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**Prove It** Hm, well first notice that $\displaystyle \displaystyle \begin{align*} \left(3x^2 - 12x + 5\right)^2 \end{align*}$, being a square, can never be negative. So it makes sense that the minimum value would be 0. Let's try to evaluate where the function will be 0...

$\displaystyle \displaystyle \begin{align*} \left(3x^2 - 12x + 5\right)^2 &= 0 \\ 3x^2 - 12x + 5 &= 0 \\ x &= \frac{12 \pm \sqrt{(-12)^2 - 4(3)(5)}}{2(3)} \\ x &= \frac{12 \pm \sqrt{144 - 60}}{6} \\ x &= \frac{12 \pm \sqrt{84}}{6} \\ x &= \frac{12 \pm 2\sqrt{21}}{6} \\ x &= \frac{6 \pm \sqrt{21}}{3}\end{align*}$

So the minimum value is 0, which occurs when $\displaystyle \displaystyle \begin{align*} x = \frac{6 - \sqrt{21}}{3} \end{align*}$ or $\displaystyle \displaystyle \begin{align*} \frac{6 + \sqrt{21}}{3} \end{align*}$