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18x^2+21x-9 I need to figure out how to work this problem

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18x^2 + 21x - 9. Factor out a 3. 

3(6x^2 + 7x - 3). Factor the trinomial. 

3(3x - 1)(2x + 3). Answer 
[tex]18x^2+21x-9=3(6x^2)+3(7x)-3(3)\\\\=3(6x^2+7x-3)=3(6x^2+9x-2x-3)\\\\=3[3x(2x)+3x(3)-1(2x)-1(3)]\\\\=3[3x(2x+3)-1(2x+3)]\\\\=\boxed{3(2x+3)(3x-1)}[/tex]

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