Question

Bài 2 Phân tích đa thức sau thành nhân tử 

a. x4 + 2x3 − 4x − 4 

b. x2(1 − x2) − 4 − 4x2 

c. x2 + y2 − x2y2 + xy − x − y 

d* a3 + b3 + c3 − 3abc 

 

Answer 1

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

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

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

 

Answer 2

a) Ta có: \(x^4+2x^3-4x-4\)

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

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

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

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

Answer 3

d) Ta có: \(a^3+b^3+c^3-3abc\)

\(=\left(a+b\right)^3-3ab\left(a+b\right)+c^3-3abc\)

\(=\left(a+b+c\right)\left[\left(a+b\right)^2-c\left(a+b\right)+c^2\right]-3ab\left(a+b+c\right)\)

\(=\left(a+b+c\right)\left(a^2+2ab+b^2-ac-bc+c^2-3ab\right)\)

\(=\left(a+b+c\right)\left(a^2+b^2+c^2-ab-ac-bc\right)\)

Answer 4

c) Ta có: \(x^2+y^2-x^2y^2+xy-x-y\)

\(=\left(x^2-x\right)+\left(y^2-y\right)+\left(-x^2y^2+xy\right)\)

\(=x\left(x-1\right)+y\left(y-1\right)+xy\left(-xy+1\right)\)

\(=\left(x+y+xy\right)\left(x-1+y-1+xy+1\right)\)