Solution :
T :
22
is given by, T(p) = [ p(1)+p(-1) p(2) ]t
q
ker(T) and q
0.
Thus,T(q) = [0 0]t and hence, [q(1) + q(-1)
q(2)]t = [0 0]t
Thus, q(1)+q(-1) = 0 and q(2) = 0.
Since q
2,
hence there exist a,b,c in
such that q(x) = ax2 + bx + c.
Now, q(1)+q(-1) = 0 implies that a+b+c+a-b+c = 0 and hence, a =
-c
Also, q(2) = 0 implies that 4a + 2b + c = 0. Using the relation a =
-c, we obtain 3a = -2b.
Thus, we have 3a = -3c = -2b
Thus, q(x) = ax2 + bx + c = ax2 - 3ax/2 - a
.
Hence, the roots of q are given by, ax2 - 3ax/2 - a = 0
which implies, 2x2 - 3x - 2 = 0. Thus, the roots
of q are -1/2 and 2.
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