Question

\(\left(x^2+3x+1\right)\cdot\left(x^2+3x+2\right)-6\)

phân tích đa thức sau thành nhân tử

Answer 1

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

Đặt \(x^2+3x+1=a,\)ta được:

\(a\left(a+1\right)-6\)

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

\(=a\left(a+3\right)-2\left(a+3\right)=\left(a+3\right)\left(a-2\right)\)

Thay \(a=x^2+3x+1,\)ta được:

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