So I was doing some pre-calculus excercises and ran into the following: Factor out completely $$p^5 - 5p^3 + 8p^2 - 40 $$
Trying some values I found -2 to be a root and took it out with Ruffini, leaving me with $$(p+2)(p^4 - 2p^3 - p^2 + 10p - 20) $$
At this point and after trying many other values I decided to look at the graph of the second one and noticed there were no nice roots. I looked at the answer at the back of the book and it was $$(p+2)(p^2 - 5)(p^2 - 2p + 4) $$
Since none of thoose have rational roots, I asked a teacher how I was supposed to solve it, and with a confused look on his face he said it must have been by trying square roots as solutions, since he didn't know any other way of doing it.
I then asked another teacher and he told me that for any fourth degree polynomial with main coefficient of 1 you can always factor it out into 2 second degree polynomials of standard form $(p^2 + ap + b)(p^2 + cp + d) $and that by applying the distributive law I could get a 4x4 system of equations, he then solved it in about 5 seconds and smirked.
When I got home I tried to solve it on my own, since he only wrote the answers (which did math with the book's) and wasn't able to do it. It is the following:
\begin{cases} a+c=-2 \\ b+d+ac=-1 \\ ad+bc=10 \\ bd=-20 \end{cases}
I showed it to a 3rd teacher and he told me he had never seen that way of solving it, and after a few minutes of trying he also failed to solve it. Ps: I tried only with substitution method. Thanks in advance.
Let $f(p) := p^4 - 2p^3 - p^2 + 10p - 20$.
Assume $f$ has symmetric roots, i.e if $f(a) = 0$ then $f(-a) = 0$. It is easy to see that it can't have two symmetric roots. So if $f$ has one symmetric root then
$$f(p) = (p - a)(p+a)(p^2 + bp + c) = p^4 +bp^3 + (c - a^2)p^2 - a^2bp - a^2c$$
Equating the coefficients,
$$\begin{cases}b = -2\\c - a^2 =-1\\-a^2b = 10 \\ -a^2c = -20\end{cases}$$.
Putting the value of $b$ from first equation into third equation gives $a^2 = 5$, then from fourth equation we get $c = 4$.
Hence, $$f(p) = \big(p -\sqrt{5}\big)\big(p + \sqrt{5}\big)\big(p^2 -2p +4 \big)$$.