# 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?

Aug 14, 2015

Use the fact that the derivative of ${x}^{7} - 3 {x}^{5} + {x}^{4} - 4 {x}^{2} + 4 x + 4$ 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 $f \left(x\right) = {x}^{7} - 3 {x}^{5} + {x}^{4} - 4 {x}^{2} + 4 x + 4 = 0$ has at least one repeated root, then its derivative will also have that root.

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

So $f \left(x\right)$ and $f ' \left(x\right)$ will have a common factor - call it $p \left(x\right)$.

We would like to find this common $p \left(x\right)$ without factoring $f \left(x\right)$ or $f ' \left(x\right)$ first.

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

Let

${f}_{7} \left(x\right) = {x}^{7} - 3 {x}^{5} + {x}^{4} - 4 {x}^{2} + 4 x + 4$

${f}_{6} \left(x\right) = 7 {x}^{6} - 15 {x}^{4} + 4 {x}^{3} - 8 x + 4$

If we just divide ${f}_{7} \left(x\right)$ by ${f}_{6} \left(x\right)$ then we will get fractions. To avoid that, first multiply ${f}_{7} \left(x\right)$ by $7$.

Then:

$\frac{7 {f}_{7} \left(x\right)}{f} _ 6 \left(x\right) = x$ with remainder $- 6 {x}^{5} + 3 {x}^{4} - 20 {x}^{2} + 24 x + 28$.

Reverse the signs of this remainder to give us:

${f}_{5} \left(x\right) = 6 {x}^{5} - 3 {x}^{4} + 20 {x}^{2} - 24 x - 28$.

If we divide ${f}_{7} \left(x\right)$ by ${f}_{5} \left(x\right)$ then the remainder will also be divisible by $p \left(x\right)$ and it will have degree $4$ or lower.

Again, we first multiply ${f}_{7} \left(x\right)$ by a scalar factor to avoid fractions and find:

$\frac{72 {f}_{7} \left(x\right)}{f} _ 5 \left(x\right) = 12 {x}^{2} + 6 x - 33$ with remainder

$- 267 {x}^{4} + 168 {x}^{3} + 852 {x}^{2} - 336 x - 636$

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

${f}_{4} \left(x\right) = 89 {x}^{4} - 56 {x}^{3} - 284 {x}^{2} + 112 x + 212$

This time we need to multiply ${f}_{7} \left(x\right)$ by ${89}^{4} = 62742241$ to avoid fractions when dividing by ${f}_{4} \left(x\right)$, but we get:

$\frac{62742241 {f}_{7} \left(x\right)}{f} _ 4 \left(x\right) = 704969 {x}^{3} + 443576 {x}^{2} + 413761 x + 1493617$

with remainder

$2016736 {x}^{3} + 32838920 {x}^{2} - 4033472 x - 65677840$.

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

${f}_{3} \left(x\right) = 8132 {x}^{3} + 132415 {x}^{2} - 16264 x - 264830$

This time, to avoid getting fractions when we divide this into ${f}_{7} \left(x\right)$ we need to multiply ${f}_{7} \left(x\right)$ by ${8132}^{5} = 35562055043425682432$ and we get:

$\frac{35562055043425682432 {f}_{7} \left(x\right)}{f} _ 3 \left(x\right) =$

$4373100718571776 {x}^{4} - 71208083085302720 {x}^{3} + 1155122511890916624 {x}^{2} - 18804720284922811004 x + 306192315840871515953$

with remainder

$- 40544526626179088636281359 {x}^{2} + 81089053252358177272562718$.

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

${f}_{2} \left(x\right) = {x}^{2} - 2$

Then we find:

${f}_{7} \frac{x}{f} _ 2 \left(x\right) = {x}^{5} - {x}^{3} + {x}^{2} - 2 x - 2$

with no remainder.

So $p \left(x\right) = {f}_{2} \left(x\right) = {x}^{2} - 2$.

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

$f \left(x\right) = \left({x}^{2} - 2\right) \left({x}^{2} - 2\right) \left({x}^{3} + x + 1\right)$

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