Question

Phân tích đa thức thành nhân tử:

( x + 2 ) ( x + 4 ) ( x + 6 ) ( x + 8 ) + 16

 

Answer 1

Ta có: (x+2)(x+4)(x+6)(x+8)+16

=[(x+2)(x+8)]+[(x+4)(x+6)]+16

\(=\left[x^2+10x+16\right]\left[x^2+10x+24\right]+16\) (1)

Đặt \(x^2+10x+16=t\), khi đó (1) trở thành:

\(t\left(t+8\right)+16=t^2+8t+16=\left(t+4\right)^2\)

Thay \(x^2+10x+16=t\), ta có: \(\left(x^2+10x+16+4\right)^2=\left(x^2+10x+20\right)^2\)

Có gì đó sai sai á nhờ :vv?

Answer 2

( x + 2 )( x + 4 )( x + 6 )( x + 8 ) + 16

= [ ( x + 2 )( x + 8 ) ][ ( x + 4 )( x + 6 ) ] + 16

= ( x² + 10x + 16 )( x² + 10x + 24 ) + 16 (*)

Đặt t = x² + 10x + 20 

(*) <=> ( t - 4 )( t + 4 ) + 16

      = t² - 16 + 16

      = t² = ( x² + 10x + 20 )²

Answer 3

Đặt \(A=\left(x+2\right)\left(x+4\right)\left(x+6\right)\left(x+8\right)+16\)

\(\Rightarrow A=\left(x+2\right)\left(x+8\right)\left(x+4\right)\left(x+6\right)+16\)

\(=\left(x^2+10x+16\right)\left(x^2+10x+24\right)+16\)

Đặt \(x^2+10x+20=t\)

\(\Rightarrow A=\left(t-4\right)\left(t+4\right)+16=t^2-16+16=t^2\)

\(=\left(x^2+10x+20\right)^2\)