For the following two (02) items: Let f(x) = ax^2 + bx + c be a...

Given $f(1) = f(4) = 2$, the quadratic polynomial can be written as $f(x) = a(x-1)(x-4) + 2$.

Since $2$ is a root of $f(x) = 0$, we have $f(2) = 0$.

Substituting $x = 2$ in the expression for $f(x)$:

$a(2-1)(2-4) + 2 = 0$

$a(1)(-2) + 2 = 0$

$-2a + 2 = 0 \Rightarrow a = 1$

Thus, the polynomial is $f(x) = 1(x-1)(x-4) + 2 = x^2 - 5x + 6$.

To find the roots of $f(x) = 0$, we solve $x^2 - 5x + 6 = 0$.

$(x-2)(x-3) = 0$

The roots are $x = 2$ and $x = 3$.

The other root is $3$.

Answer: $3$