Question

Phân tích đa thức thành nhân tử (x+1)(x+2)(x+3)(x+4) – 24

Answer 1

Ta có : \(\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)-24\)

\(=\left(\left(x+1\right)\left(x+4\right)\right)\left(\left(x+2\right)\left(x+3\right)\right)-24\)

\(=\left(x^2+5x+4\right)\left(x^2+5x+6\right)-24\)

- Đặt \(x^2+5x+5=a\)

\(=\left(a-1\right)\left(a+1\right)-24=a^2-1-24=a^2-25\)

\(=\left(a-5\right)\left(a+5\right)\)

Answer 2

Ta có: \(\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)-24\)

\(=\left(x^2+5x+4\right)\left(x^2+5x+6\right)-24\)

\(=\left(x^2+5x\right)^2+10\left(x^2+5x\right)+24-24\)

\(=\left(x^2+5x\right)^2+10\left(x^2+5x\right)\)

\(=\left(x^2+5x\right)\left(x^2+5x+10\right)\)

\(=x\left(x+5\right)\left(x^2+5x+10\right)\)