Find $T(1)$, $T(x)$ and $T(x^{2})$ and $T(ax^2+bx+c)$ Let $$T:P_{3}\rightarrow P_{3}$$ be a linear transformation such that 
$$T(2x^{2})=2x^{2}+3x, T(\frac{1}{2}x+2)=2x^{2}+4x-3, T(2x^{2}-1)=3x-1.$$
Find $$T(1)$$, $$T(x)$$ and $$T(x^{2})$$
and $$T(ax^{2}+bx+c)$$.
I will be completely honest and say that I have no idea where to even begin with this problem.
Thank you for the help!
 A: Here's one way:
given that
$T(2x^{2})=2x^{2}+3x, T(\frac{1}{2}x+2)=2x^{2}+4x-3, T(2x^{2}-1)=3x-1, \tag{0}$
start with $T(2x^2)$.  We have, by linearity,
$T(2x^2) = 2T(x^2), \tag{1}$
and since we are given
$T(2x^2) = 2x^2 + 3x, \tag{2}$
we see by (1) that
$2T(x^2) = 2x^2 + 3x, \tag{3}$
or
$T(x^2) = x^2 + \dfrac{3}{2}x; \tag{4}$
that covers $T(x^2)$; next, we use
$T(2x^{2}-1)=3x-1, \tag{5}$
together with
$T(2x^2 - 1) = 2T(x^2) -T(1), \tag{6}$
which again follows from the linearity of $T$, to obtain
$2T(x^2) - T(1) = 3x - 1, \tag{7}$
and if we insert the value of $T(x^2)$ from (4) we see that
$2(x^2 + \dfrac{3}{2}x) - T(1) = 3x -1, \tag{8}$
which may be solved for $T(1)$:
$T(1) = 2x^2 + 1. \tag{9}$
Finally, we employ
$T(\dfrac{1}{2}x+2)=2x^{2}+4x-3 \tag{10}$
and the linearity of $T$ once again to see that
$\dfrac{1}{2}T(x) + 2T(1) = 2x^{2}+4x-3; \tag{11}$
substituting $T(1)$ from (9) yields
$\dfrac{1}{2}T(x) + 2(2x^2 + 1) = 2x^2+4x-3, \tag{12}$
and some algebraic fiddling around with (12) further yields
$T(x) = -4x^2 + 8x - 10. \tag{13}$
We have now found $T(1)$, $T(x)$ and $T(x^2)$; finding $T(ax^2 + bx + c)$ is simply one more application of linearity:
$T(ax^2 + bx + c) = aT(x^2) + bT(x) + cT(1)$
$= a(x^2 + \dfrac{3}{2}x) + b(-4x^2 + 8x - 10) + c(2x^2 + 1)$
$=(a - 4b + 2c)x^2 + (\dfrac{3}{2}a + 8b)x + (c - 10b).  \tag{14}$
The above shows how it can be done in this specific case, in which it is easy to successively isolate $T(1)$, $T(x)$, $T(x^2)$ due to the form of the givens in (0); in the more general case, of which this is but a specific instance, a general linear system in the "variables" $T(1)$, $T(x)$, $T(x^2)$ could be set up and solved using standard linear algebraic techniques.  It boils down to a system of three variables in three unkowns.
Hope this helps.  Cheerio,
and as always,
Fiat Lux!!!
A: Here's a setup for you to expand on. $T$ maps polynomials to polynomials and is linear. That means if $p,q$ are two polynomials, then $T(ap+bq)=aT(p)+bT(q)$, where $a,b$ are any constants. You're given $T$'s action on three polynomials: $2x$, $x/2+2$ and $2x^2-1$. Any extra information that you want to deduce from $T$, such as $T(1)$, you need to recreate $1$ in terms of the three polynomials above. In other words, find constants $a,b,c$ such that $a2x+b(x/2+2)+c(2x^2-1)=1$. Once you find those constants, then $T(1)=aT(2x)+bT(x/2+2)+cT(2x^2-1)$, all of which you know how to evaluate. The same thing goes for $T(x^2)$. 
A: Hint: You are told how to find $T$ at three particular values. Since $T$ is linear, see if you can write the other four expressions as various linear combinations of the ones you are given. Linearity then means you can write the answers as the same linear combinations of the given values of $T$. Understand?
For example: if you knew what $T(5x-2)$ and $T(3)$ were, you could figure out what $T(5x)$ was by writing $5x = (5x-2) + \frac23\cdot(3)$; then $T(5x)=\underbrace{T(5x-2)}_{\textrm{known}}+\frac23\cdot\underbrace{T(3)}_{\textrm{known}}$.
Now apply this reasoning to your specific information.
