Answer:
The actual root of f(x) is -5/2 (first option)
Explanation:
A root of f(x) is a value of the variable "x" that makes f(x)=0
1) With x=-5/2
f(-5/2)=6(-5/2)^4+5(-5/2)^3-33(-5/2)^2-12(-5/2)+20
f(-5/2)=6(5^4/2^4)+5(-5^3/2^3)-33(5^2/2^2)+6(5)+20
f(-5/2)=6(625/16)-5(125/8)-33(25/4)+30+20
f(-5/2)=3(625/8)-625/8-825/4+50
f(-5/2)=1,875/8-625/8-825/4+50
f(-5/2)=[1,875-625-2(825)+8(50)]/8
f(-5/2)=(1,875-625-1,650+400)/8
f(-5/2)=0/8
f(-5/2)=0 then x=-5/2 is a root of f(x)
2) With x=-2
f(-2)=6(-2)^4+5(-2)^3-33(-2)^2-12(-2)+20
f(-2)=6(2^4)+5(-2^3)-33(2^2)+24+20
f(-2)=6(16)-5(8)-33(4)+44
f(-2)=96-40-132+44
f(-2)=-32 different of 0, then x=-2 is not a root of f(x)
3) With x=1
f(1)=6(1)^4+5(1)^3-33(1)^2-12(1)+20
f(1)=6(1)+5(1)-33(1)-12+20
f(1)=6+5-33+8
f(1)=-14 different of 0, then x=1 is not a root of f(x)
4) With x=10/3
f(10/3)=6(10/3)^4+5(10/3)^3-33(10/3)^2-12(10/3)+20
f(10/3)=6(10^4/3^4)+5(10^3/3^3)-33(10^2/3^2)-4(10)+20
f(10/3)=6(10,000/81)+5(1,000/27)-33(100/9)-40+20
f(10/3)=2(10,000/27)+5,000/27-11(100/3)-20
f(10/3)=20,000/27+5,000/27-1,100/3-20
f(10/3)=[20,000+5,000-9(1,100)-27(20)]/27
f(10/3)=(20,000+5,000-9,900-540)/27
f(10/3)=14,560/27 different of 0, then x=10/3 is not a root of f(x)