If #a# and #b# are nonzero integers, is it possible for #x^3ax^2+(a^2b)x+a(2ba^2)=0# to have more than one Real root?
See https://socratic.org/s/aB5FuCr4 to understand where this cubic comes from if it helps.
See https://socratic.org/s/aB5FuCr4 to understand where this cubic comes from if it helps.
1 Answer
Yes, for example with
#x^3x^220x+39=0#
which has three real roots.
Explanation:
The discriminant
#Ax^3+Bx^2+Cx+D#
is given by the formula:
#Delta = B^2C^24AC^34B^3D27A^2D^2+18ABCD#
In our example:
#{ (A=1), (B=a), (C=a^2b), (D=a(2ba^2)) :}#
So:
#Delta = (a)^2(a^2b)^24(a^2b)^34(a)^3(a(2ba^2))27(a(2ba^2))^2+18(a)(a^2b)(a(2ba^2))#
#color(white)(Delta) =16a^6 + 72a^4b  83a^2b^2 + 4b^3#
Note that:

The sign of
#a# does not affect the discriminant. 
The term of highest degree in
#a# is#16a^6# . So when#a# is large compared with#b# , the discriminant will be negative and the cubic will have only one real root. 
The term of highest degree in
#b# is#4b^3# . So when#b# is large and positive compared with#a# , the discriminant will be positive and the cubic will have three real roots.
So try
Then
So our example cubic is:
#x^3x^220x+39=0#
which does indeed have three real roots.
graph{x^3x^220x+39 [10, 10, 10, 75]}