For what real $m$ does $X^3-(m+4)X^2+(m^2+8)X-8$ have only real roots?

Question

So, if I have a polynomial, in my case has degree $3$, and has a real parameter $m$, how can I find $m$ to show that all his roots are real? I was thinking of looking at the sum of all squared roots and say that this is greater than $0$. This is my idea, but is this enough? My polynomial in this case is $$f=X^3-(m+4)X^2+(m^2+8)X-8$$ The problem is to find $m$ if all roots of $f$ are real numbers.

Answer 1

Hint : Lagrange's Mean value theorem states that between two roots of the function there is atleast one root of its derivative. In our case $$f'(x)=3x^2+2(m+4)x+m^2-8$$ now if all the roots of the cubic are real it means both the roots of the derivative must be real for this $\text{D}^2\geq 0\implies b^2-4ac=4(m+4)^2-4(3)(m^2-8)=4(m^2+8m+16-3m^2+24)=4(-2m^2+8m +24)=\geq0 $ can you continue from here.

Answer 2

If a cubic polynomial has three real roots, then its derivative needs to have two real roots. That is one condition.

These roots of the derivative are the local extrema of the cubic. You need that the minimum is negative and the maximum is positive, then the cubic will have 3 roots. In fact you only need that the values at the extrema have opposing signs, which is a condition that can be evaluated without computing the roots.

Answer 3

A cubic polynomial has three real roots if its discriminant $\Delta$ is nonnegative. Substituting in the formula in the linked article gives that the discriminant of the polynomial $$f_m(X) := X^3 - (m + 4) X^2 + (m^2 + 8) X - 8$$ in $X$ is \begin{align*} \Delta &:= - 3 m^6 + 8 m^5 - 64 m^4 + 240 m^3 - 256 m^2 + 128 m - 192 \\ &\phantom{:}= -(3 m^4 + 4 m^3 + 68 m^2 + 16 m + 48) (m - 2)^2 . \end{align*} One can readily write the quartic polynomial in $m$ as a sum of squares, so only for $m = 2$ do we have $\Delta \geq 0$---and hence among the $f_m$ only $f_2(X) = (X - 2)^3$ has only real roots.