Question

phân tích đa thức sau thành nhân tử:

(x²+x+1)(x²+x+2)-12

 

Answer 1

(x²+x+1)(x²+x+2)-12

=\(\left(x^2+x+\frac{3}{2}-\frac{1}{2}\right)\left(x^2+x+\frac{3}{2}+\frac{1}{2}\right)-12\)
=\(\left(x^2+x+\frac{3}{2}\right)^2-\frac{1}{4}-12\)
=\(\left(x^2+x+\frac{3}{2}\right)^2-\frac{49}{4}\)
=\(\left(x^2+x+\frac{3}{2}-\frac{7}{2}\right)\left(x^2+x+\frac{3}{2}+\frac{7}{2}\right)\)

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

Answer 2

\(A=\left(x^2+x+1\right)\left(x^2+x+2\right)-12\)

Đặt:   \(x^2+x+1=t\)    Khi đó ta có:

\(A=t\left(t+1\right)-12\)

\(=t^2+t-12=\left(t-3\right)\left(t+4\right)\)

Thay trở lại đc:

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

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