How do you factor #6y^2 -5yz-6z^2#?

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Answer 1 · 2016-03-20T23:18:31

#6y^2-5yz-6z^2 = (2y-3z)(3y+2z)#

Notice the symmetry/anti-symmetry of the coefficients.

As a result, if #ay+bz# is a factor then #by-az# is a factor.

If the factors are rational, that basically gives us two possible patterns:

#(y+-6z)(6y+-z)#

#(2y+-3z)(3y+-2z)#

where the signs in the first and second binomial factors are opposite to one another.

Given that #3^2-2^2 = 9-4 = 5#, we can quickly deduce:

#6y^2-5yz-6z^2 = (2y-3z)(3y+2z)#

#color(white)()#
A more pedestrian, standard approach is to use an AC method.

Look for a pair of factors of #AC = 6*6 = 36# which differ by #B=5#.

The pair #9, 4# works.

Use this pair to split the middle term, then factor by grouping as follows:

#6y^2-5yz-6z^2#

#=6y^2-9yz+4yz-6z^2#

#=(6y^2-9yz)+(4yz-6z^2)#

#=3y(2y-3z)+2z(2y-3z)#

#=(3y+2z)(2y-3z)#