<s>[INST] prove factored polynomial has no real roots the polynomial $6x^3-18x^2-6x-6$ can be factored as $6(x-r)(x^2+ax+b)$ for some $a,b \in \Bbb{R}$ and where $r$ is a real root fo the polynomial. How would you prove that the polynomial $x^2+ax+b$ has no real roots. I know that you can do polynomial long division to get concrete values for $a$ and $b$ but $r$ is a decimal so the long division would be messy and inaccurate, but other then that method I have no idea how i would prove that it has no real roots.
 [/INST] A non-calculus approach is possible, although not necessarily the most efficient.
Consider $$f(x) = x^3 - 6x^2 - x - 1.$$  Let $r$ be a real root of $f$, which we know must exist.  Then long division yields $$\frac{f(x)}{x-r} = x^2 + (r-6)x + (r^2 - 6r - 1) + \frac{r^3 - 6r^2 - r - 1}{x-r},$$ and since $r$ is a root, the remainder term is zero, yielding the factorization $$f(x) = (x-r)(x^2 + (r-6)x + (r^2 - 6r - 1)).$$  Then the quadratic term has discriminant $$(r-6)^2 - 4(r^2 - 6r - 1) = -3r^2 + 12r + 40.$$  This discriminant is nonnegative when $L = \frac{2}{3}(3 - \sqrt{39}) \le r \le \frac{2}{3}(3 + \sqrt{39}) = U$, and negative otherwise.  To determine where $r$ lies relative to the boundaries $L, U$, we evaluate $$f(L) =  \frac{-171 - 26\sqrt{39}}{9}, \quad f(U) = \frac{-171+26\sqrt{39}}{9}.$$  Both are negative.  Since $U < 2(3+\sqrt{49})/3 = 20/3$, we evaluate $f(20/3) = 593/27 > 0$, meaning there is a sign change in the interval $(U, 20/3)$, and $r$ must be in this region; thus the discriminant is negative and the quadratic factor has no real roots.  We knew to search for $r > U$ because once we determined $f(U) < 0$, the sign of the cubic term of $f$ being positive assures that for some $r > U$, $f(r) > 0$.
The proof that $f(U) < 0$ amounts to showing that $39 < (\frac{171}{26})^2$, which is elementary arithmetic.
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