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

2 Answers
May 13, 2016

Answer:

#m(2 - m)(1 + m)#

#= (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"#

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)(1 + m)#

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

May 13, 2016

We have,

#2(t-s)+4(t-s)^2-(t-s)^3#

First let's factor out one #(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)#

let's expand the square
#(t-s)*(2+4(t-s)-(t^2-2t*s+s^2))#

Now we get every thing out of brackets
#(t-s)*(2+4t-4s-t^2+2t*s-s^2)#

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/