\(=12x^2-6x=6x\left(2x-1\right)\)

a: =3x^2-3x-8x+8=(x-1)(3x-8)

b: =x^2-x-5x+5=(x-1)(x-5)

c: =x^2-6x+2x-12=(x-6)(x+2)

x 3 - 3 x 2 - 4 x + 12 = x 3 - 3 x 2 - 4 x - 12 = x 2 x - 3 - 4 x - 3 = x - 3 x 2 - 4 = x - 3 x + 2 x - 2

1a) \(=-\left(x^3-3x^2+3x-1\right)=-\left(x-1\right)^3\)

b) \(=-\left(x^3-3x^2+3x-1\right)=-\left(x-1\right)^3\)

\(a,=-\left(x-1\right)^3\left[=\left(1-x\right)^3\right]\\ b,=\left(1-x\right)^3\)

a. \(=-\left(x^3-3x^2+3x-1\right)=-\left(x-1\right)^3\)

b. \(=\left(1-x\right)^3\)

a) \(A=x^2-6x+9-9y^2\)

\(=\left(x-3\right)^2-\left(3y\right)^2\)

\(=\left(x-3-3y\right)\left(x-3+3y\right)\)

b) \(B=x^3-3x^2+3x-1+2\left(x^2-1\right)\)

\(=\left(x-1\right)^3+\left(2x+2\right)\left(x-1\right)\)

\(=\left(x-1\right)\left[\left(x-1\right)^2+2x+2\right]\)

\(=\left(x-1\right).\left(x^2+3\right)\)

a, \(A=\left(x-3\right)^2-9y^2=\left(x-3-3y\right)\left(x-3+3y\right)\)

b, \(B=\left(x-1\right)^3+2\left(x-1\right)\left(x+1\right)=\left(x-1\right)\left[\left(x-1\right)^2+2\left(x+1\right)\right]\)

\(=\left(x-1\right)\left(x^2-2x+1+2x+2\right)=\left(x-1\right)\left(x^2+3\right)\)

ý D nhé

3x² + 6x + 12

= 3(x² + 2x + 4)

= 3(x + 2)²

sorry, mik sửa:

3x² + 6x + 12

= 3(x² + 2x + 4)

\(=x\left(2x^2+3x-2\right)=x\left(2x^2+4x-x-2\right)=x\left[2x\left(x+2\right)-\left(x+2\right)\right]=x\left(2x-1\right)\left(x+2\right)\)

2x³ + 3x² - 2x

= x ( 2x² + 3x - 2 )

= x ( 2\(x^2\) + 4\(x-x-2\) )

= x [ ( 2\(x^2\) + 4x ) - ( x + 2 )]

= x  [  2x ( x + 2 ) - ( x + 2 )]

= x ( 2x - 1 ) ( x + 2 )

\(-9x^4+3x^2+2\\ =-9x^4+6x^2-3x^2+2\\ =-3x^2\left(3x^2-2\right)-\left(3x^2-2\right)\\ =-\left(3x^2-2\right)\left(3x^2+1\right)\)

\(-9x^4+3x^2+2\)

\(=-9x^4+6x^2-3x^2+2\)

\(=-3x^2\left(3x^2-2\right)-\left(3x^2-2\right)\)

\(=\left(3x^2-2\right)\left(-3x^2-1\right)\)