Bài 6 : Phân tích các đa thức sau thành nhân tử :
a, ( x^2 + x )^2 - 14 ( x^2 + x ) + 24
b, ( x^2 + x )^2 + 4x^2 + 4x - 12
c, x^4 + 2x^3 + 5x^2 + 4x - 12
d, ( x + 1 ) ( x + 2 ) ( x+ 3 ) ( x + 4 ) + 1
e, ( x + 1 ) ( x + 3 ) ( x + 5 ) ( x + 7 ) + 15
f, ( x + 1 ) ( x + 2 ) ( x + 3 ) ( x + 4 ) - 24
Giúp mk vs ạ mk đang cần gấp ạ
a, ( x2 + x )2 - 14 ( x2 + x ) + 24
= (x2 + x)2 - 2(x2 + x) -12(x2 + x) + 24
= (x2 + x).(x2 + x -2) - 12(x2 + x -2)
= (x2 + x -2).(x2 + x -12)
= (x2 + 2x - x - 2).(x2 + 4x - 3x - 12)
=[x.(x+2)-(x+2)].[x.(x+4)-3(x+4)]
= (x+2).(x-1).(x+4).(x-3)
= x4 + 2x3 - 13x2 - 14x + 24
b, ( x2 + x )2 + 4x2 + 4x - 12
= x4 + 2x3 + x2 + 4x2 + 4x -12
= x4 + 2x3 + 5x2 + 4x -12
c, x4 + 2x3 + 5x2 + 4x - 12
= x4 - x3 + 3x3 - 3x2 + 8x2 - 8x +12x -12
= x3(x-1) + 3x2(x-1) + 8x(x-1) + 12(x-1)
= (x-1) . (x3 + 3x2 + 8x +12)
= (x-1) . ( x3 +2x2 + x2 + 2x + 6x +12)
= (x-1). [x2(x+2) + x(x+2) + 6(x+2)]
= (x-1).(x+2).(x2 + x+ 6)
