How do you find the reverse distributive property of #t^2 - ts + 3t#?

1 Answer
May 29, 2016

Answer:

#t(t-s+3)#

Explanation:

First ask yourself what number or variable is present in each term. Here we see that #t# occurs in each term. This is the variable that we can factor out of each term.

When we factor out #t# we must then ask ourselves what we must multiply #t# by to get the original problem.

What must we multiply #t# by to get #t^2#?

That would be #t#. So we start out with:

#t(t#

Then we ask ourselves what we multiply #t# by to obtain #-ts#.

That would be #-s# giving us:

#t(t-s#

Now for the last term we ask ourselves what we must multiply #t# by to obtain #3t#.

That of course would be #3# giving us our answer of:

#t(t-s+3)#