Polynomial's Common Multiple

Suppose $g(x) = f(x) - 1$ . Then $g(x)$ is also a cubic monic polynomial.

Then $g(x) = p(x)(x-1) = q(x)(x-2) = r(x)(x-3)$ .

Hence, $(x-1)$ , $(x-2)$ , & $(x-3)$ are all factors of $g(x)$ , and since $g(x)$ is a cubic monic polynomial, the leading coefficient must be one.

Thus, $g(x) = (x-1)(x-2)(x-3)$ . Then $f(x) = (x-1)(x-2)(x-3) + 1$ .

As a result, $f(4) = (4-1)(4-2)(4-3) + 1 = \boxed{7}$ .

From the equations, we can easily see that f(1)=1, f(2)=1, and f(3)=1. By Method of Differences, because this is a monic cubic polynomial, the third common difference is 3!x1=6. Again, by Method of Differences, f(4)=7.

This may seem a long answer but while working on this process it doesn't took much time.

We know that: $p(x)(x-1) + 1 = q(x)(x-2) + 1$

or, $p(x)(x-1) = q(x)(x-2)$

Now since $p(x)$ and $q(x)$ are two different quadratic polynomials both must have different factors.

So $p(x)$ must have $(x-2)$ as one of the factor since in the equality above there is $(x-2)$ in right term so there must be $(x-2)$ in left term. (when written in the factored form)

Using similar intuition from this equation $p(x)(x-1) + 1 = r(x)(x-3) + 1$ we can get $(x-3)$ also a factor of $p(x)$ .

Then,

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

Now substituting the above value in $f(x)$

$f(x) = (x-2)(x-3)(x-1) + 1$

or $f(4) = (4-1)(4-2)(4-3)+1 = \boxed{7}$