Toán Lớp 8: phân tích đa thức thành n tử `h,` `x^10 +x^5 +1` `i,` `x^5 -x^4 -1`

Question

Toán Lớp 8:  phân tích đa thức thành n tử h, x^10 +x^5 +1 i,  x^5 -x^4 -1

catlantuong · Answer

Giải đáp+Lời giải và giải thích chi tiết:    h, x^10 + x^5 + 1   = x^10 - x + x^5 - x&sup2; + x&sup2; + x + 1   = (x^10 - x) + (x^5-x&sup2;) + (x&sup2; + x +1)   = x(x^9-1) + x&sup2;(x&sup3; - 1) + (x&sup2; + x + 1)   = x[(x&sup3;)&sup3; - 1&sup3;] + x&sup2; (x&sup3; - 1&sup3;) + (x&sup2; + x + 1)   = x(x&sup3;-1)(x^6 + x&sup3; + 1) + x&sup2;(x-1)(x&sup2; + x + 1) + (x&sup2; + x  + 1)   = x(x-1)(x&sup2;+x+1)(x^6 + x&sup3; + 1) + x&sup2;(x-1)(x&sup2;+x+1) + (x&sup2;+x+1)   = (x&sup2;+x+1) [x(x-1)(x^6 + x&sup3; + 1) + x&sup2;(x-1) + 1]   = (x&sup2; + x  + 1) ( x^5 + x^8 +  x&sup2; - x^7 - x^4 - x + x&sup3; - x&sup2; + 1 )   = (x&sup2; + x + 1) ( x^8 - x^7 + x^5 - x^4 + x&sup3; - x + 1)   i) x^5 - x^4 - 1   = x^5 - x&sup3; - x&sup2; -  x^4 + x&sup2; + x + x&sup3; - x - 1   = (x^5 - x&sup3; - x&sup2;) - (x^4 - x&sup2; - x) + (x&sup3; - x - 1)   = x&sup2;(x&sup3; - x - 1) - x(x&sup3; - x - 1) + (x&sup3; - x - 1)   = (x&sup3; - x - 1)(x&sup2; - x  +1)

hienchaudiem · Answer

h) 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^3-1)(x^6+x^3+1)+x^2(x-1)(x^2+x+1)+(x^2+x+1)   =(x^7+x^4+x)(x-1)(x^2+x+1)+(x^3-x^2)(x^2+x+1)+(x^2+x+1)   =(x^2+x+1)[(x^7+x^4+x)(x-1)+x^3-x^2+1]   =(x^2+x+1)(x^8-x^7+x^5-x^4+x^2-x+x^3-x^2+1)   =(x^2+x+1)(x^8-x^7+x^5-x^4+x^3-x+1)    i) 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^2-x^2-x-1)   =(x^2+x+1)(x^3-x-1)