Question

How do you factor `(r+s)(s+t)-(r+s)(s-t)+(r+s)(s+t)`?

Answer 1

`(r + s) xx {(s + t) - (s - t) + (s + t)}`

Explanation:

To factor this expression, look for a factor that all the terms have in common.

Then you can factor that one out from all the terms.

In this case, all three terms have a factor in common, namely `(r+ s)`.

Factor out `(r + s)` from each term

`(r + s) { color(white)(2/2)(s + t) - (s - t) + (s + t)color(white)(2/2)}` `larr` answer

`color(white)(mmmmm)` ―――――――――

• Be sure to keep the parentheses for the expression you factored out

It's an error to write it like this (without the beginning parentheses)

`r + s  {(s + t) - (s - t) +  (s + t)}`
This means that only the `s` is distributed to all the terms in the brackets instead of distributing the entire `(r + s)`

• Be sure to enclose the expression in brackets

It's an error to write it like this (without the brackets)

`(r + s) xx(s + t)- (s - t) +(s + t)`
This doesn't show that `(r+s)` is to be distributed to all the terms. It makes it look like `(r+s)` should multiply only `(s+t)`

`color(white)(mmmmm)` ―――――――――

Check
To check factoring, see if distributing the factor brings back the original expression

`(r + s) xx {(s + t) - (s - t) + (s + t)}`

Distribute `(r+s)` to each of the terms

`(r+ s)(s + t) - (r+s)(s-t) + (r + s)(s+ t)`

`Check`

Answer 2

`(r+s)(s+3t)`

Explanation:

`"take out the "color(blue)"common factor "(r+s)`

`rArr(r+s)[(s+t)-(s-t)+(s+t)]`

`"simplifying the terms in the bracket gives"`

`=(r+s)(cancel(s)+tcancel(-s)+t+s+t)`

`=(r+s)(s+3t)`