x^3-19x-30 
=x^3-25x+6x-30 
=x(x^2-25)+6(x-5) 
=x(x+5)(x-5)+6(x-5) 
=(x-5)(x^2+5x+6) 
=(x-5)(x^2+2x+3x+6) 
=(x-5)[x(x+2)+3(x+2)] 
=(x-5)(x+2)(x+3)

\(x^3-19x-30=x^3+2x^2-2x^2-4x-15x-30\)

\(\Rightarrow x^2\left(x+2\right)-2x\left(x+2\right)-15\left(x+2\right)\)

\(\Rightarrow\left(x+2\right)\left(x^2-2x-15\right)\)

\(\Rightarrow\left(x+2\right)\left(x+3\right)\left(x-5\right)\)

x^3 + 3x^2 - 3x^2 - 9x - 10x - 30

= x^2 ( x+3) - 3x ( x +3) -10(x+3)

=(x^2 -3x - 10)(x+3) 

=(x^2 - 5x + 2x -10)(x+3)

 =[x(x-5)+2(x-5)](x+3)

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

\(\text{x3 - 19x - 30}=x^3+2x^2-4x-15x-30\)

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

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

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

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

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

\(x^3-19x+30\)

\(=x^3-9x-10x+30\)

\(=x\left(x^2-9\right)-10\left(x-3\right)\)

\(=x\left(x-3\right)\left(x+3\right)-10\left(x-3\right)\)

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

\(=\left(x-3\right)\left(x^2-2x+5x-10\right)\)

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

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

#)Giải :

\(x^3-19x-30\)

\(=x^3-5x^2+5x^2-25x+6x-30\)

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

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

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

\(a,2x+2y-x^2-xy=2x+2y-\left(x^2+xy\right).\)

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

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

\(a,2x+2y-x^2-xy=2.\left(x+y\right)-x.\left(x+y\right)=\left(2-x\right).\left(x+y\right)\)

\(b,4x^3+9x^2-19x-30=3x^3+x^3+9x^2-9x-10x-30\)

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

Ta có: \(x^2-19x-30=\frac{4x^2-76x-120}{4}\)

                                          \(=\frac{1}{4}.\left[\left(4x^2-76x+361\right)-481\right]\)

                                          \(=\frac{1}{4}.\left[\left(2x-19\right)^2-481\right]\)

                                          \(=\frac{1}{4}.\left(2x-19-\sqrt{481}\right).\left(2x-19+\sqrt{481}\right)\)

Nghiệm xấu nên phân tích khó :) Sửa thành x³ - 19x - 30 cho dễ

x³ - 19x - 30

= x³ + 3x² - 3x² - 9x - 10x - 30

= ( x³ + 3x² ) - ( 3x² + 9x ) - ( 10x + 30 )

= x²( x + 3 ) - 3x( x + 3 ) - 10( x + 3 )

= ( x + 3 )( x² - 3x - 10 )

= ( x + 3 )( x² + 2x - 5x - 10 )

= ( x + 3 )[ x( x + 2 ) - 5( x + 2 ) ]

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

\(x^2-19x-30=\left(x-\frac{19+\sqrt{481}}{2}\right)\left(x-\frac{19-\sqrt{481}}{2}\right)\)

\(a,x^2-5=x^2-\left(\sqrt{5}\right)^2=\left(x-\sqrt{5}\right)\left(x+\sqrt{5}\right)\)

\(b,x^4+x^3+x+1=x^3.\left(x+1\right)+\left(x+1\right)\)

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

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

\(c,x^3-19x-30=x^3-25x+6x-30\)

\(=x.\left(x^2-25\right)+6.\left(x-5\right)\)

\(=x.\left(x-5\right)\left(x+5\right)+6.\left(x-5\right)\)

\(=\left(x-5\right).\left[x\left(x+5\right)+6\right]\)

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

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

\(=\left(x-5\right)\left[x.\left(x+2\right)+3.\left(x+2\right)\right]\)

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

a) 4x*(x+y)*(x+y+z)*(x+z)+y^2+z^2

=4*x*y*z^2+4*x^2*z^2+z^2+4*x*y^2*z+12*x^2*y*z+8*x^3*z+4*x^2*y^2+y^2+8*x^3*y+4*x^4

b) x^3-19x-30

=(x-5)*(x+2)*(x+3)

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

\(x^3-19x-30=x^3+6x-25x-30=x\left(x^2-25\right)+6x-30=x\left(x^2-25\right)+6\left(x-5\right)\)

\(=x\left(x-5\right)\left(x+5\right)+6\left(x-5\right)=\left(x-5\right)\left[\left(x\right)\left(x+5\right)+6\right]\)