# What is the standard form of y=-(x - 1) (x^2 +6)(x + 5)^2 ?

Feb 5, 2018

$y = - {x}^{5} - 9 {x}^{4} - 21 {x}^{3} - 29 {x}^{2} - 90 x + 150$

#### Explanation:

Given: $y = - \left(x - 1\right) \left({x}^{2} + 6\right) {\left(x + 5\right)}^{2}$

Standard form of a polynomial requires distribution and putting terms in descending order:

Note: ${\left(a + b\right)}^{2} = \left({a}^{2} + 2 a b + {b}^{2}\right)$

$y = - \left(x - 1\right) \left({x}^{2} + 6\right) {\left(x + 5\right)}^{2}$
$= - \left(x - 1\right) \left({x}^{2} + 6\right) \left({x}^{2} + 10 x + 25\right)$

$y = - \left(x - 1\right) \left({x}^{4} + 10 {x}^{3} + 25 {x}^{2} + 6 {x}^{2} + 60 x + 150\right)$

$y = - \left(x - 1\right) \left({x}^{4} + 10 {x}^{3} + 31 {x}^{2} + 60 x + 150\right)$

Distribute again:

$y = - \left({x}^{5} + 10 {x}^{4} + 31 {x}^{3} + 60 {x}^{2} + 150 x - {x}^{4} - 10 {x}^{3} - 31 {x}^{2} - 60 x - 150\right)$

$y = - \left({x}^{5} + 9 {x}^{4} + 21 {x}^{3} + 29 {x}^{2} + 90 x - 150\right)$
$y = - {x}^{5} - 9 {x}^{4} - 21 {x}^{3} - 29 {x}^{2} - 90 x + 150$