Question

[N.T.M.D]

Phân tích đa thức thành nhân tử:

(x^2-3x)^2+(2x^2-6x)-24

Answer 1

[l҉o҉n҉g҉ d҉z҉]

( x² - 3x )² + ( 2x² - 6x ) - 24

= ( x² - 3x )² + 2( x² - 3x ) - 24 (*)

Đặt t = x² - 3x 

(*) trở thành :

t² + 2t - 24

= t² - 4t + 6t - 24

= t( t - 4 ) + 6( t - 4 )

= ( t - 4 )( t + 6 )

= ( x² - 3x - 4 )( x² - 3x + 6 )

= ( x² + x - 4x - 4 )( x² - 3x + 6 )

= [ x( x + 1 ) - 4( x + 1 ) ]( x² - 3x + 6 )

= ( x + 1 )( x - 4 )( x² - 3x + 6 )

Answer 2

[FL.Han_]

\(\left(x^2-3x\right)^2+\left(2x^2-6x\right)-24\)

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

Đặt \(a=x^2-3x\)

(1)=\(a^2+2a-24\)

\(=a^2-4a+6a-24\)

\(=a\left(a-4\right)+6\left(a-4\right)\)

\(=\left(a-4\right)\left(a+6\right)\)

\(=\left(x^2-3x-4\right)\left(x^2-3x+6\right)\)

\(=\left(x^2-4x+x-4\right)\left(x^2-3x+6\right)\)

\(=\left[x\left(x-4\right)+\left(x-4\right)\right]\left(x^2-3x+6\right)\)

\(=\left(x+1\right)\left(x-4\right)\left(x^2-3x+6\right)\)