Using Factor Theorem (cont.) – selected details for a CSMC question

This posts follows a previous post. For the convenience of readers, the question is repeated:

Suppose that A₁=-6, B₁ = 10 and C₁ = -5. For each positive integer n \geq 2 , let \\ A_n = 2 B_{n-1} – (A_{n-1})^2, \\ B_n = (B_{n-1})^2 – 2 A_{n-1} C_{n-1}, \\ C_n = -(C_{n-1})^2 .

Prove that the polynomial f_{100}(x) = x^3 + A_{100} x^2 + B_{100} x + C_{100} has three distinct positive real roots.Selected Details for the solution of c.

Now we start to work out the problem just-shown.

We find that 1+A_1 + B_1 + C_1 = 0 , then f_1 (x) = x^3 + A_1 x^2 + B_1 x + C_1 = (x-1) (x^2 + (A_1+1) x - C_1) ; (using polynomial division \frac{x^3 + A_1 x^2 + B_1 x + C_1}{x -1} to find the quadratic cofactor). By bringing A₁=-6, C₁= -5 into the factorized form of f_1(x) :

f_1 (x) = (x-1) (x^2 -5 x +5).

Obviously, x^2 – 5x +5 = 0 has two distinct roots [as the discriminant is (-5)² – 4(5) = 5 >0; and note for this quadratic, none of the roots is 1]. Thus f_1(x) has three distinct roots.

In general, if 1 + A_n + B_n + C_n = 0 (this is indeed the case), then

f_n (x) = x^3 + A_n x^2 + B_n x + C_n \\ = (x-1) (x^2 + (A_n+1) x – C_n).

To show that 1 + A_n + B_n + C_n = 0 , we proceed as follows: 1 + A_n + B_n + C_n \\ = 1 + (2B_{n-1} - A^2_{n-1}) + (B^2_{n-1} - 2 A_{n-1} C_{n-1}) + (- C^2_{n-1}) \\ = (1+B_{n-1})^2 - (C_{n-1} + A_{n-1})^2 = 0 (the last equality holds if assuming 1 + A_{n-1} + B_{n-1} + C_{n-1} =0 ). Since 1 + A_1 + B_1 + C_1 =0 , from math induction, we obtain for any n, 1 + A_n + B_n + C_n = 0 .

It is clear now that f_n(x) always has 1 as one of its roots. Let us denote D_n as the discriminant of the quadratic (x^2 + (A_n+1) x – C_n) . Then we should have (if f_n(x) has three distinct roots)

• (claim 1) D_n >0 ;
• (claim 2) At x = 1, x^2 + (A_n+1) x - C_n = 2 + A_n - C_n \neq 0 .

Yet the two claims are not confirmed up to here. So more effort is needed! The following table (that lists feature related to function fn(x)) is made:

By inspection, we find an interesting pattern that A_n <0, B_n >0, C_n <0 , or equivalently -A_n >0, B_n >0, -C_n >0 . From this we can deduce claim 2: 2 + A_n - C_n >2 >0 . We observe another interesting pattern: B_n > - C_n - 2 C_{n-1} >0 which implies B_n + C_n > -2 C_{n-1}>0, \\ -1 - A_n > -2 C_{n-1} >0. Note that (– C_n )= (C_n-1)², and squaring on both sides of the previous line, to obtain: D_n = (1+A_n)^2 -4 (-C_n) >0 which is exactly Claim 1. (There is one last thing: it remains to be shown formally, the two patterns mentioned above .. ..)