Question

Phân tích đa thức thành nhân tử:

a, x⁴-4x³-8x²+8x

b, x³+x²-4x-4

bạn nào giúp mk vs

 

Answer 1

Hướng dẫn thôi :

a) x ( x + 2 ) ( x^2 - 6x + 4 )

b) ( x + 1 ) ( x + 2 ) ( x - 2 )

Answer 2

cách làm cơ

Answer 3

b) x^3 + x^2 - 4x - 4

= x^2 ( x + 1 ) - 4 ( x + 1 )

= ( x + 1 ) ( x^2 - 2^2 )

= ( x + 1 ) ( x + 2 ) ( x - 2)

Answer 4

\(x^4-4x^3-8x^2+8x\)

\(=x\left(x^3-4x^2-8x+8\right)\)

\(=x.\left[\left(x^3+2x^2\right)-\left(6x^2+12x\right)+\left(4x+8\right)\right]\)

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

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

Tham khảo nhé~

Answer 5

\(x^3+x^2-4x-4\)

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

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

\(=\left(x+1\right)\left(x-2\right)\left(x+2\right)\)