a) ( x 2  – 4x + 1)( x 2  – 2x + 3).

b) ( x 2  + 5x – 1)( x 2  + x – 1).

\(2x^3-12x^2+18x=2x\left(x^2-6x+9\right)=2x\left(x-3\right)^2\)

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

2x³- 12x² + 18x = 2x (x² - 6x + 9 ) = 2x ( x - 3 )²

\(3x^3+5x^2-14x+4\\ =\left(3x^3-x^2\right)+\left(6x^2-2x\right)-\left(12x-4\right)\\ =x^2\left(3x-1\right)+2x\left(3x-1\right)-4\left(3x-1\right)\\ =\left(x^2+2-4\right)\left(3x-1\right)\)

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

\(=5x\left(x-y\right)^2\)

Phân tích thành nhân tử:

\(5x^3-10x^2y+5xy^2=5x\left(x^2-2xy+y^2\right)=5x\left(x-y\right)^2\)

Chúc bạn học tốt!!!

\(5x^3-10x^2y+5xy^2\)

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

\(=5x\left(x-y\right)^2\)

\(=-12x^2\left(y-x\right)+18x^3\left(y-x\right)\)

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

\(x^4-14x^3+71x^2-154x+120\)

\(=x^4-2x^3-12x^3+24x^2+47x^2-94x-60x+120\)

\(=x^3\left(x-2\right)-12x^2\left(x-2\right)+47x\left(x-2\right)-60\left(x-2\right)\)

\(=\left(x-2\right)\left(x^3-12x^2+47x-60\right)\)

\(=\left(x-2\right)\left(x^3-3x^2-9x^2+27x+20x-60\right)\)

\(=\left(x-2\right)\left[x^2\left(x-3\right)-9x\left(x-3\right)+20\left(x-3\right)\right]\)

\(=\left(x-2\right)\left(x-3\right)\left(x^2-9x+20\right)\)

\(=\left(x-2\right)\left(x-3\right)\left(x^2-4x-5x+20\right)\)

\(=\left(x-2\right)\left(x-3\right)\left[x\left(x-4\right)-5\left(x-4\right)\right]\)

\(=\left(x-2\right)\left(x-3\right)\left(x-4\right)\left(x-5\right)\)

1. 

\(\left(12x^2+6x\right)\left(y+z\right)+\left(12x^2+6x\right)\left(y-z\right)\\ =\left(12x^2+6x\right)\left(y+z+y-z\right)\\ =2y\left(12x^2+6x\right)\\ =2y.6x\left(2x+1\right)\\ =12xy\left(2x+1\right)\)

2. 

\(x\left(x-6\right)+10\left(x-6\right)=0\\ \Leftrightarrow\left(x-6\right)\left(x+10\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=6\\x=-10\end{matrix}\right.\)

Vậy \(x\in\left\{6;-10\right\}\) là nghiệm của pt

Bài 1:

Ta có: \(\left(12x^2+6x\right)\left(y+z\right)+\left(12x^2+6x\right)\left(y-z\right)\)

\(=\left(12x^2+6x\right)\left(y+z+y-z\right)\)

\(=6x\left(2x+1\right)\cdot2y\)

\(=12xy\left(2x+1\right)\)

Bài 2: 

Ta có: \(x\left(x-6\right)+10\left(x-6\right)=0\)

\(\Leftrightarrow\left(x-6\right)\left(x+10\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=6\\x=-10\end{matrix}\right.\)