Notice you can factorize
[tex]x^5+5x^4-5x^3-25x^2+4x+20[/tex]
by grouping the terms as
[tex](x^5-5x^3+4x)+(5x^4-25x^2+20)=x(x^4-5x^2+4)+5(x^4-5x^2+4)[/tex]
[tex]\implies f(x)=(x+5)(x^4-5x^2+4)[/tex]
Then you know right away that [tex]x=-5[/tex] is a (real) root, so we eliminate C and D.
The remaining quartic can be factored easily:
[tex]x^4-5x^2+4=(x^2)^2-5x^2+4=(x^2-4)(x^2-1)=(x-2)(x+2)(x-1)(x+1)[/tex]
which admits four more (also real) roots, [tex]x=\pm2[/tex] and [tex]x=\pm1[/tex], so the answer is B.
