How do you factor the trinomial #6a^2-54a+108#?

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Answer 1 · 2016-04-07T13:30:33

Factors o #6a^2-54a+108# are #6*(a-6)*(a-3)#

In #6a^2-54a+108#, each of the coefficients is divisible by #6#, so taking #6# common we get

#6a^2-54a+108=6xx(a^2-9a+18)#

Now to factorize #a^2-9a+18# let us split middle term into components whose product is #18# and these are #6# and #3#.

Hence, #6a^2-54a+108# can be written as

#6(a^2-6a-3a+18)# or

#6*{a(a-6)-3(a-6}# or

#6*(a-6)*(a-3)#