# 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?

##### 1 Answer

Use the fact that the derivative of

#### Explanation:

Here's the idea I was thinking of:

If

So

So

We would like to find this common

If we divide

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

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

Again, we first multiply

#(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

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

This time we need to multiply

#(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

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

This time, to avoid getting fractions when we divide this into

#(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

#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

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.