Phân tích đa thức thành nhân tử bằng cách đặt nhân tử chung

4 ( x - 2)

2( x - 1)

8( x - 1)

6( x + 2)

5( x³ - 2)

5( x³ + 3)

5( x³ - 3)

15( x³ - 1)

$$4x\left(x-3\right)$$  

$$4x\left(x+3\right)$$  

$$4\left(x^2-3\right)$$  

$$4\left(x-3x\right)$$  

$$2\left(x-5\right)$$  

$$2\left(x^2-3\right)$$  

$$2\left(x^2+10\right)$$  

$$2\left(x^2+5\right)$$  

$$4\left(y-1\right)\left(1-8x\right)$$  

$$4\left(y-1\right)\left(1+8x\right)$$  

$$4\left(y-1\right)\left(1+2x\right)$$  

$$4\left(y-1\right)\left(1-2x\right)$$  

$$\left(x-2\right)\left(5+x-1\right)$$  

$$\left(x-2\right)\left(x-3\right)$$  

$$\left(2-x\right)\left(x+3\right)$$  

$$\left(4-x\right)\left(x+2\right)$$  

$$x\left(4x^4+2\right)$$  

$$x\left(4x-10\right)$$  

$$2x\left(2x^4-5\right)$$  

$$2x\left(2x^4+5\right)$$  

$$2\left(x-12\right)$$  

$$2\left(x+12\right)$$  

$$4\left(x-6\right)$$  

$$2\left(x-10\right)$$  

$$9\left(x-18\right)$$  

$$9\left(x-2\right)$$  

$$9\left(x-\sqrt[]{2}\right)\left(x+\sqrt[]{2}\right)$$  

$$9\left(x^2-2\right)$$  

6(x²-2)

4(x - 3)

6( x - 2)

6 ( x + 2)