# What is the standard form of y= (3x – 4) (2x – 1) (x – 2)?

Mar 30, 2018

$y = 6 {x}^{3} - 23 {x}^{2} + 26 x - \frac{32}{3}$

#### Explanation:

$y = \left(3 x - 4\right) \left(2 x - 1\right) \left(x - 2\right)$

$= 3 \times 2 \times 1 \times \left(x - \frac{4}{3}\right) \left(x - \frac{1}{2}\right) \left(x - 2\right)$

$= 6 \left({x}^{3} - \left(\frac{4}{3} + \frac{1}{2} + 2\right) {x}^{2} + \left(\frac{4}{3} \times \frac{1}{2} + \frac{1}{2} \times 2 + 2 \times \frac{4}{3}\right) x - \frac{4}{3} \times \frac{1}{2} \times 2\right)$

$\frac{4}{3} + \frac{1}{2} + 2 = \left(\frac{8}{6} + \frac{3}{6} + \frac{12}{6}\right) = \frac{8 + 3 + 12}{6} = \frac{23}{6} = \frac{69}{18}$

$\frac{4}{3} \times \frac{1}{2} + \frac{1}{2} \times 2 + 2 \times \frac{4}{3} = \frac{2}{3} + 1 + \frac{8}{3} = \frac{2}{3} + \frac{3}{3} + \frac{8}{3}$
$= \frac{2 + 3 + 8}{3} = \frac{13}{3} = \frac{26}{6} = \frac{78}{18}$

$\frac{4}{3} \times \frac{2}{3} \times 2 = \frac{4 \times 2 \times 2}{3 \times 3} = \frac{16}{9} = \frac{32}{18}$

$\left(3 x - 4\right) \left(2 x - 1\right) \left(x - 2\right) = 6 \left({x}^{3} - \frac{69}{18} {x}^{2} + + \frac{78}{18} x - \frac{32}{18}\right)$

$y = 6 {x}^{3} - 23 {x}^{2} + 26 x - \frac{32}{3}$