# What is the standard form of y=-(x - 1) (x^2 +6)(x/3 + 2) ?

Feb 24, 2018

$y = - \frac{1}{3} {x}^{4} - \frac{5}{3} {x}^{3} - 10 x + 12$

#### Explanation:

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

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

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

$y = - \frac{1}{3} \left({x}^{4} - {x}^{3} + 6 {x}^{2} - 6 x + 6 {x}^{3} - 6 {x}^{2} + 36 x - 36\right)$

$y = - \frac{1}{3} \left({x}^{4} + 5 {x}^{3} + 30 x - 36\right)$

$y = - \frac{1}{3} {x}^{4} - \frac{5}{3} {x}^{3} - 10 x + 12$
Which fits the standard form of $y = A {x}^{4} + B {x}^{3} + C {x}^{2} + D x + E$