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

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

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

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

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

c) \(x^3-6x^2+11x-6=x^3-x^2-5x^2+5x+6x-6=x^2\left(x-1\right)-5x\left(x-1\right)+6\left(x-1\right)=\left(x-1\right)\left(x-2\right)\left(x-3\right)\)

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

b)x³-19x-30=x³+2x²-2x²-4x-15x-30=x²(x+2)-2x(x+2)-15(x+2)=(x²-2x-15)(x+2)=(x²+3x-5x-15)(x+2)=[x(x+3)-5(x+3)](x+2)=(x-5)(x+3)(x+2)=(x-5)(x+2)(x+3)

c)x³-6x²+11x-6=x³-x²-5x²+5x+6x-6=x²(x-1)-5x(x-1)+6(x-1)=(x²-5x+6)(x-1)=(x²+x-6x+6)(x-1)=[x(x+1)-6(x+1)](x-1)=(x-6)(x+1)(x-1)=(x-6)(x-1)(x+1)

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

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

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

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

x³+6x²-11x+6=(x+1)(x+2)(x+3)

\(x^3+6x^2+11x+6=x^3+x^2+5x^2+5x+6x+6\)

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

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

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

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

Trần Vương Quốc Đạt,bn giải rõ ra đi ,như thế kia thì mk cx lm đc

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)