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 $$f(1)=(1+1)(1-2)(1+3)(1-4)-1^2=\boxed{23}$$