Bài 1: Phân tích đa thức thành nhân tử
a) Ta có: \(8a^3-6a^2-1+3a\)
\(=\left[\left(2a\right)^3-1^3\right]-3a\left(2a-1\right)\)
\(=\left(2a-1\right)\left(4a^2+2a+1\right)-3a\left(2a-1\right)\)
\(=\left(2a-1\right)\left(4a^2+2a+1-3a\right)\)
\(=\left(2a-1\right)\left(4a^2-a+1\right)\)
b) Ta có: \(x^3-2x^2y+xy^2-9x\)
\(=x\left(x^2-2xy+y^2-9\right)\)
\(=x\left[\left(x^2-2xy+y^2\right)-9\right]\)
\(=x\left[\left(x-y\right)^2-3^2\right]\)
\(=x\left(x-y-3\right)\left(x-y+3\right)\)
c) Ta có: \(5x^2-45\)
\(=5\left(x^2-9\right)\)
\(=5\left(x-3\right)\left(x+3\right)\)
d) Ta có: \(2x^3-4x^2+2x\)
\(=x\left(2x^2-4x+2\right)\)
\(=x\left(2x^2-2x-2x+2\right)\)
\(=x\left[2x\left(x-1\right)-2\left(x-1\right)\right]\)
\(=x\left(x-1\right)\left(2x-2\right)\)
\(=2x\left(x-1\right)^2\)
e) Ta có: \(6x\left(3x-2\right)-12\left(2-3x\right)\)
\(=6x\left(3x-2\right)+12\left(3x-2\right)\)
\(=\left(3x-2\right)\left(6x+12\right)\)
\(=6\left(3x-2\right)\left(x+2\right)\)
f) Ta có: \(4x^2-8xy+4y^2-10\)
\(=\left(2x\right)^2-2\cdot2x\cdot2y+\left(2y\right)^2-10\)
\(=\left(2x-2y\right)^2-10\)
\(=\left(2x-2y-\sqrt{10}\right)\left(2x-2y+\sqrt{10}\right)\)
g) Ta có: \(2x^2-8x+8\)
\(=2\left(x^2-4x+4\right)\)
\(=2\left(x-2\right)^2\)
h) Ta có: \(\left(2x+1\right)^2-\left(x-1\right)^2\)
\(=\left[\left(2x+1\right)-\left(x-1\right)\right]\left[\left(2x+1\right)+\left(x-1\right)\right]\)
\(=\left(2x+1-x+1\right)\left(2x+1+x-1\right)\)
\(=3x\left(x+2\right)\)
