Question

(x^2+3x+1)(x^2+3x+2)-6

phân tích đa thức thành nhân tử

Answer 1

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

Đặt \(\left(x^2+3x+1\right)=a\), 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 \(a=\left(x^2+3x+1\right)\), ta được:

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

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