How do you simplify #(-7c)/(21c^2-35c)#?

2 Answers
Jim G.
Sep 25, 2016

#-1/(3c-5)#

Explanation:

When dealing with algebraic fractions the first step is to factorise the numerator/denominator, if possible.

Here, the denominator has a #color(blue)"common factor"# of 7c which can be taken out.

#rArr(-7c)/(7c(3c-5))#

Now cancel, common factors on numerator/denominator.

#rArr(-cancel(7c)^1)/(cancel(7c)^1(3c-5))#

#=-1/(3c-5)#

EZ as pi
Sep 25, 2016

#1/(5-3c)#

Explanation:

With algebraic fractions, check first if they can be factored,

#(-7c)/(21c^2-35c) = " " (-7c)/(7c(3c-5))" "larr# with factors - cancel!

#=(-cancel(7c))/(cancel(7c)(3c-5)) " "larr# the minus sign can be anywhere ...

#= (color(red)(-)1)/(3c-5) = color(red)(-)1/(3c-5) = 1/(color(red)(-)(3c-5)#

#=1/(5-3c)" "larr# note the switch-round

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