Polynomials all around.
Let there be a polynomial p ( x ) = x 3 − 3 x 2 + 4 x − 1 such that p ( a ) = p ( b ) = p ( c ) = 0 and a = b = c .
Find the value of ( 2 − a ) ( 2 − b ) ( 2 − c ) .
The answer is 3.
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2 solutions
Its okay , Cheers! xD
But why I get 29 instead? Actually, I thought that by vieta 's theorem, We shall get 8-4(-3)+2(4)-(-1) Which is 29
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Because sum of roots is 3 and not − 3 and product of roots is 1 and not − 1 .
Amazing solution!
Forget Vieta! Just plug and chug!
From Vieta's formula ,
a + b + c = 3 a b + a c + b c = 4 a b c = 1
The expression we want is
( 2 − a ) ( 2 − b ) ( 2 − c ) = ( 4 − 2 a − 2 b + a b ) ( 2 − c ) = 8 − 4 a − 4 b + 2 a b − 4 c + 2 a c + 2 b c − a b c = 8 − 4 ( a + b + c ) + 2 ( a b + a c + b c ) − a b c = 8 − 4 ( 3 ) + 2 ( 4 ) − 1 = 8 − 1 2 + 8 − 1 = 3
Since p ( x ) is a cubic polynomial and a,b and c are the zeroes of p ( x ) , We can write
p ( x ) = ( x − a ) ( x − b ) ( x − c )
If we replace x by 2, p ( x ) = ( 2 − a ) ( 2 − b ) ( 2 − c )
p ( 2 ) = ( 2 − a ) ( 2 − b ) ( 2 − c ) = 8 − 1 2 + 8 − 1 = 3
Answer= 3