How do you factor the trinomial # 9(a + b)^2 - 60(a + b) + 100#?

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Answer 1 · 2016-07-23T11:46:30

#{3(a+b)-10}^2#

#9(a+b)^2-60(a+b)+100#

=# 9(a+b)^2-30(a+b) -30(a+b)+100#

=# 3(a+b){3(a+b)-10} - 10{3(a+b)-10} #

=# {3(a+b)-10}{3(a+b)-10}#

=#{3(a+b)-10}^2#

Answer 2 · 2016-07-23T13:27:08

#(3(a+b)-10)(3(a+b)-10)#

=# (3(a+b)-10)^2#

This is simply a disguised quadratic which looks a whole lot worse than it is.

Let #(a + b)# be #x#

The expression can now be written as:

#9x^2 -60x +100#

Find factors of 9 and 100 which add to give 60.

9 and 100 are both square numbers ... Let's work with that first.
Use factors of 9 and 100 and find the cross-product.

#" 3 10" rArr 3xx10 = 30#
#" 3 10" rArr 3xx10 = 30 " "30+30=60#

#(3x-10)(3x-10)#

But #x = (a+b)#

#(3(a+b)-10)(3(a+b)-10)#