Question

Phân tích đa thức thành nhân tử

1, \(x^5+x^4+1\)

2, \(x\left(x+4\right)\left(x+6\right)\left(x+10\right)+128\)

3, \(x^2-4xy+4y^2-2x+4y-35\)

4, \(x^4+6x^3+7x^2-6x+1\)

5, \(x^2\left(y-z\right)+y^2\left(z-x\right)+z^2\left(x-y\right)\)

6, \(x\left(y-z\right)^3+y\left(z-x\right)^3+x\left(x-y\right)^3\)

7, \(x^{10}+x^5+1\)

Answer 1

1 , \(x^5+x^4+1=\left(x^5+x^4+x^3\right)-\left(x^3+x^2+x\right)+\left(x^2+x+1\right)\)

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

2 , \(x\left(x+4\right)\left(x+6\right)\left(x+10\right)+128=\left(x^2+10x\right)\left(x^2+10x+24\right)+128\)(*)

Đặt x² + 10 = a , a>0 (1)

=> (*) <=> a(a+24)+128=a² + 24a+128=(a+8)(a+16) (**)

Thay (1) vào (**) ta được :

(*) <=> \(\left(x^2+10+8\right)\left(x^2+10+16\right)\)