Question

Phân tích đa thức sau thành nhân tử : x(x+1)(x+2)(x+3)+1

Answer 1

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

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

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

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

Answer 2

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

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

=(x²+3x)(x²+3x+2)+1

Đặt x²+3x+1=y, ta có: 

(y-1)(y+1)+1

=y²-1+1

=y²

Thay y=x²+3x+1, lại có: 

(x²+3x+1)²