Question

Giúp mình phân tích đa thức này thành nhân tử: x.(x+1).(x+2).(x+3)+1

Answer 1

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

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

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

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

gọi \(\left(x^2+3x\right)=a\)

\(\Rightarrow\left(t+2\right)t+1\)

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

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

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