How do you combine #t/(6+5t+t^2)-2/(2+3t+t^2)#?

1 Answer
maganbhai P.
Mar 8, 2018

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

Explanation:

We know that, #t^3+5t+6=t^2+3t+2t+6=t(t+3)+2(t+3)=(t+3)(t+2)#And, #t^2+3t+2=t^2+2t+t+2=t(t+2)+1(t+2)=(t+2)(t+1)#We have, #t/(t^2+5t+6)-2/(t^2+3t+2)##=t/((t+3)(t+2))-2/((t+2)(t+1))#
#=(t(t+1)-2(t+3))/((t+1)(t+2)(t+3))=(t^2+t-2t-6)/((t+1)(t+2)(t+3))##=(t^2-t-6)/((t+1)(t+2)(t+3))#
#=(t^2-3t+2t-6)/((t+1)(t+2)(t+3))=((t-3)cancel((t+2)))/((t+1)cancel((t+2))(t+3))=(t-3)/((t+1)(t+3))=(t-3)/(t^2+4t+3)#

Related questions
See all questions in Addition and Subtraction of Rational Expressions
Impact of this question
1321 views around the world
You can reuse this answer
Creative Commons License