解:设x^4-x^3+4x^2+3x+5=(x^2+ax+1)(x^2+bx+5)=x^4+(a+b)x^3+(ab+b)x^2+(5a+b)x+5 ∴a+b=-1,ab+b=4,5a+b=1 解得：a=1,b=-2 ∴x^4-x^3+4x^2+3x+5=(x^2+x+1)(x^2-2x+5). 设x^4+x^3-7x^2-x+6=(x^2+ax-1)(x^2+bx-6)=x4+(a+b)x^3+(ab-7)x^2-(6a+b)x+6 ∴a+b=1,ab-7=-7,6a+b=-1 解得：a=0,b=1 ∴x^4+x^3-7x^2-x+6=(x^2-1)(x^2+x-6)=(x+1)(x-1)(x+3)(x-2)
X^4+X^3-7X^2-X+6=X^3(X+1)-(7X^2+X-6) =X^3(X+1)-(X+1)(7X-6) =(X+1)(X^3-X-6X+6) =(X+1)(X(X^2-1)-6(X-1)) =(X+1)(X-1)(X^2+X-6) =(X+1)(X-1)(X-2)(X+3)
X^4+X^3-7X^2-X+6=X^3(X+1)-(7X^2+X-6)
=X^3(X+1)-(X+1)(7X-6)
=(X+1)(X^3-7X+6)
答：1.(x^2+4x+8)^2+3x(x^2+4x+8)+2x^2 =[(x^2+4x+8)+x][(x^2+4x+8)+2x] =(x^2+5x+8)(x^2+...详情>>
答：x^4+4=x^4+4x^2+4-4x^2=(x^2+2x+2)(x^2-2x+2) =(x+1+i)(x+1-i)(x-1+i)(x-1-i)详情>>
