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## Homework Statement

Show that for negative c (a,b,c - real) equation [itex]x^3+ax^2+bx+c=0[/itex] has at least one positive root.

**2. The attempt at a solution**

Considering the equivalent form of the equation above for large |x|:

[itex]x^3(1+\frac{a}{x}+\frac{b}{x^2}+\frac{c}{x^3})=0[/itex] we can conclude that there exists at least one real root, because the function changes signes if x does the same.

Let r,s,t be some roots of this equation. At least one of them is real. The cubic polynomial on LHS can be written in the form:

[itex](x-r)(x-s)(x-t)[/itex]. Thus the equation admits the following form: [itex]x^3-(r+s+t)x^2+(rs+st+rt)x-rst=0[/itex]. Comparing yields: -(r+s+t)=a, rs+st+rt=b and -rst=c. Now if for example r is real and so are s and t, then in order for c to be negative the product rst must be positive. This in turn is true only if at least one the factors is positive.

Is it correct?

Now the task says nothing about whether all the roots are assumed to be real or not. In case only one root is real, for example r, is it correct to say that the product st is real?

Is there another way to solve the problem?