Question 27388
 If f, g be two polynomials in Q, then

  T(f+g)(x) = (x + 1)(f+g)(x) − {{{(f+g)(1)x^2}}}
            = (x + 1)(f(x)+g(x)) − {{{(f(1)+g(1))x^2 }}}
            = (x+1) f(x) - {{{f(1) x^2 }}}+ (x+1)g(x) - {{{g(1) x^2}}}
            = T(f)(x) + T(g)(x)
            = (T(f)+T(g))(x),
 Hence, T(f+g) = T(f)+T(g). 
and for a in Q,
 T(af)(x) = (x + 1)(af)(x) - {{{(af)(1)x^2}}}
          = a(x+1) f(x) - {{{af(1)x^2}}}
          = a((x+1)f(x) - {{{f(1)x^2}}})
          = a T(f(x))
  Hence, T(af) = aT(f).

 This shows T is linear from Q[x]=V to Q[x].
 (Or check T(af+bg)(x) = a T(f)(x) + b Tg(x) directly.

 Kenny
 

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