How do you simplify #(-49y^4 + 42y^3)/ (7y)#?

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Answer 1 · 2015-05-09T03:55:28

First write it as:
#(-49y^4)/(7y)+(42y^3)/(7y)=-7y^4/y+6y^3/y=#
Then use the fact that: #a^m/a^m=a^(m-n)#
So:#-7y^(4-1)+6y^(3-1)=#
#=-7y^3+6y^2#

Answer 2 · 2015-05-09T04:41:52

The answer is #-y^2(7y-6)# or multiply times #-1# to get #y^2(6-7y)# .

Simplify #(-49y^4+42y^3)/(7y)# .

Factor out #7y^3# .

#(-7y^3(7y-6))/(7y)#

Simplify.

#(-cancel7y^3)/(cancel7y)(7y-6)# =

#-y^(3-1)(7y-6)# =

#-y^2(7y-6)#

You can multiply times #-1# to get #y^2(6-7y)#. (optional)