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.

• @Coolwater I've checked. It only has one root at approximately $3.383$. – Dylan Nov 29 '17 at 6:01
• Actually, $r$ is not a decimal; if anything, you expressed (an approximation of) it as decimal. – celtschk Nov 29 '17 at 6:59
• @Dylan. Concerning your deleted answer, the fact that $f'(x)$ has two roots is not a problem. – Claude Leibovici Nov 29 '17 at 7:02
• @ClaudeLeibovici I realized that after. The function is negative at both critical points so it still only has one root\ – Dylan Nov 29 '17 at 8:20

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.

Without removing the common factor $6$, consider the function and its derivatives $$f(x)=6x^3-18x^2-6x-6$$ $$f'(x)=18 x^2-36 x-6$$ $$f''(x)=36x-36$$ The first derivative cancels at two points $$x_1=1-\frac{2}{\sqrt{3}} \qquad \text{and} \qquad x_2=1+\frac{2}{\sqrt{3}}$$ $$f(x_1)=\frac{32}{\sqrt{3}}-24 <0\qquad \text{and} \qquad f(x_2)=-24-\frac{32}{\sqrt{3}} <0$$ $$f''(x_1)=-24 \sqrt{3}\qquad \text{and} \qquad f''(x_2)=24 \sqrt{3}$$ So $x_1$ is a maximum and $x_2$ a minimum. So, only one possible real root to the cubic.

Now, consider $$f(x)=6x^3-18x^2-6x-6+k$$ then $$f(x_1)=\frac{32}{\sqrt{3}}-24 +k$$ So, if $k=24-\frac{32}{\sqrt{3}}$, there will a double root to the cubic and if $k>24-\frac{32}{\sqrt{3}}$ three real roots.

• how can you determine that there is only one real root, only through the maximum and minimum? – acevj Nov 29 '17 at 7:07
• @acevj. If $x_1$ corresponds to a maximum and $f(x_1)<0$, how ould the curve intersect the $x$ axis ? – Claude Leibovici Nov 29 '17 at 7:13