# What is the standard form of  y= (x-5)(2x-2)(3x-1)?

Oct 26, 2017

Seems to me the standard form follows this pattern: $A {x}^{3} + B {x}^{2} + C x + D = 0$

#### Explanation:

So, let's start multiplying out the factors in parentheses:

$y = \left(x - 5\right) \cdot \left(2 \cdot x - 2\right) \cdot \left(3 x - 1\right)$.

FOIL the first two parentheses and we get:

$y = \left(2 {x}^{2} - 2 x - 10 x + 10\right) \cdot \left(3 x - 1\right)$

OR

$y = \left(2 {x}^{2} - 12 x + 10\right) \cdot \left(3 x - 1\right)$

FOIL these parentheses:

$y = 6 {x}^{3} - 38 {x}^{2} + 42 x - 10$

OR

$6 {x}^{3} - 38 {x}^{2} + 42 x - 10 = 0$.