Quadratic Equations

Let f(x)=3x3+bx2+bx+3
(A) All the roots are negative. So, the number of sign change in f(−x) should be 3 , which is possible only when b is positive. According to options (A) ⟶ IV.
(B) Clearly for any value of b,f(x)=0 has root x=−1.
‌∴f(x)=0⇒(x+1)(3x2+(b−3)x+3)=0
‌⇒3x2+(b−3)x+3≠0
[ ∵ two roots are complex]
‌⇒‌‌(b−3)2−4(3)(3)<0
‌⇒‌‌b2−6b−27<0
‌⇒‌‌(b+3)(b−9)<0
‌⇒‌‌−3<b<9
Hence, B ⟶ II.
(C) We know that one root is x=−1
(Negative).
Two roots are complex if −3<b<9.
If b=−3, then f(x)=0
⇒‌‌(x+1)(x−1)2=0
In this case two roots are positive and equal.(D) All roots will be real and distinct if and only if 3x2+(b−3)x+3=0 have two distinct real solutions other, than -1 .
‌(b−3)2−4(3)(3)>0
⇒b2−6b−27>0
⇒(b+3)(b−9)>0
⇒b∈(−∞,−3)∪(9,∞)
Thus, D ⟶ III
Therefore, A⟶IV,B⟶II,C⟶V,D⟶III