Show that the cubic eq:
$$x^3+ax^2+bx+c = 0 \quad a,b,c\in \mathbb{R}$$
has at least one real root.
I know that the above equation can be broken down into $(x-a)(x-b)(x-c) = 0$ , but I have no idea what to do next. I can't use IVT to do this because I don't have a specified range.
(edit): For others reading this, the equation CANNOT be broken down to $(x-a)(x-b)(x-c) = 0$
Answer
All odd polynomials have at least one real root because of the intermediate value theorem. To prove this just plug in a very large positive number and a very large negative number for $x$ (e.g. $10^{23}$ and $-10^{23})$ and note that corresponding $y$ values will have opposite signs. Then the IVT tells you that there is at least one value of $x$ between the two large numbers for which $y=0$ i.e., the polynomial has a root.
Another thing to note is that the only irreducible polynomials over the reals are quadratic and linear. Since this is a polynomial of degree three it has to be decomposible either as three linear factors or one irreducible quadratic factor with a linear factor.
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