Let $f(x)$ be a polynomial with integer coefficients. There exist distinct integers $p,$ $q,$ $r,$ $s,$ $t$ such that
\[f(p) = f(q) = f(r) = f(s) = 10\]and $f(t) > 10.$ What is the smallest possible value of $f(t)?$

We are given a polynomial $ f(x) $ with integer coefficients such that:

\[
f(p) = f(q) = f(r) = f(s) = 10
\]
for distinct integers $ p, q, r, s $, and we are also told that $ f(t) > 10 $ for some integer $ t $. The task is to find the smallest possible value of $ f(t) $.

### Step 1: Setting up the polynomial
Since $ f(p) = f(q) = f(r) = f(s) = 10 $, the polynomial $ f(x) - 10 $ has roots at $ p, q, r, s $. Therefore, we can express $ f(x) $ as:
\[
f(x) = a(x - p)(x - q)(x - r)(x - s) + 10
\]
where $ a $ is an integer coefficient because $ f(x) $ has integer coefficients.

### Step 2: Finding the smallest value of $ f(t) $
We need to find the smallest possible value of $ f(t) $ where $ f(t) > 10 $. First, substitute $ t $ into the expression for $ f(x) $:
\[
f(t) = a(t - p)(t - q)(t - r)(t - s) + 10
\]
For $ f(t) > 10 $, we require:
\[
a(t - p)(t - q)(t - r)(t - s) > 0
\]
This implies that the product $ (t - p)(t - q)(t - r)(t - s) $ must have the same sign as $ a $.

### Step 3: Testing the smallest case
Let's assume $ a = 1 $ to minimize the value of $ f(t) $. Then we want:
\[
(t - p)(t - q)(t - r)(t - s) > 0
\]
We now choose specific values for $ p, q, r, s $ to find the smallest $ t $ such that $ f(t) > 10 $.

#### Case: $ p = 1, q = 2, r = 3, s = 4 $
In this case, the polynomial becomes:
\[
f(x) = (x - 1)(x - 2)(x - 3)(x - 4) + 10
\]
We now compute $ f(t) $ for various values of $ t $:

- For $ t = 0 $:
  \[
  f(0) = (0 - 1)(0 - 2)(0 - 3)(0 - 4) + 10 = (-1)(-2)(-3)(-4) + 10 = 24 + 10 = 34
  \]
  So, $ f(0) = 34 $, which satisfies $ f(0) > 10 $.

Thus, the smallest possible value of $ f(t) $ is $ \boxed{34} $.

### Step 4: Verifying if smaller values are possible
We already set $ a = 1 $, which minimizes the polynomial's growth. Changing the values of $ p, q, r, s $ would either increase the value of $ f(t) $ or not result in a smaller $ f(t) > 10 $. Therefore, the smallest possible value of $ f(t) $ is indeed $ \boxed{34} $.

Would you like more details on any part of this solution, or do you have any further questions?

---

Here are 5 related questions you might find interesting:
1. How would the solution change if we allowed non-integer coefficients for $ f(x) $?
2. Can you generalize this approach to find the smallest value of $ f(t) $ for different values of $ a $?
3. What if the condition were $ f(t) \geq 10 $ instead of $ f(t) > 10 $?
4. How does the degree of the polynomial affect the solutions to this type of problem?
5. How would the solution differ if the number of points where \(

Let $f(x)$ be a polynomial with integer coefficients. There exist distinct integers $p,$ $q,$ $r,$ $s,$ $t$ such that
\[f(p) = f(q) = f(r) = f(s) = 10\]and $f(t) > 10.$ What is the smallest possible value of $f(t)?

Explore how to find the smallest value of a polynomial with integer coefficients where f(p) = f(q) = f(r) = f(s) = 10 and f(t) > 10. This problem highlights polynomial root multiplicity, inequalities, and the Fundamental Theorem of Algebra, suitable for high school students in Grades 10-12.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation