Let be the unique polynomial of minimal degree with the following properties:
- has a leading coefficient ,
- is a root of ,
- is a root of ,
- is a root of , and
- is a root of .
The roots of are integers, with one exception. The root that is not an integer can be written as , where and are relatively prime integers. What is ?
- A)
- B)
- C)
- D)
- E)
Answer
D
Key insight
Each clue is a value of P: P(1) is 1, and P is 0 at 0, 4 and 9. Those roots alone make P(1) equal 24, so add another root.
Solution
Each clue looks strange, but each one just says what equals at some number. To decode a clue, plug the given number in for .
1. " is a root of ." Plug in : , so .
2. " is a root of ." Plug in : , so .
3. " is a root of ." Plug in : , so .
4. " is a root of ." Plug in : , so .
5. Build . So has roots , and , and its leading coefficient is . The simplest try is
But then , not . So three factors are not enough.
6. Add one more root. Try degree with an unknown fourth root :
Now , and this must equal :
7. The roots are and . The one that is not a whole number is , so .
The answer is .
Why this works
" is a root of " means: put into the something, and of the result is . Once every clue is read this way, you have three roots and one extra value. A polynomial with leading coefficient is a product of factors, and the one extra value fixes the one unknown root. The only danger is misreading a clue, so decode them slowly, one at a time.
The trap
Reading '2 is a root of P(x - 2)' as P(2) = 0 instead of P(0) = 0, or '3 is a root of P(3x)' as P(3) = 0 instead of P(9) = 0.
Concise solution
Unpack each condition as a statement about values of :
- is a root of means , so .
- is a root of means .
- is a root of means .
- is a root of means , so .
Thus is monic with roots . The smallest candidate is the cubic , but it gives . So the degree must be at least , and with one more root ,
Hence and . This degree- polynomial is unique, and its only non-integer root is , so .
The answer is .
Review in the AoPS books
- Intermediate Algebra, Chapter 6: Polynomial Division (§6.2, §6.4)
- Introduction to Algebra, Chapter 16: Functions
- Intermediate Algebra, Chapter 20: Some Advanced Strategies (§20.2)
- Intermediate Algebra, Chapter 7: Polynomial Roots Part I (§7.1, §7.2, §7.3)
Techniques
Set up the equation/formula and compute; no special trick needed · Substitute to simplify (u = x+1/x, shifting, scaling)