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

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 

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

Đa thức này không phân tích được thành nhân tử.

Bạn coi lại đề.

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

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

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

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

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

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

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

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

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

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

\(x^2\left(x+4\right)^2-\left(x+4\right)^2-\left(x^2-1\right)\\ =\left(x+4\right)^2\left(x^2-1\right)-\left(x^2-1\right)\\ =\left(x^2-1\right)\left[\left(x+4\right)^2-1\right]\\ =\left(x-1\right)\left(x+1\right)\left(x+4-1\right)\left(x+4+1\right)\\ =\left(x-1\right)\left(x+1\right)\left(x+3\right)\left(x+5\right)\)

\(= (x+4)^2(x^2-1)-(x^2-1)=[(x+4)^2-1](x^2-1)\)

\(=(x+4-1)(x+4+1)(x-1)(x+1)\)

\(=(x+3)(x+5)(x-1)(x+1)\)

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

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

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

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

\(a,\)\(x^3-13x-12\)

\(=x^3-x-12x-12\)

\(=x\left(x^2-1\right)-12\left(x+1\right)\)

\(=x\left(x-1\right)\left(x+1\right)-12\left(x+1\right)\)

\(=\left(x+1\right)\left(x^2-x-12\right)\)

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

\(=\left(x+1\right)\left[x\left(x-4\right)+3\left(x+4\right)\right]\)

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

a) \(x^3-13x-12\)

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

\(=x^2\left(x+1\right)-x\left(x+1\right)-12\left(x+1\right)\)

\(=\left(x+1\right)\left(x^2-x-12\right)\)

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

\(=\left(x+1\right)\left[x\left(x-4\right)+3\left(x-4\right)\right]\)

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

b) \(2x^4+3x^3-9x^2-3x+2\)câu này hình như sai đề rồi, bạn xem lại nhen

c) \(x^4-3x^3-6x^2+3x+1\)câu này cx thế, bạn xem lại nha

b) \(2x^4+3x^3-9x^2-3x+2\)

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

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

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

x²-2x-15=(x²-5x)+(3x-15)=x(x-5)+3(x-5)=(x-5)(x+3)

\(x^2-3x-15=\left(x^2-2.\dfrac{3}{2}x+\dfrac{9}{4}\right)-\dfrac{69}{4}=\left(x-\dfrac{3}{2}\right)^2-\left(\dfrac{\sqrt{69}}{2}\right)^2\)

\(=\left(x-\dfrac{3}{2}-\dfrac{\sqrt{69}}{2}\right)\left(x-\dfrac{3}{2}+\dfrac{\sqrt{69}}{2}\right)\)

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

\(\left(x^2+5x-3\right)\left(x^2+5x-5\right)-15=\left(x^2+5x-3\right)\left(x^2+5x-3-2\right)-15=\left(x^2+5x-3\right)^2-2\left(x^2+5x-3\right)+1-16=\left(x^2+5x-3-1\right)^2-4^2=\left(x^2+5x-4\right)^2-4^2=\left(x^2+5x-8\right)\left(x^2+5x\right)=x\left(x+5\right)\left(x^2+5x-8\right)\)

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

\(=\left(x^2+5x\right)^2-8\left(x^2+5x\right)-15\)

\(=x\left(x+5\right)\left(x^2+5x-8\right)\)

b) \(a^6-b^3\)

\(=\left(a^2\right)^3-b^3\)

\(=\left(a^2-b\right)\left(a^4+a^2b+b^2\right)\)

c) \(x^4-1\)

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

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

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