a) \(\left(x-2\right)^2+4x=x^2-4x+4+4x=x^2+4\)

b) \(a^3-27=\left(a-3\right)\left(a^2+3a+9\right)\)

a) \(4a^2+2ab=2a\left(2a+b\right)\)

c)\(x^2-xy+2x-2y=\left(x^2-xy\right)+\left(2x-2y\right)=x\left(x-y\right)+2\left(x-y\right)=\left(x+2\right)\left(x-y\right)\)

\(x^6+2x^5+x^4-2x^3-2x^2+1=\left(x^3+x^2-1\right)^2\)

\(2x^4+7x^3-2x^2-13x+6\)

\(=2x^4+6x^3+x^3+3x^2-5x^2-15x+2x+6\)

\(=2x^3\left(x+3\right)+x^2\left(x+3\right)-5x\left(x+3\right)+2\left(x+3\right)\)

\(=\left(2x^3+x^2-5x+2\right)\left(x+3\right)\)

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

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

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

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

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

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

a. \(=6.\left(x^2+2x+1\right)=6\left(x+1\right)^2\)

b. \(=8x^2-6x+4x-3=2x\left(4x-3\right)+\left(4x-3\right)=\left(2x+1\right)\left(4x-3\right)\)

c. \(=8x^2+6x-4x-3=2x\left(4x+3\right)-\left(4x+3\right)=\left(2x-1\right)\left(4x+3\right)\)

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

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

c) \(=x^2\left(x-1\right)-x\left(x-1\right)+4\left(x-1\right)=\left(x-1\right)\left(x^2-x+4\right)\)

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

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

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

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

2x^4+3x^3-12x^2-7x+6 = (2x^4-x^3)+(4x^3-2x^2)-(10x^2-5x)-(12x-6)

= x^3.(2x-1)+2x^2.(2x-1)-5x.(2x-1)-6.(2x-1) = (2x-1).(x^3+2x^2-5x-6) 

= (2x-1).[ (x^3+x^2)+(x^2+x)-(6x+6) ] = (2x-1).(x+1).(x^2+x-6) = (2x-1).(x-1).[(x^2-2x)+(3x-6)]

= (2x-1).(x+1).(x-2).(x+3)

k mk nha

\(2x^4+3x^3-12x^2-7x+6\)

\(=\left(2x^4+2x^3\right)+\left(x^3+x^2\right)-\left(13x^2+13x\right)+\left(6x+6\right)\)

\(=\left(x+1\right)\left(2x^3+x^2-13x+6\right)\)

2x³ + 3x² - 11x - 6 

Thử với x = 2 ta có :

2.2³ + 3.2² - 11.2 - 6 = 0

Vậy x = 2 là nghiệm của đa thức. Theo hệ quả của định lí Bézout thì đa thức trên chia hết cho x - 2

Thực hiện phép chia 2x³ + 3x² - 11x - 6 cho x - 2 ta được 2x² + 7x + 3

=> 2x³ + 3x² - 11x - 6 = ( x - 2 )( 2x² + 7x + 3 )

Lại có : 2x² + 7x + 3 = 2x² + 6x + x + 3 = 2x( x + 3 ) + ( x + 3 ) = ( x + 3 )( 2x + 1 )

=> 2x³ + 3x² - 11x - 6 = ( x - 2 )( x + 3 )( 2x + 1 )

Lại Bezout, thế này thì ...

2x³ + 3x² - 11x - 6

= ( 2x³ + 2x² - 12x ) + ( x² + x - 6 )

= 2x ( x² + x - 6 ) + ( x² + x - 6 )

= ( 2x + 1 ) ( x² + x - 6 )

= ( 2x + 1 ) [ ( x² + 3x ) - ( 2x + 6 ) ]

= ( 2x + 1 ) [ x ( x + 3 ) - 2 ( x + 3 ) ]

= ( 2x + 1 ) ( x - 2 ) ( x + 3 )