Problem: If a2βξ =0 and r and s are the roots of a0β+a1βx+a2βx2=0, then the equality a0β+a1βx+a2βx2=a0β(1βrxβ)(1βsxβ) holds:
Answer Choices:
A. for all values of x,a0βξ =0
B. for all values of x
C. only when x=0
D. only when x=r or x=s
E. only when x=r or x=s,a0βξ =0
Solution:
a2βx2+a1βx+a0β=a2β(x2+a2βa1ββx+a2βa0ββ)=a2β(x2β(r+s)x+rs) =a2β(rβx)(sβx)=a2βrs(1βrxβ)(1βsxβ),rsξ =0 Since rs=a2βa0ββ and rsξ =0 and a2βξ =0, then a0βξ =0
β΄a2βx2+a1βx+a0β=a2β(a2βa0ββ)(1βrxβ)(1βsxβ)=a0β(1βrxβ)(1βsxβ) for all values of x provided a0βξ =0