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

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

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

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

\(\left(x^2-3x\right)^2-14x^2+42x+40\\ =\left(x^2-3x-7\right)^2-9\\ =\left(x^2-3x-10\right)\left(x^2-3x-4\right)\)

\(\left(x^2-3x\right)^2-14x^2+42x+40\)

\(=\left(x^2-3x-4\right)\left(x^2-3x-10\right)\)

\(=\left(x-4\right)\left(x+1\right)\left(x-5\right)\left(x+2\right)\)

\(4\left(x^2y^2+z^2t^2+2xyzt\right)-\left(x^2+y^2-z^2-t^2\right)^2\)

\(=\left(2xy-2tz\right)^2-\left(x^2+y^2-z^2-t^2\right)\)

\(=\left(2xy-2tz-x^2-y^2+z^2+t^2\right)\left(2xy-2tz+x^2+y^2-z^2-t^2\right)\)

\(=\left[-\left(x-y\right)^2+\left(z-t\right)^2\right]\left[\left(x+y\right)^2-\left(t+z\right)^2\right]\)

\(=-\left(x-y-z+t\right)\left(x-y+z-t\right)\left(x+y-t-z\right)\left(x+y+t+z\right)\)

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

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

\(=\left(x^2+4x+8\right)\left(x^2+6x+8\right)+x\left(x^2+6x+8\right)\)

\(=\left(x^2+4x+8+x\right)\left(x^2+6x+8\right)=\left(x^2+5x+8\right)\left(x^2+6x+8\right)\)

Ta có: \(\left(x^2+4x+8\right)^2+3x\left(x^2+4x+8\right)+2x^2\)

\(=\left(x^2+5x+8\right)\left(x^2+6x+8\right)\)

\(=\left(x^2+5x+8\right)\left(x+2\right)\left(x+4\right)\)

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

\(3x^3+6x+3-3y^2=3\left[\left(x^2+2x+1\right)-y^2\right]=3\left[\left(x+1\right)^2-y^2\right]=3\left(x-y+1\right)\left(x+y+1\right)\)

`16x^2z^2+y^2-z^2-16x^2y^2`

`=16x^2(z^2-y^2)+(y^2-z^2)`

`=16x^2(z-y)(y+z)+(y-z)(y+z)`

`=(y+z)[16x^2(z-y)+y-z]`

`=(y+z)(16x^2z-16x^2y+y-z)`

\(16x^2z^2+y^2-z^2-16x^2y^2\\ =16x^2\left(z^2-y^2\right)-\left(z^2-y^2\right)\\ =\left(z^2-y^2\right)\left(16x^2-1\right)\\ =\left(z-y\right)\left(z+y\right)\left(4x+1\right)\left(4x-1\right)\)

\(x^2-2x-24\)

\(=x^2-6x+4x-24\)

\(=x(x-6)+4(x-6)\)

\(=(x+4)(x-6)\)

\(x^2-2x-24\\ =x^2-2x+1-25\\ =\left(x-1\right)^2-5^2\\ =\left(x-1-5\right)\left(x-1+5\right)\\ =\left(x-6\right)\left(x+4\right)\)

\(x^2-2x-24=\left(x-6\right)\left(x+4\right)\)

-x² - 5x + 24

= -x² + 3x - 8x + 24

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

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

=(3x-x²)+(24-8x)=3x(1-x)+8(1-x)=(1-x)(3x+8)

\(-x^2-5x+24\)

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

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

(1 + x²)² - 4x(1 - x²)

= (1 + x²)(1 + x²) - 4x(1 - x²)

= (1 + x² - 4x)(1 + x² - 1 + x²)

= 2x²(x² - 4x + 1)

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

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

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

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