Question

Phân tích đa thức thành nhân tử

​a, x⁷+x²​+1

​b, x⁸​+x+1

​c, x⁸+x⁷​+1

d, x⁸​+3x⁴+1

​e, x¹⁰+x⁵​+1

​

Answer 1

a)

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

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

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

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

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

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

b)

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

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

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

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

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

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

Answer 2

c)

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

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

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

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

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

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

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

Answer 3

d) Biểu thức không phân tích được thành nhân tử.

e)

\(x^{10}+x^5+1=x^{10}-x+x^5-x^2+x^2+x+1\)

\(=x(x^9-1)+x^2(x^3-1)+(x^2+x+1)\)

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

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

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

\(=(x-1)(x^2+x+1)(x^7+x^4+x^2+x)+(x^2+x+1)\)

\(=(x^2+x+1)[(x-1)(x^7+x^4+x^2+x)+1]\)

\(=(x^2+x+1)(x^8-x^7+x^5-x^4+x^3-x+1)\)