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)

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) \(3x^2-6xy=3x\left(x-2y\right)\)

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

c) \(=x\left(x-2y\right)-3\left(x-2y\right)=\left(x-2y\right)\left(x-3\right)\)

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

\(a,=3x\left(x-2y\right)\\ b,=x\left(x-3\right)^2\\ c,Sửa:x^2-2xy-3x+6y=x\left(x-2y\right)-3\left(x-2y\right)=\left(x-2y\right)\left(x-3\right)\\ d,=\left(3x-5\right)\left(2x-3\right)\)

a, `3x^2-6xy=3x(x-2y)`

b, `x^3-6x^2+9x=x(x^2-6x+9)=x(x-3)^2`

c, `x^2-2xy-3y+6y=x^2-2xy+3y`

d, `6x^2-19x+15=6x^2-9x-10x+15=3x(2x-3)-5(2x-3)=(3x-5)(2x-3)`

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

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

\(a,=-\left(x-1\right)^3\left[=\left(1-x\right)^3\right]\\ b,=\left(1-x\right)^3\)

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

b. \(=\left(1-x\right)^3\)

\(6x^2-19x+15=6x^2-9x-10x+15\)

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

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

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

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

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

\(5x^2-19x-4=5x^2-20x+x-4\)

\(=\left(5x^2-20x\right)+\left(x-4\right)\)

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

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

= 5x^2 + x - 20x - 4

= (5x^2 + x) - (20x + 4)

= x(5x+1) - 4 (5x + 1)

= (5x+1) (x - 4)

\(5x^2-19x-4\)

\(=5x^2-20x+\left(x-4\right)\)

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

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