PolynomiaL(s) Again

if a is a roo tto the first one, -a is also a root

so the roots of the above if real are : -3a, -a, a, 3a (for real roots to be in AP)

hence 9a⁴=m⁴ => 3a²=±m²

and -9a²-a²=-3m-2 =>10a²=3m+2

so m²-3m-2=0 so m = 3±√9+82 = 3±√172

or m²+3m+2=0 so m=-1 or -2

There is only one +ve root above...

2) Let P(x) be a Polynomial of degree 4 such tht -
P(1)=P(3)=P(5)=P'(7) =0.

If the real number x (≠1,3,5) is such tht P(x)=0 can be expressed as x=u/v (u & v being relatively prime), what is the value of (u+v) ?

P(x) = (x-1)(x-3)(x-5)(x-a)

P'(x) =(x-1)(x-3)(x-5)(x-a)[1/(x-1)+1/(x-3)+1/(x-5)+1/(x-a)] at x=7 , it is zero...

hence 0=1/6+1/4+1/2+1/(7-a)

so 1/(a-7) = 11/12

so a=8+1/11 = 89/11

From here, i did not understand what is meant by this last step.. if P(x) can be expressed as u/v.. because all rational numbers can be!