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

2 Answers
Mar 3, 2018

Answer:

#(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#

Mar 3, 2018

Answer:

#(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)#