# Polynomial Help

1. Oct 2, 2005

### an_mui

If m, n, and 1 are non-zero roots of the equation $$x^3 - mx^2 + nx - 1 = 0$$, then find the sum of the roots

This is what I did..

m, n, 1 are the roots. m and n not equal to 0

$$x^3 - mx^2 + nx - 1 = 0$$
f(m) = 0 --> m^3 - m^3 + mn - 1
1 = mn (1)

f(1) = 1 - m + n - 1 = 0
... m = n (2)

Sub (2) --> (1)

$$m^2 = 1$$
[tex]m = +/- 1

since m and n are equal... the roots must be either 1, 1, 1 or -1, -1, 1. The answer on the sheet says the answer is -1. So my question is, how do we determine which roots are the answers. Thanks

2. Oct 2, 2005

### hotvette

This might work. If m, n, and 1 are the roots, that means (x-m)*(x-n)*(x-1) = 0. If you multiply this out and substitute the 2 different sets of values for m & n, I'm guessing that only 1 set is valid.

Last edited: Oct 2, 2005
3. Oct 2, 2005

### an_mui

oops nvm it didn't work

Last edited: Oct 2, 2005
4. Oct 8, 2005

### hotvette

looks like there are 2 possible solutions.

5. Oct 8, 2005

### AKG

There is only one solution. I used long division. If you divide your polynomial by (x-1), you know you should get a quadratic polynomial, but you get:

x² + (1-m)x + (n+1-m) + (n-m)/(x-1)

So you know n - m = 0. This is one useful fact. Now you're left with:

x² + (1-m)x + (n+1-m)

You know that m is a root of this polynomial, so dividing by (x-m) should give a linear polynomial. When you do the long division, you get:

x + 1 + (n+1)/(x-m)

So you know n + 1 = 0. This gives n = -1. Having already got that n - m = 0, you now get m = -1. And you have the other root being 1.

(-1) + (-1) + 1 = -1.

6. Oct 8, 2005

### hotvette

Good one AKG!

7. Oct 9, 2005

### HallsofIvy

Staff Emeritus
In your very first post, you note that the solutions must be either {1, 1, 1} or {1, -1, -1} and ask how can you tell that only {1, -1, -1} is correct.

If {1, 1, 1} were the roots, then the equation could be written
(x-1)(x-1)(x-1)= 0 but (x-1)(x-1)(x-1)= x3- 3x2+ 3x- 1, not x3- x2+ x- 1 as it should be with m=n= 1.

If {1, -1, -1}, on the other hand, were the roots then the equation could be written (x-1)(x+1)(x+1)= (x2- 1)(x+1)= x3+x2- x- 1 as required.