P ( x ) is a polynomial with integral coefficients such that the absolute value of the constant term of P ( x ) is smaller than 1000. Given further that P ( 1 9 ) = P ( 9 4 ) = 1 9 9 4 , find the constant term of the polynomial P ( x ) .
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But how can u consider p(x) to be a quadratic function. There can be a million functions which satisfy the above conditions. For example 1: P(x)= (x-19)(x-94)(x-1) + 1994 and 2: P(x)=(x-19)(x-94)(x-1)(x-2) + 1994
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You're right. But if P ( x ) = h ( x ) ( x − 1 9 ) ( x − 9 4 ) + 1 9 9 4 , the constant term of P ( x ) will be 1 7 8 6 h ( 0 ) + 1 9 9 4 , so h ( 0 ) = − 1 is forced and the constant term of P ( x ) must be 2 0 8 . But yes, h ( x ) can be any polynomial whose constant term is − 1 .
Let the constant term be 'a'. Since P(19)=1994 ,so (1994 - a) is divisible by 19. P(94)=a,so (1994 - a) is divisible by 94. (1994 - a ) is divisible by 19×94 = 1786. Since |a| is less than 1000, (1994 -a) must lie between 994 and 2994. So (1994 -a) = 1786 and a=208.
p(x)=18 modulo 19 and p(x)=20 modulo 98,gives p(x)=208 modulo 1786.Now,|constant term|<1000 implies 208 is indeed the answer.
the polynomial is k(x-19)(x-94) + 1994 where k is integer.
Next constant term of polynomial is
k(19)94
Also k(19)94<1000
It seems answer is 208 that is k=-1. But i have no idea why k = -1 OR A/C k<= - 1 should work.I believe the question also has the condition constant term>0
But how can u consider p(x) to be a quadratic function. There can be a million functions which satisfy the above conditions. For example 1: P(x)= (x-19)(x-94)(x-1) + 1994 and 2: P(x)=(x-19)(x-94)(x-1)(x-2) + 1994
the questions says, absolute value of constant term is less than 1000. The constant term is k(19)(94)+1994, not just k(19)(94)
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P ( x ) = a ( x − 1 9 ) ( x − 9 4 ) + 1 9 9 4
If x = 0 , then a ( − 1 9 ) ( − 9 4 ) + 1 9 9 4 = 1 7 8 6 a + 1 9 9 4
For constant term of absolute value smaller than 1000, then a must be − 1 .
Therefore, 1786(-1) + 1994 = 208.