\(x^3-2x^2-x\)

\(=x\cdot x^2-x\cdot2x-x\cdot1\)

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

=(4x+1)(3x+2)(12x-1)(x+1)-4

=(12x²+11x+2)(12x²+11x-1)-4

đặt a=12x²+11x+2

khi đó đa thức trở thành:

a(a-3)-4

=a²-3a-4

=a²+a-4a-4

=a(a+1)-4(a+1)

=(a+1)(a-4)

thay x vào là ok

\(x^3-4x^2-12x+27\)

\(=x^3+3x^2-7x^2-21x+9x+27\)

\(=x^2\left(x+3\right)-7x\left(x+3\right)+9\left(x+3\right)\)

\(=\left(x^2-7x+9\right)\left(x+3\right)\)

\(x^8+3x^4+4\)

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

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

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

\(4x^4+4x^3+5x^2+2x+1\)

\(=\left(4x^4+2x^3+2x^2\right)+\left(2x^3+x^2+x\right)+\left(2x^2+x+1\right)\)

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

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

nhóm x^4+3x^3 thành 1 nhóm, 9x-27 thành 1 nhóm pn iu

x⁴+3x³-9x-27

= x(x+3)-9(x+3)

= (x+3)(x-9)

a) ( 4x+1) (12x-1) (3x+2) (x+1) -4

=(4x+1)(3x+2)(12x-1)(x+1)-4

=(12x²+11x+2)(12x²+11x-1)-4

Đặt t=12x²+11x+2 ta được:

t.(t-3)-4

=t²-3t-4

=t²+t-4t-4

=t.(t+1)-4.(t+1)

=(t+1)(t-4)

thay t=12x²+11x+2 ta được:

(12x²+11x+3)(12x²+11x-2)

Vậy ( 4x+1) (12x-1) (3x+2) (x+1) -4=(12x²+11x+3)(12x²+11x-2)

b) (x²+2x)²+9x²+18x+20

=(x²+2x)²+9.(x²+2x)+20

Đặt y=x²+2x ta được:

y²+9y+20

=y²+4y+5y+20

=y.(y+4)+5.(y+4)

=(y+4)(y+5)

thay y=x²+2x ta được:

(x²+2x+4)(x²+2x+5)

Vậy (x²+2x)²+9x²+18x+20=(x²+2x+4)(x²+2x+5)

A=x⁴+3x³-9x-9

=x⁴-9+3x³-9x

=(x²-3)(x²+3)+3x.(x²-3)

=(x²-3)(x²+3+3x)

bạn phân tích chưa hết kìa

x⁴+x³+3x³+3x²-20x-20

=x³ (x+1)+3x² (x+1)-20(x+1)

=(x+1)(x³+3x²-20)