Answer
See below
Work Step by Step
a) Obtain: $Ker (T)=\{p \in P_2(R): T(p)=0\}\\
=\{ax^2+bx+c:ax^2+(a+2b+c)x+(3a-2b-c)=0\}$
We have the system:
$a=0\\
a+2b+c=0\\
3a-2b+c=0$
thus, $a=0\\
c=-2b$
Thus, $Ker (T)=\{+bx-2b:b\in R\}\\
=\{b(x-2):b\in R\}$
$\dim [Ker(T)]=1$
b) Then, $Rng(T)= \{T(p):p \in P_2(R)\}\\
=\{ax^2+x(a+2b+c)+(3a-2b-c):a,b,c \in R\}\\
=\{ a(x^2+x+3)+2b(x-1)+c(x-1): a,b,c \in R\}\\
=\{a(x^2+x+3)+(2b+c)(x-1): a,b,c \in R\}\\
=span \{(x^2+x+3);(x-1)\}\\
\rightarrow \dim [Rng(T)]=2$
