=3x(x+2y)

=3x(x+2y)

3x² + 6xy

= 3x(x+2y)

\(9x^4+36x^3+29x^2-14x\)

\(=x\left(9x^3+36x^2+29x-14\right)\)

\(=x\left(9x^3+18x^2+18x^2+36x-7x-14\right)\)

\(=x\left[9x^2\left(x+2\right)+18x\left(x+2\right)-7\left(x+2\right)\right]\)

\(=x\left(x+2\right)\left(9x^2+18x-7\right)\)

\(=x\left(x+2\right)\left(9x^2-3x+21x-7\right)\)

\(=x\left(x+2\right)\left[3x\left(3x-1\right)+7\left(3x-1\right)\right]\)

\(=x\left(x+2\right)\left(3x+7\right)\left(3x-1\right)\)

Đa thức này không phân tích được thành nhân tử

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

Đa thức này không phân tích được thành nhân tử.

Bạn coi lại đề.

Ta có: \(4\left(x+5\right)\left(x+6\right)\left(x+10\right)\left(x+12\right)+3x^2\)

\(=4\left(x^2+60+17x\right)\left(x^2+60x+16x\right)+3x^2\)

\(=4\left[\left(x^2+60\right)^2+33x\left(x^2+60\right)+272x^2\right]+3x^2\)

\(=4\left(x^2+60\right)^2+132x\left(x^2+60\right)+1091x^2\)

\(\left(x^2-5x\right)^2-3x^2+15x-18\)

\(=\left(x^2-5x\right)^2-3\left(x^2-5x\right)-18\)

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

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

\(=\left(x^2-5x\right)^2-3\left(x^2-5x\right)-18\\ =\left(x^2-5x\right)^2-6\left(x^2-5x\right)+3\left(x^2-5x\right)-18\\ =\left(x^2-5x\right)\left(x^2-5x-6\right)+3\left(x^2-5x-6\right)\\ =\left(x^2-5x+3\right)\left(x^2-5x-6\right)\\ =\left(x-6\right)\left(x+1\right)\left(x^2-5x+3\right)\)

\(=x^4-10x^3+25x^2-3x^2+15x-18=x^4-10x^3+22x^2+15x-18=x^4+x^3-11x^3-11x^2+33x^2+33x-18x-18=x^3\left(x+1\right)-11x^2\left(x+1\right)+33x\left(x+1\right)-18\left(x+1\right)=\left(x+1\right)\left(x^3-11x^2+33x-18\right)=\left(x+1\right)\left(x^3-6x^2-5x^2+30x+3x-18\right)=\left(x+1\right)\left[x^2\left(x-6\right)-5x\left(x-6\right)+3\left(x-6\right)\right]=\left(x+1\right)\left(x-6\right)\left(x^2-5x\right)=\left(x+1\right)\left(x-6\right)x\left(x-5\right)\)

\(4\left(x^2+15x+50\right)\left(x^2+18x+72\right)-3x^2\\ =4\left(x+5\right)\left(x+10\right)\left(x+6\right)\left(x+12\right)-3x^2\\ =4\left(x^2+16x+60\right)\left(x^2+17x+60\right)-3x^2\)

Đặt \(x^2+16x+60=a\)

\(=4a\left(a+x\right)-3x^2\\ =4a^2+4ax-3x^2\\ =\left(2a-x\right)\left(2a+3x\right)\\ =\left[2\left(x^2+16x+60\right)-x\right]\left[2\left(x^2+16x+60\right)+3x\right]\\ =\left(2x^2+31x+120\right)\left(2x^2+35x+120\right)\)

Đặt 

\(\left(x-5\right)\left(x-1\right)\left(x+3\right)\left(x+7\right)+60\)

\(=\left(x^2+2x-35\right)\left(x^2+2x-3\right)+60\)

\(=\left(x^2+2x\right)^2-38\left(x^2+2x\right)+105+60\)

\(=\left(x^2+2x\right)^2-3\left(x^2+2x\right)-35\left(x^2+2x\right)+165\)

\(=\left(x^2+2x-3\right)\left(x^2+2x-35\right)\)

\(=\left(x+3\right)\left(x-1\right)\left(x+7\right)\left(x-5\right)\)

Ta có: (x²+6x-5)(x²+6x+3)-20

      = [(x²+6x-1)-4][(x²+6x-1)+4]-20

      = (x²+6x-1)²-16-20

      = (x²+6x-1)²-36

      = (x²+6x-7)(x²+6x-5)

      = (x+7)(x-1)(x²+6x-5)

\(\left(x^2+6x-5\right)\left(x^2+6x+3\right)\\ =\left(x^2+6x-1\right)^2-16-20\\ =\left(x^2+6x-1\right)^2-36\\ =\left(x^2+6x-1-6\right)\left(x^2+6x-1+6\right)\\ =\left(x^2+6x-7\right)\left(x^2+6x+5\right)\\ =\left(x-1\right)\left(x+7\right)\left(x+1\right)\left(x+5\right)\)

\(\left(x^2+6x-5\right)\left(x^2+6x+3\right)-20\)

\(=\left(x^2+6x\right)^2-2\left(x^2+6x\right)-35\)

\(=\left(x^2+6x-7\right)\left(x^2+6x+5\right)\)

\(=\left(x+7\right)\left(x-1\right)\left(x+1\right)\left(x+5\right)\)