证明:多项式x^2+px+1的根为m、n,多项式x^2+qx+1的根为g、h,则有(m-g)(n-g)(m+h)(n+h)=q^2-p^2.
问题描述:
证明:多项式x^2+px+1的根为m、n,多项式x^2+qx+1的根为g、h,则有(m-g)(n-g)(m+h)(n+h)=q^2-p^2.
答
mn=1 m+n=-p gh=1 g+h=-q(m-g)(n-g)(m+h)(n+h)=(mn-g(m+n)+g^2)(mn+h(m+n)+h^2)=(1-g(-p)+g^2)(1+h(-p)+h^2) =1-g(-p)+g^2+h(-p)-gh(-p)^2+g^2h(-p)+h^2-gh^2(-p)+g^2h^2=1-g(-p)+g^2+h(-p)- (p)^2+g(-p)+h^2-h(-p)+...