Let F(x)=f(x)ex, ∴ f' (x) = exf' (x)+exf (x) = ex (2ax+b+ax2+bx+c),

And ∵x=- 1 is an extreme point of f(x)ex,

∴F'(- 1)=e2(-a+c)=0, that is, a=c,

∴δ=b2-4ac=b2-4a2，

When δ = 0, b = 2a, that is, the equation of the straight line where the symmetry axis is located is x =1;

When δ > 0, b2a> 1, that is, the axis of symmetry is on the left of the straight line x=- 1 or on the right of the straight line x= 1.

F (- 1) = A-B+C = 2A-B < 0, so D is wrong, so D is chosen.

Another solution: let F(x)=f(x)ex, ∴ f ′ (x) = exf ′ (x)+exf (x), and for d, there are f ′ (-1) > 0 and f (- 1) > 0.

20 1 1 Zhejiang science T 10:

When a=b=c=0, S= 1 and

| T | = 0； When a≠0, c≠0 and B2-4c; 0, S=3 and |T|=3.

Supplement: f(x)=(x+a)(x2+bx+c)=0 must have roots -a, g (x) = (ax+1) (cx2+bx+1) = 0, and only if a≠0 must have roots.

The number of roots of cx2+bx+ 1=0 cannot exceed the number of roots of equation x2+bx+c=0.