Question

If you are told that `x^7-3x^5+x^4-4x^2+4x+4 = 0` has at least one repeated root, how might you solve it algebraically?

Answer 1

Use the fact that the derivative of `x^7-3x^5+x^4-4x^2+4x+4` must be `0` at repeated roots and try to find a common factor by division of polynomials.

Explanation:

Here's the idea I was thinking of:

If `f(x)=x^7-3x^5+x^4-4x^2+4x+4=0` has at least one repeated root, then its derivative will also have that root.

So `f'(x)=7x^6-15x^4+4x^3-8x+4=0` has a root for each of the repeated roots in `f(x)=0`.

So `f(x)` and `f'(x)` will have a common factor - call it `p(x)`.

We would like to find this common `p(x)` without factoring `f(x)` or `f'(x)` first.

If we divide `f(x)` by `f'(x)` then the remainder will also be divisible by `p(x)` and it will have a lower degree than `f'(x)`.

Let

`f_7(x)=x^7-3x^5+x^4-4x^2+4x+4`

`f_6(x)=7x^6-15x^4+4x^3-8x+4`

If we just divide `f_7(x)` by `f_6(x)` then we will get fractions. To avoid that, first multiply `f_7(x)` by `7`.

Then:

`(7f_7(x)) / f_6(x) = x` with remainder `-6x^5+3x^4-20x^2+24x+28`.

Reverse the signs of this remainder to give us:

`f_5(x)=6x^5-3x^4+20x^2-24x-28`.

If we divide `f_7(x)` by `f_5(x)` then the remainder will also be divisible by `p(x)` and it will have degree `4` or lower.

Again, we first multiply `f_7(x)` by a scalar factor to avoid fractions and find:

`(72f_7(x))/f_5(x) = 12x^2+6x-33` with remainder

`-267x^4+168x^3+852x^2-336x-636`

All of these coefficients are divisible by `3`, so divide by `-3` to reverse the sign to get:

`f_4(x)=89x^4-56x^3-284x^2+112x+212`

This time we need to multiply `f_7(x)` by `89^4 = 62742241` to avoid fractions when dividing by `f_4(x)`, but we get:

`(62742241 f_7(x))/f_4(x)=704969x^3+443576x^2+413761x+1493617`

with remainder

`2016736x^3+32838920x^2-4033472x-65677840`.

All of these coefficients are divisible by `248` so divide by that to get:

`f_3(x)=8132x^3+132415x^2-16264x-264830`

This time, to avoid getting fractions when we divide this into `f_7(x)` we need to multiply `f_7(x)` by `8132^5 = 35562055043425682432` and we get:

`(35562055043425682432 f_7(x))/f_3(x) =`

`4373100718571776x^4-71208083085302720x^3+1155122511890916624x^2-18804720284922811004x+306192315840871515953`

with remainder

`-40544526626179088636281359x^2+81089053252358177272562718`.

Both of these coefficients are divisible by `40544526626179088636281359`, so to get a positive leading coefficient, divide through by `-40544526626179088636281359` to get:

`f_2(x) = x^2-2`

Then we find:

`f_7(x) / f_2(x) = x^5-x^3+x^2-2x-2`

with no remainder.

So `p(x) = f_2(x) = x^2-2`.

I leave it as an exercise for the reader to find the factorisation

`f(x) = (x^2-2)(x^2-2)(x^3+x+1)`

So basically the idea works, but the numbers get horrendous.