Question

Phân tích đa thức thành nhân tử:

( x + 1)( x+ 2)( x + 3)( x + 4) - 24

Answer 1

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

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

. Đặt \(x^2+5x+4=t\) (1)

(*) <=> \(t\left(t+2\right)-24=t^2+2t-24=\left(t-4\right)\left(t+6\right)\) (2)

Thay (1) vào (2) ta suy ra : \(\left(x+1\right)\left(x+2\right)\left(x+3\right)\) \(\left(x+4\right)-24=\)\(\left(x^2+5x+4-4\right)\left(x^2+5x+4+6\right)\) = \(\left(x^2+5x\right)\left(x^2+5x+10\right)\) = \(x\left(x+5\right)\left(x^2+5x+10\right)\)

Answer 2

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

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

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

\(=\left(x^2+5x+4\right)^2+2.\left(x^2+5x+4\right)+1-25\)

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

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

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

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

Answer 3

Ta có 

<=>(x+1)(x+4)(x+2)(x+3)-24

<=>(X^2+5x4)(x^2+5x+6)-24

Đặt x^2+5x+5=x    (1)

Ta có 

<=>(x+1)(x-1)-24

<=>x^2-25

Thay 1 vào x  ta có

(x^2+5x+5)^2-5^2

<=>(x^2+10)(x^2+5x)(dpcm)