# Factorising 4th degree polynomial

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:

$$a+c=-2 \ b+d+ac=-1 \ ad+bc=10 \ bd=-20$$

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.

• Well, I think you should never ask the 3rd teacher a math related question again. :P – Cornman Oct 11 '17 at 15:08
• The first and third ones are not professors, just engineering students that teach math at highschool. They've been teaching their programs exceptionally well, so I don't think a question that they don't expect should condemn them. – Otomeram Oct 11 '17 at 15:12
• If they teach math, they should be able to factor basic polynomials... – Sir_Math_Cat Oct 11 '17 at 15:13
• I would call basic one with evident roots, and they obviously can do that. This is not as easy. – Otomeram Oct 11 '17 at 15:15
• The method of coefficient comparision should be known by every math teacher. But besides that my first comment was more of a "joke". I do not want to condemn your teachers. – Cornman Oct 11 '17 at 15:16

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, $$f(p) = \big(p -\sqrt{5}\big)\big(p + \sqrt{5}\big)\big(p^2 -2p +4 \big)$$.

• So you mean I had to make that assumption? – Otomeram Oct 12 '17 at 13:14
• Also, is there not a way to solve the system without assuming a=0? – Otomeram Oct 12 '17 at 13:14
• @Otomeram As far as I know, it is a standard procedure to check for symmetric roots after checking for integer roots. We check for symmetric roots because coefficient equations obtained are easy to solve for symmetric roots. You can use this assumption for solving other polynomials also. – user8277998 Oct 13 '17 at 8:58