If a_2≠0 and r, s are the roots of a_0+a_1x+a_2x^2=0, then the equality a_0+a_1x+a_2x^2 &=a_0(1- xr) & ×(1- xs) holds:

(A) for all values of x, a_0≠0

(E) only when x=r or x=s, a_0≠0

1965 AMC 12 Problem 22

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22.

a_2\ne0

and

r,

s

are the roots of

a_0+a_1x+a_2x^2=0,

then the equality

\begin{aligned} a_0+a_1x+a_2x^2 &=a_0\left(1-\frac xr\right)\\ &\quad\cdot\left(1-\frac xs\right) \end{aligned}

holds:

for all values of

x,

a_0\ne0

for all values of

x

only when

x=0

only when

x=r

x=s

only when

x=r

x=s,

a_0\ne0

The right side requires both roots to be nonzero

Use

rs=\frac{a_0}{a_2}

to compare the leading coefficients

a_0\ne0,

then

rs=\frac{a_0}{a_2}\ne0

and

\begin{gathered} a_0\left(1-\frac xr\right) \left(1-\frac xs\right)\\ =\frac{a_0}{rs}(x-r)(x-s)\\ =a_2(x-r)(x-s), \end{gathered}

which is the original polynomial for every

x.

a_0=0,

at least one denominator on the right is zero.

Therefore, the correct answer is A.