Answer
See below
Work Step by Step
a) Obtain: $Ker (T)=\{p \in P_2(R): T(p)=0\}\\
=\{ax+bc:(b-a)+x(2b-3a)+bx^2=0\}$
We have the system:
$b-a=0\\
2b-3a=0\\
b=0$
thus, $a=b=0$
Thus, $Ker (T)=\{0x+0:x\in R\}\\
=\{0\}$
$\dim [Ker(T)]=0$
Apply Rank Nullity Theorem:
$Rng(T)=\{T(p): p \in P_2(R)\}\\
=\{(b-a)+x(2b-3a)+bx^2: a,b \in R\}\\
=\{b-a+2bx-3ax+bx^2: a,b \in R\}\\
=\{-a(1+3x)+b(1+2x+x^2):a,b \in R\}\\
=span\{1+3x,1+2x+x^2\}$
Since $\{1+3x,1+2x+x^2\}$ is the basic for $Rng(T)$,
then $\dim [Rng(T)]=2$
