\((x-1)(x-3)(x+8)\)

\(x^3+4x^2-29x+24\)

\(=x^2\left(x+8\right)-4x\left(x+8\right)+3\left(x+8\right)\)

\(=\left(x+8\right)\left(x^2-4x+3\right)\)

\(=\left(x+8\right)\left[x\left(x-1\right)-3\left(x-1\right)\right]\)

\(=\left(x+8\right)\left(x-1\right)\left(x-3\right)\)

\(x^3+9x^2+26x+24\)

\(=x^3+3x^2+6x^2+18x+8x+24\)

\(=\left(x^3+3x^2\right)+\left(6x^2+18x\right)+\left(8x+24\right)\)

\(=x^2\left(x+3\right)+6x\left(x+3\right)+8\left(x+3\right)\)

\(=\left(x+3\right)\left(x^2+6x+8\right)\)

\(=\left(x+3\right)\left(x^2+2x+4x+8\right)\)

\(=\left(x+3\right)\left[\left(x^2+2x\right)+\left(4x+8\right)\right]\)

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

\(=\left(x+3\right)\left(x+2\right)\left(x+4\right)\)

\(15^3+29x^2-8x-12=15x^3+30x^2-x^2-2x-6x-12\)

= \(15x^2.\left(x+2\right)-x.\left(x+2\right)-6.\left(x+2\right)\)= \(\left(x+2\right).\left(15x^2-x-6\right)\)

= \(\left(x+2\right).\left(15x^2-10x+9x-6\right)\)= \(\left(x+2\right).\left(3x-2\right).\left(5x+3\right)\)

\(x^3+9x^2+26x+24=x^3+3x^2+6x^2+18x+8x+24\)\(=x.^2\left(x+3\right)+6x.\left(x+3\right)+8.\left(x+3\right)\)\(=\left(x+3\right).\left(x^2+6x+8\right)\)\(\left(x+3\right).\left(x^2+2x+4x+8\right)=\left(x+2\right).\left(x+3\right).\left(x+4\right)\)

Ta có : 15x³ + 29x² - 8x - 12

= 15x³ + 30x² - x² - 8x - 12

= 15x(x + 2) - (8x + 16) - (x² - 4)

= 15x(x + 2) - 8(x + 2) - (x - 2)(x + 2)

= (x + 2)(15x - 8 - x + 2)

= (x + 2) (14x - 6) 

 click zô nha >_<                 

Ta có : 15x³ + 29x² - 8x - 12

= 15x³ + 30x² - x² - 8x - 12

= 15x(x + 2) - (8x + 16) - (x² - 4)

= 15x(x + 2) - 8(x + 2) - (x - 2)(x + 2)

= (x + 2)(15x - 8 - x + 2)

= (x + 2) (14x - 6)

(x+2)(14x-6)

#)Giải :

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

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

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

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

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

\(x^4+2x^3+5x^2+4x-12\)

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

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

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

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

Câu 1.

Đoán được nghiệm là 2.Ta giải như sau:

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

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

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

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

\(1,=x\left(x^2-2x+1-y^2\right)=x\left[\left(x-1\right)^2-y^2\right]=x\left(x-y-1\right)\left(x+y-1\right)\\ 2,=\left(x+y\right)^3\\ 3,=\left(2y-z\right)\left(4x+7y\right)\\ 4,=\left(x+2\right)^2\\ 5,Sửa:x\left(x-2\right)-x+2=0\\ \Leftrightarrow\left(x-2\right)\left(x-1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)