An algebra problem by Paul Ryan Longhas

$(\color{#D61F06}{x^2+1})(\color{#3D99F6}{x^2+2})(\color{#3D99F6}{x^2+3})(\color{#D61F06}{x^2+4})-120=0$

Multiplying the similar color brackets and substituting $x^4+5x^2=\color{#EC7300}{t}$ to get:

$(\color{#EC7300}{t}+4)(\color{#EC7300}{t}+6)-120=0$

$\implies t^2+10t-96=0$

$\implies (\color{#EC7300}{t}+16)(\color{#EC7300}{t}-6)=0$

$\implies\color{#EC7300}{t}=x^4+5x^2=-16,6$

$\underbrace{x^4+5x^2-6=0}_{x^2=-6,1}\text{ and } \underbrace{x^4+5x^2+16=0}_{D<0~\therefore~ \text{no real roots}}$

$\implies x=\pm 1$

Sum of real roots = $\boxed 0$ .

We don't even have to calculate the roots because if +a is a root of the equation then -a will also be a root. So sum of roots will always be 0.

(5)(4)(3)(2) then equate lol

$\text{This is an monic polynomial of degree 8 \& it's sum of roots will be given by the co-efficient of }x^7$

$\text{The equation can be written as :}$

$x^8+10x^6+35x^4+50x^2+24-120=0$

$\text{By Vieta's we have sum of roots = coefficient of }x^7=0$