We could express any vector [tex]p\in P_2(\mathbb R)[/tex] as
[tex]p=ax^2+bx+c[/tex]
So any vector [tex]u\in U[/tex] would have coefficients [tex]a,b,c[/tex] that satisfy
[tex]16a+4b+c=0\implies c=-16a-4b[/tex]
from which we can show that any such vector is a linear combination of some other vectors:
[tex]u=ax^2+bx+(-16a-4b)=a(x^2-16)+b(x-4)[/tex]
More explicitly, we've shown that [tex]u[/tex] is a linear combination of the vectors [tex]x^2-16[/tex] and [tex]x-4[/tex]; in other words, [tex]u\in\mathrm{span}\{x^2-16,x-4\}[/tex], and in fact these two vectors form a basis for [tex]U[/tex]. But this set does not span all of [tex]P_2(\mathbb R)[/tex] because there's no combination of these two vectors that can be used to obtain a constant. We want the transformation to be usable for any vector in [tex]P_2(\mathbb R)[/tex], so we need to add an additional vector extend the basis. We can do this simply by appending [tex]1[/tex] into the spanning set. (Do check that the vectors remain linearly independent.)
Now we want the transformation to map those polynomials [tex]p(x)[/tex] for which [tex]p(4)=0[/tex] to the zero vector. We know which vectors belong to the basis of [tex]U[/tex], so we need
[tex]f(x^2-16)=0[/tex]
[tex]f(x-4)=0[/tex]
[tex]f(1)=1[/tex]
where the choice of the assignment for [tex]f(1)[/tex] is arbitrary, so long as it's non-zero.
