# How do you factor 2(t-s)+4(t-s)^2-(t-s)^3?

May 13, 2016

$m \left(2 - m\right) \left(1 + m\right)$

$= \left(t - s\right) \left(2 - t + s\right) \left(1 + t - s\right)$

#### Explanation:

Note that there is a common bracket in each term. Start by dividing this out.

$\left(t - s\right) \left(2 + 4 \left(t - s\right) - {\left(t - s\right)}^{2}\right) \text{ note that this a disguised quadratic}$

Let (t-s) = m

=$m \left(2 + m - {m}^{2}\right) \Rightarrow \text{find the factors of 2 and 1 which subtract to give 1}$

$m \left(2 - m\right) \left(1 + m\right)$

However, m = (t - s) rArr (t - s)(2 - (t - s)(1 + (t - s))
$= \left(t - s\right) \left(2 - t + s\right) \left(1 + t - s\right)$

May 13, 2016

We have,

$2 \left(t - s\right) + 4 {\left(t - s\right)}^{2} - {\left(t - s\right)}^{3}$

First let's factor out one $\left(t - s\right)$ because it is common to all, this will make thing easier to handle. We are left with

$\left(t - s\right) \cdot \left(2 + 4 \left(t - s\right) - {\left(t - s\right)}^{2}\right)$

let's expand the square
$\left(t - s\right) \cdot \left(2 + 4 \left(t - s\right) - \left({t}^{2} - 2 t \cdot s + {s}^{2}\right)\right)$

Now we get every thing out of brackets
$\left(t - s\right) \cdot \left(2 + 4 t - 4 s - {t}^{2} + 2 t \cdot s - {s}^{2}\right)$

I'm not sure you can go any further, I've played with the right bracket and put it through a factor calculator and got nothing/