#### Tackling Polynomial Problems

Two of the recent daily problems are related to polynomial:

 (2853) Let $P(x) = x^n + a_{n-1}x^{n-1} + \cdots + a_1x + a_0$ be a polynomial with integral coefficients. Suppose that there exist four distinct integers $a, b, c, d$ with $P(a) = P(b) = P(c) = P(d) = 5$. Prove that there is no integer $k$ satisfying $P(k) = 8$.

 (120) Let $f(x) = x^4 + ax^3 + bx^2 + cx + d$. If $f(-1) = -1$, $f(2)=-4$, $f(-3) = -9$, and $f(4) = -16$. Find $f(1)$.

The given polynomial $P(x)$ in the first problem is completely abstract. For the second problem, it is possible to determine the exact form of $f(x)$ by solving $a$, $b$, $c$, and $d$. But that is not the best solution. So, a natural question to ask is how to solve such problems?

When a polynomial appears in a competition problem, two of the most frequently tested topics are roots and Vieta's formula (which is also about roots). Let's first review a well-known conclusion:

If distinct numbers $x_1$, $x_2$, $\cdots$, $x_k$ satisfy $f(x_i)=0$ where $i=1, 2,\cdots, k$, then $f(x)$ must be in the form of $$f(x)=g(x)(x-x_1)(x-x_2)\cdots(x-x_k)$$

where $g(x)$ is a polynomial or a constant depending on the degree of $f(x)$ and $k$.

In order to apply this, one critical step is to construct a polynomial such that the given numbers are its roots. For example, in problem #2853, we need to find a polynomial whose roots are $a$, $b$, $c$, and $d$. With this in mind, it is easy to see that the following polynomial satisfy the desired feature $$P(x) - 5$$

Therefore, we can write the following relation $$P(x) - 5 = g(x)(x-a)(x-b)(x-c)(x-d)$$

$$\implies P(x) = g(x)(x-a)(x-b)(x-c)(x-d)+5$$

If there exists an integer $k$ such that $P(k)=8$, then we have $$P(k)=g(k)(k-a)(k-b)(k-c)(k-d) + 5 = 8$$

This means that $$g(k)(k-a)(k-b)(k-c)(k-d) = 3$$

Note that all the five terms on the left are integers, and $(k-a)$, $(k-b)$, $(k-c)$, $(k-d)$ must be distinct. However it is not possible to write $3$ as a product of five integers among which four are distinct. Therefore, we conclude no such $k$ exists.

For problem #120, the given conditions can be rewritten as $f(-1)+1=0$, $f(2)+4=0$, $f(-3)+9=0$, and $f(4)+16=0$. Therefore, we find $-1$, $2$, $-3$, and $4$ are the roots of $f(x)+x^2$. It follows that $$f(x)+x^2 = k(x+1)(x-2)(x+3)(x-4)$$

where $k$ is a constant because we know that the degree of $f(x)$ is $4$. Comparing the coefficients of $x^4$ results in $k=1$. Hence, $$f(x)=(x+1)(x-2)(x+3)(x-4)-x^2$$

Then, setting $x=1$ immediately leads to $$f(1)=(1+1)(1-2)(1+3)(1-4)-1^2=\boxed{23}$$