Phân tích đa thức thành nhân tử bằng phương pháp đặt nhân tử chung
(3x + 2)^2 + (3x - 2)^2 - 2(9x^2 - 4)
\(=\left(3x+2\right)^2-2\left(3x+2\right)\left(3x-2\right)+\left(3x-2\right)^2\)
\(=\left(3x+2-\left(3x-2\right)\right)^2\)
\(=\left(3x+2-3x+2\right)^2\)
\(=4^2\)
\(=16\)
\(\left(3x+2\right)^2+\left(3x-2\right)^2-2\left(9x^2-4\right)\)
\(=\left(3x+2\right)^2+\left(3x-2\right)^2-2.\left(3x-2\right)\left(3x+2\right)\)
\(=\left(3x+2-3x+2\right)^2\)
\(=4^2=16\)
=\(\left(3x+2\right)^2-\left(9x^2-4\right)+\left(3x+2\right)^2-\left(9x^2-4\right)\)
= \(\left(3x+2\right)\left(3x+2-3x+2\right)+\left(3x-2\right)\left(3x-2-3x-2\right)\)
= \(4\left(3x+2\right)-4\left(3x-2\right)\)
=\(16\)