How do you multiply #(t^2+5t)/(t+1)div(t+5)#?

1 Answer

#t/(t+1)#

Explanation:

With the statement #(t^2+5t)/(t+1) -: (t+5)#, let's first see that we can factor the numerator on the left-hand side. The other thing to notice is that when we have something in the form of #a/b -: c/d#, it can be restated as #a/b xx d/c#, and so:

#(t^2+5t)/(t+1) -: (t+5)/1#

#(t(t+5))/(t+1) xx 1/(t+5)#

And now we can cancel:

#(t(cancel(t+5)))/(t+1) xx 1/cancel(t+5)#

#t/(t+1)#