factorise 3a(3a+2c)-4b(b+c)
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(3a*3a+3a*2c)-(4b*b+4b*c)
=9a^2+6ac-4b^2+4bc
=(9a^2-4b^2)- (6ac+4bc)
=(3a+2b)(3a-2b)-2c(3a+2b)
=(3a+2b)(3a-2b-2c)
=9a^2+6ac-4b^2+4bc
=(9a^2-4b^2)- (6ac+4bc)
=(3a+2b)(3a-2b)-2c(3a+2b)
=(3a+2b)(3a-2b-2c)
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