Question

Phân tích đa thức thành nhân tử:

a) (x-1)(x-2)(x-3)(x-4)+1

b) (x²+3x+2)(x²+7x+12)+1

Answer 1

a, \(\left(x-1\right)\left(x-2\right)\left(x-3\right)\left(x-4\right)+1\)

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

\(=\left(x^2-5x+4\right)\left(x^2-5x+6\right)+1\)

Đặt \(x^2-5x+5=y\) ta được:

\(\left(y-1\right)\left(y+1\right)+1=y^2-1+1=y^2\)

Thay \(y=x^2-5x+5\) ta được:

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

Answer 2

b, \(\left(x^2+3x+2\right)\left(x^2+7x+12\right)+1\)

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

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

\(=\left(x^2+5x+4\right)\left(x^2+5x+6\right)+1\)

Đặt \(t=x^2+5x+5\) ta có:

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

Thay \(t=x^2+5x+5\) ta được:

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