Too little information!
A monic quintic polynomial function f ( x ) with real coefficients leaves no remainder if and only if divided by ( x − 3 ) , ( x − 5 ) , ( x + 2 ) , ( x − 4 ) . Find the sum of all possible values of f ( 0 ) .
The answer is 1200.
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1 solution
Also, does anybody know what this theorem/lemma/something is?
If x 1 , x 2 , . . . , x n are the roots of f ( x ) , then for some function g ( x ) , then g ( x 1 ) , g ( x 2 ) , . . . , g ( x n ) are the roots of f ( g − 1 ( x ) ) .
Nice problem. The wording isn't quite right though. f ( x ) also leaves a remainder when divided by a quadratic factor.
Since f ( x ) is divisible by ( x − 3 ) , ( x − 5 ) , ( x + 2 ) , ( x − 4 ) , then f ( x ) must be of the form a ( x − 3 ) ( x − 5 ) ( x + 2 ) ( x − 4 ) ( x − r ) for some r . Also, since f ( x ) is monic, then a must be equal to 1.
Also, if r is not equal to 3 , 5 , − 2 , 4 , then the polynomial will become divisible by some other factor other than those given, so, clearly, r must be equal to 3 , 5 , − 2 or 4 .
So, f ( 0 ) = ( 0 − 3 ) ( 0 − 5 ) ( 0 + 2 ) ( 0 − 4 ) ( 0 − r ) = 1 2 0 r . Since r can be equal to 3 , 5 , − 2 or 4 , we substitute these to get
f ( 0 ) = 1 2 0 r
f ( 0 ) = ( 3 6 0 , 6 0 0 , 4 8 0 , − 2 4 0 ) Adding these values together, we get
3 6 0 + 6 0 0 + 4 8 0 − 2 4 0 = 1 2 0 0