Prove no real roots exist for factored cubic function

Question

$6x^3-18x^2-6x-6$ can be expressed as $6(x-3.383)(x^2+ax+b)$ where $a,b \in \Bbb{R}$, how would you prove that $(x^2+ax+b)$ has no real roots?

Answer 1

To show that $x^2+ax+b$ has no real root, it suffices to show that the discriminant $a^2-4b < 0$.

Given the factor, $6(x-3.383)$, we should be able to recover $a$ and $b$ and compute the discriminant.

Answer 2

The sum of the roots equal $3$ and $a$ is the sum of the unidentified roots.

$a = 3-3.383 = 0.383$

and the product or the roots equals $1$

$b = \frac {1}{3.383} \approx 0.29$

Answer 3

You can use also the following way. $$6x^3-18x^2-6x-6=6(x^3-3x^2-3x-1)=$$ $$=6(x^3-3x^2+3x-1-4x+4-4)=6((x-1)^3-4(x-1)-4).$$

Now, we know that the equation $x^3+px+q=0$ has an unique real root iff $\Delta=\frac{p^3}{27}+\frac{q^2}{4}>0.$

In our case $$\Delta=\frac{(-4)^3}{27}+\frac{(-4)^2}{4}>0$$ and we are done!

Answer 4

Leaving the $6$, we even do not need to know the first root. Consider $$f(x)=6x^3-18x^2-6x-6$$ The first derivative $$f'(x)=18 x^2-36 x-6$$ cancels at $$x_1=1-\frac{2}{\sqrt{3}} \qquad \text{and} \qquad x_2= 1+\frac{2}{\sqrt{3}}$$ $x_1$ corresponds to a maximum and $x_2$ to a minimum $$f(x_1)=-24+\frac{32}{\sqrt{3}} <0 \qquad \text{and} \qquad f(x_2)=-24-\frac{32}{\sqrt{3}}<0$$ So, there is only one real root which is $x > x_2$.