How do you factor #27c ^ { 2} - 6a t - r ^ { 2}# completely?

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Answer 1 · 2018-06-27T03:31:52

The only factored form I see is #3(9c^2-2at)-r^2#

Notice that among the terms, the only common factor within any of the terms is a 3 within the 27 and the 6, so we could factor this way:

#3(9c^2-2at)-r^2#

Answer 2 · 2018-06-27T03:51:01

Please see the explanation below.

Just in case it was meant to be:

#27 c^2 − 6rc − r^2 #

Then:

Using AC method for factoring the quadratics in the form of:

#Ax^2+Bx+C#

We need to find two numbers that multiply to -27, and add to -6,

those numbers are -9 and 3 thus:

#27c^2-6rc-r^2#

#=27c^2-9rc+3rc-r^2#

#=9c(3c-r)+r(3c-r)#

#=(9c+r)(3c-r)#