\(x^3-6x^2+11x-6\)

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

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

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

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

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

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

Vì x-1;x-2;x-3 là ba số nguyên liên tiếp 

nên \(\left(x-1\right)\left(x-2\right)\left(x-3\right)⋮3!=6\)

=>\(x^3-6x^2+11x-6⋮6\)

a) x³ - 6x² + 11x - 6

= ( x³ - 2x² ) - ( 4x² - 8x ) + ( 3x - 6 )

= x²( x - 2 ) - 4x( x - 2 ) + 3( x - 2 )

= ( x - 2 )( x² - 4x + 3 )

= ( x - 2 )( x² - x - 3x + 3 )

= ( x - 2 )[ x( x - 1 ) - 3( x - 1 ) ]

= ( x - 2 )( x - 1 )( x - 3 )

b) x³ - 6x² - 9x + 14

= ( x³ - x² ) - ( 5x² - 5x ) - ( 14x - 14 )

= x²( x - 1 ) - 5x( x - 1 ) - 14( x - 1 )

= ( x - 1 )( x² - 5x - 14 )

= ( x - 1 )( x² + 2x - 7x - 14 )

= ( x - 1 )[ x( x + 2 ) - 7( x + 2 ) ]

= ( x - 1 )( x + 2 )( x - 7 )

c) x³ + 6x² + 11x + 6

= ( x³ + 2x² ) + ( 4x² + 8x ) + ( 3x + 6 )

= x²( x + 2 ) + 4x( x + 2 ) + 3( x + 2 )

= ( x + 2 )( x² + 4x + 3 )

= ( x + 2 )( x² + x + 3x + 3 )

= ( x + 2 )[ x( x + 1 ) + 3( x + 1 ) ]

= ( x + 2 )( x + 1 )( x + 3 )

e) x⁶ - 9x³ + 8

Đặt t = x³

bthuc <=> t² - 9t + 8 

            = t² - t - 8t + 8

            = t( t - 1 ) - 8( t - 1 )

            = ( t - 1 )( t - 8 )

            = ( x³ - 1 )( x³ - 8 )

            = ( x - 1 )( x² + x + 1 )( x - 2 )( x² + 2x + 4 )

b) nhẩm đưuọc nghiệm x=1

\(\Leftrightarrow\left(x-1\right)\left(x^2-5x+6\right)=0\Rightarrow\orbr{\begin{cases}x=1\\x^2-5x+6\left(2\right)\end{cases}}\)

\(\left(2\right)\Leftrightarrow\left(x-2\right)\left(x-3\right)\Rightarrow\orbr{\begin{cases}x=2\\x=3\end{cases}}\) KL x=1,2,3

c)

(x^2+3x+1)^2=x^4+9x^2+1+6x^3+2x^2+6x  (nhân pp dẽ hơn ghép)

\(\orbr{\begin{cases}x=\frac{3-\sqrt{5}}{2}\\x=\frac{3+\sqrt{5}}{2}\end{cases}}\)

Bài 1: 

a: \(x^3-6x^2+11x-6\)

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

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

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

b: \(x^3-6x^2-9x+14\)

\(=x^3-7x^2+x^2-7x-2x+14\)

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

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

c: \(x^3+6x^2+11x+6\)

\(=x^3+3x^2+3x^2+9x+2x+6\)

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

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

Phân tích đa thức thành nhân tử:

\(a,x^3-6x^2+11x-6\)

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

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

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

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

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

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

Phương trình nào?

\(PT\Leftrightarrow\left(x^3+6x^2+11x-2\right)^2+13\left(x^3+6x^2+11x-2\right)+40=0\\ \Leftrightarrow\left(x^3+6x^2+11x-2+5\right)\left(x^3+6x^2+11x-2+8\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x^3+6x^2+11x+3=0\left(\text{vô nghiệm}\right)\\x^3+6x^2+11x+6=0\end{matrix}\right.\\ \Leftrightarrow x^3+6x^2+11x+6=0\\ \Leftrightarrow x^3+x^2+5x^2+5x+6x+6=0\\ \Leftrightarrow\left(x+1\right)\left(x^2+5x+6\right)=0\\ \Leftrightarrow\left(x+1\right)\left(x^2+2x+3x+6\right)=0\\ \Leftrightarrow\left(x+1\right)\left(x+2\right)\left(x+3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=-1\\x=-2\\x=-3\end{matrix}\right.\)

`@` `\text {Ans}`

`\downarrow`

`4x^3 - 4x^2 - 9x + 9`

`= (4x^3 - 4x^2) - (9x - 9)`

`= 4x^2(x - 1) - 9(x - 1)`

`= (4x^2 - 9)(x - 1)`

____

`x^3 + 6x^2 + 11x + 6`

`= x^3 + x^2 + 5x^2 + 5x + 6x + 6`

`= (x^3 + x^2) + (5x^2 + 5x) + (6x + 6)`

`= x^2*(x + 1) + 5x(x + 1) + 6(x + 1)`

`= (x^2 + 5x + 6)(x+1)`

____

`x^2y - x^3 - 9y + 9x`

`= (x^2y - 9y) - (x^3 - 9x)`

`= y(x^2 - 9) - x(x^2 - 9)`

`= (y - x)(x^2 - 9)`

b: =x^3+x^2+5x^2+5x+6x+6

=(x+1)(x^2+5x+6)

=(x+1)(x+2)(x+3)

c: =x^2(y-x)-9(y-x)

=(y-x)(x^2-9)

=(y-x)(x-3)(x+3)

a: =(4x^3-4x^2)-(9x-9)

=4x^2(x-1)-9(x-1)

=(x-1)(4x^2-9)

=(x-1)(2x-3)(2x+3)