How do I solve the equation $x^5 - 5x^4 - 5x^3 +25x^2 +4x -20 = 0$ given that its roots are of the form $+a, -a, +b, -b, c$? I understand that it is an easy problem, but I am not able to solve it at all! Any clue on how to approach this problem will be amazing! Thank you. (PS: I am just beginning to learn math).
 A: Let $P(x)=x^5 - 5x^4 - 5x^3 +25x^2 +4x -20$. 
Note a equation that would have $-a,a,-b,b,c$ as it's roots would be of the form $(x-a)(x+a)(x-b)(x+b)(x-c)=(x^2-a^2)(x^2-b^2)(x-c)$. 
So we get
$$P(x)=x^5 - 5x^4 - 5x^3 +25x^2 +4x -20 = (x^2-a^2)(x^2-b^2)(x-c)=(x^4-(a^2+b^2)x^2+a^2b^2)(x-c)=x^5-cx^4+\dots$$
This tells us that $c=5$.
By polynomial division, if we divide $(x^5 - 5x^4 - 5x^3 +25x^2 +4x -20)$ by $(x-5)$, note that we get  $x^4-5x^2+4$. 
However, note $$x^4-5x^2+4=(x^2)^2-5x^2+4=(x^2-1)(x^2-4)=(x-1)(x+1)(x-2)(x+2)$$
A: Hint: use general theory of equations ie sum of root is $a-a+b-b+c=-(-5)=5$ thus $c=5$ then product of roots. Then summation of two roots grouped together... You have $c$ you have $4$ next equations to generate from there you can get $a,-a,b,-b$ or substitute $x$ as the value of roots. Add the equations of $a,-a$ you will get a quadratic after plugging $a^2=u$ which can also be done for $b$ as signs of odd powers will just cancel out after adding and you already know $c$ that's it.
A: if you know that $a, -a, b, -b$ are roots then $(x^2 - a^2)(x^2-b^2)$ are factors.
however this is first thing that I see.
$$(x^5 - 5x^4) - 5(x^3-5x^2) + 4(x - 5) = 0$$
