First set of questions is all roots, or solutions, of the equations and second set is to find where the functions are zero, the rational roots only. The methods for roots and zeroes are basically the same.
1. 3x^2+10x+3=3x^2+9x+x+3=3x(x+3)+(x+3)=(3x+1)(x+3)=0; x=-1/3, -3
2. 5x^2+5x-x-1=5x(x+1)-(x+1)=(5x-1)(x+1)=0; x=1/5, -1
3. Cubic equation. Trial shows that x=-1 is a solution, so divide cubic by factor (x+1). Use synthetic division:
-1 | 3 11 5 -3
| 3 -3 -8 3
3 8 -3 | 0 and (x+1)(3x^2+8x-3)=
(x+1)(3x^2+9x-x-3)=(x+1)(3x(x+3)-(x+3))=(x+1)(3x-1)(x+3)=0; x=-1, 1/3, -3
4. 3x^4-2x^2-5=(3x^2-5)(x^2+1)=0; 3x^2=5, so x=+/-sqrt(5/3)
5. Trial shows that 3x=-1 is a solution, i.e., x=-1/3. Use synthetic division:
-1/3 | 3 10 30 9
| 3 -1 -3 -9
3 9 27 | 0
(3x+1)(x^2+3x+9)=0; x=-1/3 is the only real solution
6. (5x^2+7)(x^2-2)=0; x=+/-sqrt(2)
7. No real roots
8. (5x^2+2)(x^2-8)=0; x=+/-sqrt(8) or +/-2sqrt(2)
9. (5x^2+9)(x^2-4)=(5x^2+9)(x+2)(x-2)=0; x=+/-2
10. (5x^2+3)(x^2-8)=0; x=+/-sqrt(8) or +/-2sqrt(2)
11. f(x)=5x(x^2-1)-(x^2-1)=(5x-1)(x-1)(x+1); x=1/5, 1, -1
12. Try x=1/2, f(x)=0. So f(x)=(2x-1)(x^2-11x-1) has no more rational roots
13. f(x)=3x(x^2-1)-(x^2-1)=(3x-1)(x-1)(x+1); zeroes at x=1/3, 1, -1
14. f(x)=(x+2)(x^2+7x+3); only rational zero at x=-2
15. (x-3)(2x^2+7x-3); rational zero at x=3
16. (x-1)^2(4x-1); x=1, 1/4
17. (x+1)(2x-1)(2x+1); x=-1, +/-1/2
18. (x+1)^2(3x-1); x=-1, 1/3
19. (x-5)(3x^2-7x-75); x=5, no other rational roots
20. (x-3)(2x^2+11x-4); x=3, other roots irrational