How to get $\frac{2}{3}(3x-5)^{2}+\frac{19}{3}$ from this expression $6x^{2}-20x+23$.

Question

I'm trying to figure out how to get $\frac{2}{3}(3x-5)^{2}+\frac{19}{3}$ from this expression $6x^{2}-20x+23$.

Hints?

Answer 1

$$\begin{align}6x^2-20x+23&=\frac{2}{3}\left(\frac326x^2-\frac3220x+\frac3223\right)\\ &=\frac23\left(9x^2-15x+\frac{69}2\right)\\ &=\frac23\left(9x^2-15x+\frac{50}2+\frac{19}{2}\right)\\ &=\frac23\left(9x^2-15x+25+\frac{19}{2}\right)\\ &=\frac23\left((3x-5)^2+\frac{19}{2}\right)\\ &=\frac23\left(3x-5\right)^2+\frac23\cdot\frac{19}{2}\\ &=\frac23\left(3x-5\right)^2+\frac{19}{3}\end{align}$$

Answer 2

HINT:

Complete the square, which can be done by noting that:

$$6x^2 - 20x + 23 = 6(x^2 - \frac{10}{3}x + \frac{25}{9}) + \frac{19}{3}$$

Then multiply $(x^2 - \frac{10}{3}x + \frac{25}{9})$ by $9$ and divide $6$ by $9$.