Why is this not a linear transformation from P2 to P3? $L(p(x)) = x^2 + p(x)$

I haven't gotten any homework problems that aren't a linear transformation, so I found a problem where this isn't the case. I know the two rules needed to prove it is a linear transformation, and I tried them here:

$L(a(p(x)) = ax^2 + ap(x)$ = $a(x^2 + p(x))$ = $aL(p(x))$

$L(p(x) + q(x))$ = $x^2 + (p(x) + q(x))$ = $x^2 + p(x) + q(x)$

$L(p(x)) + L(q(x))$ = $2x^2 + p(x) + q(x)$, so it is not equal to the above expression and therefore it is not a linear transformation.

Did I work this out right?
The equation $L(ap(x))=ax^2+ap(x)$ isn't right.  To compute $L(ap(x))$, you look at the formula for $L(p(x))$ and plug in "$ap(x)$" for "$p(x)$", so you get $L(ap(x))=x^2+ap(x)$.  Your computation for addition is correct, however.

To show that $L$ is not linear, you need to show that it is not true that $L(p(x)+q(x))=L(p(x))+L(q(x))$ for all $p$ and $q$ (or the corresponding statement for scalar multiplication).   To show this, you need to actually produce a counterexample.  That is, can you write down any specific polynomials $p(x)$ and $q(x)$ such that $L(p(x)+q(x))\neq L(p(x))+L(q(x))$?  (This shouldn't be hard from what you've done so far!)
In a linear function (transformation) the standard projections  are of the 1st degree...here you have a quadratic !