đặt y=x²+1

=>y²=(x²+1)²

y²=x⁴+2x²+1

đặt P(x)=x^4+6x^3+11x^2+6x+1

=x⁴+2x²+1+6x³+6x+9x²

=x⁴+2x+1+6x(x²+1)+9x²

thay y²=x⁴+2x²+1 và y=x²+1 ta được 

Q(y)=y²+6xy+9x²

=(y+3x)²

thay y=x²+1 ta được:

(x²+3x+1)²

vậy x^4+6x^3+11x^2+6x+1=(x²+3x+1)²

x³+6x²+11x+6 = x³+6x²+12x-x+8-2 = (x³+6x²+12x+8) - (x+2)                                                                                                                                   = (x+2)³ - (x+2) = (x+2)[(x+2)² - 1] = (x+2)(x+2-1)(x+2+1) = (x+2)(x+1)(x+3)

noob

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&_&

\(=x^3-6x^2+12x-8-x+2\)

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

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

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

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

Phân tích đa thức thành  nhân tử :

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

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

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

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

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

_Chúc bạn học tốt_

\(x^3+6x^2+11x+6=\left(x+1\right)\left(x+2\right)\left(x+3\right)\)

x³+6x²-11x+6=(x+1)(x+2)(x+3)

x^3 + 6x^2 + 11x + 6 
= x^3 + x^2 + 5x^2 + 5x + 6x + 6 
= x^2(x + 1) + 5x(x + 1) + 6(x + 1) 
= (x + 1)(x^2 + 5x + 6) 
= (x + 1)(x^2 + 2x + 3x + 6) 
= (x + 1)[x(x + 2) + 3(x + 2) 
= (x + 1)(x + 2)(x + 3) 

6x³ - 11x² - x - 2

= 6x³ - 12x² + x² - 2x + x - 2

= ( 6x³ - 12x² ) + ( x² - 2x ) + ( x - 2 )

= 6x²( x - 2 ) + x( x - 2 ) + 1( x - 2 )

= ( x - 2 )( 6x² + x + 1 )

          Bài làm :

Ta có :

6x³ - 11x² - x - 2

= 6x³ - 12x² + x² - 2x + x - 2

= ( 6x³ - 12x² ) + ( x² - 2x ) + ( x - 2 )

= 6x²( x - 2 ) + x( x - 2 ) + 1( x - 2 )

= ( x - 2 )( 6x² + x + 1 )