How do you factor #\frac { 16c ^ { 8} f ^ { 9} - 12c ^ { 6} f ^ { 7} + 10c ^ { 3} t ^ { 4} } { 2c ^ { 3} f }#?

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Answer 1 · 2018-02-02T11:38:51

#8c^5f^8-6c^3f^6+5f^3#

#(16c^8f^9-12c^6f^7+10c^3f^4)/(2c^3f)#

Factor out everything that is common is the numerator.

#(2c^3f^4(8c^5f^5-6c^3f^3+5))/(2c^3f)#

Now this becomes:

#(2c^3f^4)/(2c^3f)*(8c^5f^5-6c^3f^3+5)#

Remove the common terms in the fraction:

#(cancel(2)cancel(c^3)f^4)/(cancel(2)cancel(c^3)cancel(f))*(8c^5f^5-6c^3f^3+5)#

#f^3(8c^5f^5-6c^3f^3+5)#

#8c^5f^8-6c^3f^6+5f^3#