Let there be a cubic polynomial equation P ( x ) with one root equal to 5, and two other roots, whose average is also 5 . If P ( 8 ) = 2 0 1 6 , find P ( 2 ) .
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No fair, no Viete's! I even labelled it as symmetry...
Just kidding + 1 . short, nice solution.
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I'm using the symmetry f ( x ) = − f ( 1 0 − x ) ;)
We just have to prove that P ( 5 + x ) = − P ( 5 − x ) , and the answer comes naturally.
Since the average of the two other roots is 5 , then the two roots can be expressed as 5 + k , 5 − k for some number k .
So, P ( x ) = a ( x − 5 ) ( x − ( 5 + k ) ) ( x − ( 5 − k ) ) for some a , by the factorization of a polynomial.
So, P ( 5 + x ) = a ( ( x + 5 ) − 5 ) ( ( x + 5 ) − ( 5 + k ) ) ( ( x + 5 ) − ( 5 − k ) )
P ( 5 + x ) = a ( x ) ( x + k ) ( x − k )
Similarly, P ( 5 − x ) = a ( ( − x + 5 ) − 5 ) ( ( − x + 5 ) − ( 5 + k ) ) ( ( − x + 5 ) − ( 5 − k ) )
P ( 5 − x ) = a ( − x ) ( − x + k ) ( − x − k )
P ( 5 − x ) = − a ( x ) ( x + k ) ( x − k )
But, since P ( 5 + x ) = a ( x ) ( x + k ) ( x − k ) ,
P ( 5 − x ) = − ( a ( x ) ( x + k ) ( x − k ) )
P ( 5 + x ) = − P ( 5 − x )
Which is exactly what we wanted to prove. So,
P ( 8 ) = P ( 5 + 3 ) = − ( P ( 5 − 3 ) = − P ( 2 )
2 0 1 6 = − P ( 2 ) , P ( 2 ) = − 2 0 1 6 ,
Which is our final answer.
Right .... Same way ( + x → 0 lim ln ( x 1 + x ) )
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Wow.... Because + 1 is too mainstream XD
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Very true... ;-P.. I thought of trying something new.
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@Rishabh Jain – How did you find this problem...? It's not even in the "New" section yet....
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@Manuel Kahayon – But in Explore section problems appear right at the moment someone posts them.. :-)
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@Rishabh Jain – Ohhh...??? Where are those sections...?
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@Manuel Kahayon – On clicking continue after solving a problem we directly go to explore section where on clicking New all the recent new problems appear.
Easier solution; by using Vieta's formulae, the sum of the roots is 1 5 and the product let's call it d , so the polynomial becomes;
P ( x ) = x 3 − 1 5 x 2 + ( 5 0 + 5 d ) x − d
Substitute 8 and 2 0 1 6 and solve for d and the answer will come.
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By Viète, the other monic factor must be x 2 − 1 0 x + b , so f ( x ) = a ( x − 5 ) ( x 2 − 1 0 x + b ) . We are told that f ( 8 ) = 3 a ( b − 1 6 ) = 2 0 1 6 so f ( 2 ) = − 3 a ( b − 1 6 ) = − 2 0 1 6 .