How do you factor 2(t-s)+4(t-s)^2-(t-s)^32(t−s)+4(t−s)2−(t−s)3?

2 Answers
EZ as pi
May 13, 2016

m(2 - m)(1 + m)m(2−m)(1+m)

= (t - s)( 2 - t + s)(1 + t - s)=(t−s)(2−t+s)(1+t−s)

Explanation:

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

(t-s)(2 + 4(t-s) - (t-s)^2) " note that this a disguised quadratic"(t−s)(2+4(t−s)−(t−s)2) note that this a disguised quadratic

Let (t-s) = m

=m(2 + m - m^2) rArr "find the factors of 2 and 1 which subtract to give 1"m(2+m−m2)⇒find the factors of 2 and 1 which subtract to give 1

m(2 - m)(1 + m)m(2−m)(1+m)

However, m = (t - s) rArr (t - s)(2 - (t - s)(1 + (t - s))⇒(t−s)(2−(t−s)(1+(t−s))
= (t - s)( 2 - t + s)(1 + t - s)=(t−s)(2−t+s)(1+t−s)

Maggie
May 13, 2016

We have,

2(t-s)+4(t-s)^2-(t-s)^32(t−s)+4(t−s)2−(t−s)3

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

(t-s)*(2+4(t-s)-(t-s)^2)(t−s)⋅(2+4(t−s)−(t−s)2)

let's expand the square
(t-s)*(2+4(t-s)-(t^2-2t*s+s^2))(t−s)⋅(2+4(t−s)−(t2−2t⋅s+s2))

Now we get every thing out of brackets
(t-s)*(2+4t-4s-t^2+2t*s-s^2)(t−s)⋅(2+4t−4s−t2+2t⋅s−s2)

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/

Related questions
See all questions in Factoring Completely
Impact of this question
1981 views around the world
You can reuse this answer
Creative Commons License