Question

phân tích thành nhân tử

a, \(^{x^5}\)+x+1

b, \(^{x^5}\)+\(x^4\)+1

c, \(x^8\)+x+1

d, \(x^8\)+\(x^7\)+1

e, 4\(x^8\)+1

Answer 1

Lời giải:

a)

\(x^5+x+1=x^5-x^2+x^2+x+1\)

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

\(=(x^2+x+1)[x^2(x-1)+1]\)

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

b) \(x^5+x^4+1=x^5-x^2+x^4-x+x^2+x+1\)

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

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

\(=(x^2+x+1)[x^2(x-1)+x(x-1)+1]\)

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

Answer 2

c) \(x^8+x+1\)

\(=x^8-x^2+x^2+x+1\)

\(=x^2(x^6-1)+(x^2+x+1)\)

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

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

\(=(x^2+x+1)[x^2(x-1)(x^3+1)+1]\)

\(=(x^2+x+1)(x^6-x^5+x^3-x^2+1)\)

e) \(4x^8+1=(2x^4)^2+1=(2x^4)^2+1+2.2x^4-2.2x^4\)

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

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

Answer 3

d) \(x^8+x^7+1\)

\(=x^8-x^2+x^7-x+x^2+x+1\)

\(=x^2(x^6-1)+x(x^6-1)+(x^2+x+1)\)

\(=(x^6-1)(x^2+x)+(x^2+x+1)\)

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

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

\(=(x^2+x+1)[(x-1)(x^3+1)(x^2+x)+1]\)

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