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

May 15, 2016

$y = - {x}^{5} + 5 {x}^{4} - 10 {x}^{3} + 10 {x}^{2} - x + 1$

#### Explanation:

Notice that ${\left(x - 1\right)}^{2} {\left(x - 1\right)}^{3} = {\left(x - 1\right)}^{5}$

The binomial theorem tells us that:

${\left(a + b\right)}^{n} = {\sum}_{k = 0}^{n} \left(\begin{matrix}n \\ k\end{matrix}\right) {a}^{n - k} {b}^{k}$

where the binomial coefficient: ((n),(k)) = (n!)/(k! (n-k)!)

The binomial coefficients can be found as rows in Pascal's triangle:

The row $1 , 5 , 10 , 10 , 5 , 1$ gives us the binomial coefficients for ${\left(a + b\right)}^{5}$:

${\left(a + b\right)}^{5} = {a}^{5} + 5 {a}^{4} b + 10 {a}^{3} {b}^{2} + 10 {a}^{2} {b}^{3} + 5 a {b}^{4} + {b}^{5}$

Then put $a = x$, $b = - 1$ to find:

${\left(x - 1\right)}^{5} = {x}^{5} - 5 {x}^{4} + 10 {x}^{3} - 10 {x}^{2} + 5 x - 1$

So:

$y - 4 x$

$= - {\left(x - 1\right)}^{2} {\left(x - 1\right)}^{3}$

$= - {\left(x - 1\right)}^{5}$

$= - \left({x}^{5} - 5 {x}^{4} + 10 {x}^{3} - 10 {x}^{2} + 5 x - 1\right)$

$= - {x}^{5} + 5 {x}^{4} - 10 {x}^{3} + 10 {x}^{2} - 5 x + 1$

Add $4 x$ to both ends to get:

$y = - {x}^{5} + 5 {x}^{4} - 10 {x}^{3} + 10 {x}^{2} - x + 1$

This has the powers of $x$ in descending order, which is the standard form.