How do you simplify #\frac { 40k ^ { 4} } { 4k } - \frac { 44k ^ { 3} } { 4k } - \frac { 60k ^ { 2} } { 4k } - \frac { 4k } { 4k } - \frac { 21} { 4k }#?

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Answer 1 · 2018-04-24T22:52:53

#(40k^4-44k^3-60k^2-4k-21)/(4k)# or

#10k^3-11k^2-15k-1-(21/4k)#

The denominator is the same for every term, so this will just be equal to

#(40k^4-44k^3-60k^2-4k-21)/(4k)#

The numerator is unfactorable, but we can divide all terms by #4k#. We get:

#10k^3-11k^2-15k-1-(21/4k)#

Answer 2 · 2018-04-24T22:57:17

#10k^3 - 11k^2 - 15k - 1 - 21/(4k)#

Notice that the variable, k, cancels out as:
#k^a/k^b = k^(a-b)#

This goes for any variable in division. The coefficients, or the constants, simplify as normal division. Such as 40/4 is 10.

#21/4# does not cancel or simply, therefore it can be left alone. We could take it to #1/4*(21/k)# if we wanted, but it's more simple as is.