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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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)

a) nhận xét hệ số : 1 + 4 - 29 + 24 = 0

=> x³ + 4x² - 29x + 24 = x²(x-1) + 5x(x-1) - 24(x-1)

= (x-1)(x²+5x-24) = (x-1)(x-3)(x+8)

b) ...

a) \(x^3+4x^2-29x+24\)=\(\left(x+8\right)\left(x^2-4x+3\right)\)=\(\left(x+8\right)\left(x^2-x-3x+3\right)\)=\(\left(x+8\right)\left(x-1\right)\left(x-3\right)\)

b) \(x^4+6x^3+7x^2-6x+1\)=\(x^4+3x^3-x^2+3x^3+9x^2-3x-x^2-3x+1\)=\(x^2\left(x^2+3x-1\right)+3x\left(x^2+3x-1\right)-\left(x^2+3x-1\right)\)=\(\left(x^2+3x-1\right)\left(x^2+3x-1\right)\)=\(\left(x^2+3x-1\right)^2\)

\(x^3+2x^2-29x-30=\left(x^3+x^2\right)+\left(x^2+x\right)-\left(30x+30\right)\)

\(=x^2\left(x+1\right)+x\left(x+1\right)-30\left(x+1\right)=\left(x+1\right)\left(x^2+x-30\right)\)

\(=\left(x+1\right)\left(x^2+6x-5x-30\right)=\left(x+1\right)\left[x\left(x+6\right)-5\left(x+6\right)\right]=\left(x+1\right)\left(x-5\right)\left(x+6\right)\)

a,\(=x^3-x^2+5x^2-5x-24x+24\)

\(=x^2\left(x-1\right)+5x\left(x-1\right)-24\left(x-1\right)\)

\(=\left(x-1\right)\left(x^2+5x-24\right)\)

\(=\left(x-1\right)\left(x^2-3x+8x-24\right)\)

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

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

a)=(x-3)(x-1)(x+8)

b)=(x²+3x-1)²