Question

A polynomial of degree 14 takes the value zero at each of the first 7 odd primes and also at their reciprocals. Find the value of the ratio \[\dfrac{{P(2)}}{{P\left( {\dfrac{1}{2}} \right)}}\].
A) ${4^7}$
B) $ - {4^7}$
C) ${2^7}$
D) 0

Answer 1

Hint: First we will write all the factors of the function using the roots given to us then multiply them all to a scalar and get the function $P(x)$. Now, we will find the value of \[\dfrac{{P(x)}}{{P\left( {\dfrac{1}{x}} \right)}}\] and then put in the value $x = 2$ in the resultant to get the required answer.

Complete step-by-step answer:
Let us first write the first 7 odd prime numbers.
Prime numbers are those numbers which do not have any other factor other than 1 and themselves only.
Hence, the first 7 odd primes are 3, 5, 7, 11, 13, 17, 19.
We eliminated 1 because 1 is neither a prime nor a unit and 2 is an even prime, therefore, we have not written that as well.
If $P(x)$ is our original polynomial, therefore, $(x - 3)(x - 5)(x - 7)(x - 11)(x - 13)(x - 17)(x - 19)$ is a factor of $P(x)$. ……….(1)
Now, their reciprocal are also the roots of the polynomial $P(x)$.
Therefore, $\left( {x - \dfrac{1}{3}} \right)\left( {x - \dfrac{1}{5}} \right)\left( {x - \dfrac{1}{7}} \right)\left( {x - \dfrac{1}{{11}}} \right)\left( {x - \dfrac{1}{{13}}} \right)\left( {x - \dfrac{1}{{17}}} \right)\left( {x - \dfrac{1}{{19}}} \right)$ is also a factor of $P(x)$…...(2)
Combining (1) and (2), we will have:-
$(x - 3)(x - 5)(x - 7)(x - 11)(x - 13)(x - 17)(x - 19)\left( {x - \dfrac{1}{3}} \right)\left( {x - \dfrac{1}{5}} \right)\left( {x - \dfrac{1}{7}} \right)\left( {x - \dfrac{1}{{11}}} \right)\left( {x - \dfrac{1}{{13}}} \right)\left( {x - \dfrac{1}{{17}}} \right)\left( {x - \dfrac{1}{{19}}} \right)$ is a factor of $P(x)$.
Now these are 14 factors all together. If we multiply, we will get a polynomial of 14 degrees.
Hence, $P(x)$ must be equal to $\lambda (x - 3)(x - 5)(x - 7)(x - 11)(x - 13)(x - 17)(x - 19)\left( {x - \dfrac{1}{3}} \right)\left( {x - \dfrac{1}{5}} \right)\left( {x - \dfrac{1}{7}} \right)\left( {x - \dfrac{1}{{11}}} \right)\left( {x - \dfrac{1}{{13}}} \right)\left( {x - \dfrac{1}{{17}}} \right)\left( {x - \dfrac{1}{{19}}} \right)$…….(3)
Rewriting it after simplifying, we will get $P(x)$ equal to:-
$\lambda (x - 3)(x - 5)(x - 7)(x - 11)(x - 13)(x - 17)(x - 19)\left( {\dfrac{{3x - 1}}{3}} \right)\left( {\dfrac{{5x - 1}}{5}} \right)\left( {\dfrac{{7x - 1}}{7}} \right)\left( {\dfrac{{11x - 1}}{{11}}} \right)\left( {\dfrac{{13x - 1}}{{13}}} \right)\left( {\dfrac{{17x - 1}}{{17}}} \right)\left( {\dfrac{{19x - 1}}{{19}}} \right)$……….(4)
Now, let us replace $x$ with $\dfrac{1}{x}$ in (3), we will get:-
Hence, $P\left( {\dfrac{1}{x}} \right)$ must be equal to $\lambda \left( {\dfrac{1}{x} - 3} \right)\left( {\dfrac{1}{x} - 5} \right)\left( {\dfrac{1}{x} - 7} \right)\left( {\dfrac{1}{x} - 11} \right)\left( {\dfrac{1}{x} - 13} \right)\left( {\dfrac{1}{x} - 17} \right)\left( {\dfrac{1}{x} - 19} \right)\left( {\dfrac{1}{x} - \dfrac{1}{3}} \right)\left( {\dfrac{1}{x} - \dfrac{1}{5}} \right)\left( {\dfrac{1}{x} - \dfrac{1}{7}} \right)\left( {\dfrac{1}{x} - \dfrac{1}{{11}}} \right)\left( {\dfrac{1}{x} - \dfrac{1}{{13}}} \right)\left( {\dfrac{1}{x} - \dfrac{1}{{17}}} \right)\left( {\dfrac{1}{x} - \dfrac{1}{{19}}} \right)$Rewriting it after simplifying, we will get $P\left( {\dfrac{1}{x}} \right)$ equal to:-
$\lambda \left( {\dfrac{{1 - 3x}}{x}} \right)\left( {\dfrac{{1 - 5x}}{x}} \right)\left( {\dfrac{{1 - 7x}}{x}} \right)\left( {\dfrac{{1 - 11x}}{x}} \right)\left( {\dfrac{{1 - 13x}}{x}} \right)\left( {\dfrac{{1 - 17x}}{x}} \right)\left( {\dfrac{{1 - 19x}}{x}} \right)\left( {\dfrac{{3 - x}}{{3x}}} \right)\left( {\dfrac{{5 - x}}{{5x}}} \right)\left( {\dfrac{{7 - x}}{{7x}}} \right)\left( {\dfrac{{11 - x}}{{11x}}} \right)\left( {\dfrac{{13 - x}}{{13x}}} \right)\left( {\dfrac{{17 - x}}{{17x}}} \right)\left( {\dfrac{{19 - x}}{{19x}}} \right)$Rewriting it to get $P\left( {\dfrac{1}{x}} \right)$ equal to:-
$\lambda \left( {\dfrac{{3x - 1}}{x}} \right)\left( {\dfrac{{5x - 1}}{x}} \right)\left( {\dfrac{{7x - 1}}{x}} \right)\left( {\dfrac{{11x - 1}}{x}} \right)\left( {\dfrac{{13x - 1}}{x}} \right)\left( {\dfrac{{17x - 1}}{x}} \right)\left( {\dfrac{{19x - 1}}{x}} \right)\left( {\dfrac{{x - 3}}{{3x}}} \right)\left( {\dfrac{{x - 5}}{{5x}}} \right)\left( {\dfrac{{x - 7}}{{7x}}} \right)\left( {\dfrac{{x - 11}}{{11x}}} \right)\left( {\dfrac{{x - 13}}{{13x}}} \right)\left( {\dfrac{{x - 17}}{{17x}}} \right)\left( {\dfrac{{x - 19}}{{19x}}} \right)$…………….(6)
Now, using (4) and (5), we will get \[\dfrac{{P(x)}}{{P\left( {\dfrac{1}{x}} \right)}}\] equal to:
$\dfrac{{\lambda (x - 3)(x - 5)(x - 7)(x - 11)(x - 13)(x - 17)(x - 19)\left( {\dfrac{{3x - 1}}{3}} \right)\left( {\dfrac{{5x - 1}}{5}} \right)\left( {\dfrac{{7x - 1}}{7}} \right)\left( {\dfrac{{11x - 1}}{{11}}} \right)\left( {\dfrac{{13x - 1}}{{13}}} \right)\left( {\dfrac{{17x - 1}}{{17}}} \right)\left( {\dfrac{{19x - 1}}{{19}}} \right)}}{{\lambda \left( {\dfrac{{3x - 1}}{x}} \right)\left( {\dfrac{{5x - 1}}{x}} \right)\left( {\dfrac{{7x - 1}}{x}} \right)\left( {\dfrac{{11x - 1}}{x}} \right)\left( {\dfrac{{13x - 1}}{x}} \right)\left( {\dfrac{{17x - 1}}{x}} \right)\left( {\dfrac{{19x - 1}}{x}} \right)\left( {\dfrac{{x - 3}}{{3x}}} \right)\left( {\dfrac{{x - 5}}{{5x}}} \right)\left( {\dfrac{{x - 7}}{{7x}}} \right)\left( {\dfrac{{x - 11}}{{11x}}} \right)\left( {\dfrac{{x - 13}}{{13x}}} \right)\left( {\dfrac{{x - 17}}{{17x}}} \right)\left( {\dfrac{{x - 19}}{{19x}}} \right)}}$
Simplifying it, we will get \[\dfrac{{P(x)}}{{P\left( {\dfrac{1}{x}} \right)}}\] equal to:-
\[\dfrac{{\left( {\dfrac{1}{3}} \right)\left( {\dfrac{1}{5}} \right)\left( {\dfrac{1}{7}} \right)\left( {\dfrac{1}{{11}}} \right)\left( {\dfrac{1}{{13}}} \right)\left( {\dfrac{1}{{17}}} \right)\left( {\dfrac{1}{{19}}} \right)}}{{\left( {\dfrac{1}{x}} \right)\left( {\dfrac{1}{x}} \right)\left( {\dfrac{1}{x}} \right)\left( {\dfrac{1}{x}} \right)\left( {\dfrac{1}{x}} \right)\left( {\dfrac{1}{x}} \right)\left( {\dfrac{1}{x}} \right)\left( {\dfrac{1}{{3x}}} \right)\left( {\dfrac{1}{{5x}}} \right)\left( {\dfrac{1}{{7x}}} \right)\left( {\dfrac{1}{{11x}}} \right)\left( {\dfrac{1}{{13x}}} \right)\left( {\dfrac{1}{{17x}}} \right)\left( {\dfrac{1}{{19x}}} \right)}}\]
Simplifying it further, we will get \[\dfrac{{P(x)}}{{P\left( {\dfrac{1}{x}} \right)}}\] equal to:
\[\dfrac{1}{{\left( {\dfrac{1}{x}} \right)\left( {\dfrac{1}{x}} \right)\left( {\dfrac{1}{x}} \right)\left( {\dfrac{1}{x}} \right)\left( {\dfrac{1}{x}} \right)\left( {\dfrac{1}{x}} \right)\left( {\dfrac{1}{x}} \right)\left( {\dfrac{1}{x}} \right)\left( {\dfrac{1}{x}} \right)\left( {\dfrac{1}{x}} \right)\left( {\dfrac{1}{x}} \right)\left( {\dfrac{1}{x}} \right)\left( {\dfrac{1}{x}} \right)\left( {\dfrac{1}{x}} \right)}}\]
Simplifying it further, we will get \[\dfrac{{P(x)}}{{P\left( {\dfrac{1}{x}} \right)}}\] equal to \[\dfrac{1}{{{{\left( {\dfrac{1}{x}} \right)}^{14}}}}\] which is equal to ${x^{14}}$. ………..(7)
Now, we need \[\dfrac{{P(2)}}{{P\left( {\dfrac{1}{2}} \right)}}\]. Therefore, putting $x = 2$ in (7), we will get:-
\[\dfrac{{P(2)}}{{P\left( {\dfrac{1}{2}} \right)}} = {2^{14}} = {\left( {{2^2}} \right)^7} = {4^7}\].

Hence, the correct option is (A).

Note: The students must keep in mind the fact that they cannot write the polynomial without assuming any scalar quantity like here, we assumed $\lambda $ to be the coefficient of the whole polynomial because we are not given the direct polynomial.