = 0. For this equation to have two distinct real roots, the discriminant must be strictly positive, meaning (m+1)^2 - 4(1)(-4) > 0. This simplifies to (m+1)^2 + 16 > 0, which is an inequality that holds true for all real numbers m.

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