An algebra problem by Rajen Kapur
f ( x ) is a cubic polynomial such that f ( a ) = f ( b ) = f ( c ) = 0 and f ( p ) = 0 . Select the correct choice of the expression, c y c l i c a , b , c ∏ ( b − p a + p )
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1 solution
Perhaps we should specify that a , b , c are the roots of the polynomial?
IE, this would fail for f ( x ) = ( x − 1 ) 2 ( x − 2 ) where we set a = 1 , b = 2 , c = 2 .
Reply to Calvin Lin: Sir, It does not fail for any chosen a, b, and c, as long as p is different which has been stated explicitly in the statement of the question.
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Please take f ( x ) = ( x − 1 ) 2 ( x − 2 ) , a = 1 , b = 2 , c = 2 and p = 3 . Plug that into your equation. What do you get?
We have
∏ ( b − p a + p ) = 2 − 3 1 + 3 × 2 − 3 2 + 3 × 1 − 3 2 + 3 = − 5 0
while f ( 3 ) = 4 , f ( − 3 ) = − 8 0 , which gives f ( 3 ) f ( − 3 ) = 4 − 8 0 = − 2 0 .
Do you see where your solution breaks down? This is why the multi-set of roots is important. Simply saying that f ( a ) = f ( b ) = f ( c ) = 0 only gives us { a , b , c } ⊂ "roots of f ".
Reply to Calvin Lin: For the chosen a, b, and c f ( x ) = ( x − 1 ) ( x − 2 ) 2 . Please correct your expression f ( x ) , then everything will turn out fine.
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You are missing my point. In the problem, you stated that " f ( x ) is a cubic polynomial such that f ( a ) = f ( b ) = f ( c ) = 0 ". The expressions that I stated satisfy the written conditions that you gave.
Your written condition (a,b,c are roots of f(x) ) are not restrictive enough to result in the conditions that you are thinking of in your head (IE that a, b, c are exactly the multi set of roots of f(x) ). That is where your solution breaks down. You made this additional assumption, which isn't a corollary of the written conditions.
The given expression ( b − p a + p ) ( c − p b + p ) ( a − p c + p ) = ( p − a ) ( p − b ) ( p − c ) ( − p − a ) ( − p − b ) ( − p − c ) = f ( p ) f ( − p ) . bold text