Problem: If a2â€‹î€ =0 and r and s are the roots of a0​+a1​x+a2​x2=0, then the equality a0​+a1​x+a2​x2=a0​(1−rx​)(1−sx​) holds:
Answer Choices:
A. for all values of x,a0â€‹î€ =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,a0â€‹î€ =0
Solution:
a2​x2+a1​x+a0​=a2​(x2+a2​a1​​x+a2​a0​​)=a2​(x2−(r+s)x+rs) =a2​(r−x)(s−x)=a2​rs(1−rx​)(1−sx​),rsî€ =0 Since rs=a2​a0​​ and rsî€ =0 and a2â€‹î€ =0, then a0â€‹î€ =0
∴a2​x2+a1​x+a0​=a2​(a2​a0​​)(1−rx​)(1−sx​)=a0​(1−rx​)(1−sx​) for all values of x provided a0â€‹î€ =0