Question 1209702
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Suppose P(x) is a polynomial of smallest possible degree such that:
* P(x) has rational coefficients.
* P(-2) = P(sqrt{5}) = P(sqrt{7}) = P(sqrt{17}) = 0.
* P(-1) = 13.
Determine the value of P(0).
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<pre>
In this problem, a polynomial P(x) with rational coefficients has the roots
 
    -2, {{{sqrt(5)}}},  {{{sqrt(7)}}},  {{{sqrt(17)}}}.


that are quadratic irrationalities.


To make a polynomial with rational coefficients, which has these roots, we shall
associate the conjugate number over Q with every quadratic irrationality.


So, P(x) has the form

    P(x) = {{{a*(x+2)*(x-sqrt(5))*(x+sqrt(5))*(x-sqrt(7))*(x+sqrt(7))*(x-sqrt(17))*(x+sqrt(17))}}} = 

         = {{{a*(x+2)*(x^2-5)*(x^2-7)*(x^2-17)}}}.


Here "a" is the coefficient, now unknown.  We will determine it from the condition P(-1) = 13.
It gives

           {{{a*((-1)+2)*((-1)^2-5)*((-1)^2-7)*((-1)^2-17)}}} = 13,  

or

           a*1*(-4)*(-6)*(-16) = 13,

           a*(-384) = 13,

           a = {{{-13/384}}}.


Now the polynomial is fully defined  P(x) = {{{(-13/384)*(x+2)*(x^2-5)*(x^2-7)*(x^2-17)}}}.


To find P(0), plug in 0 instead of x:  

    P(0) = {{{(-13/384)*2(-5)*(-7)*(-17)}}} = {{{(13*2*5*7*17)/384)}}} = {{{15470/384}}} = {{{7735/192}}}.


<U>ANSWER</U>.  P(x) = {{{(-13/384)*(x+2)*(x^2-5)*(x^2-7)*(x^2-17)}}}  and  P(0) = {{{7735/192}}}.
</pre>

Solved.