We have the linear map $f:\mathbb{R}_2[x]\rightarrow\mathbb{R}_2[x]$ with $$f(p(x))=10p(1)+5p(x)+3\lambda xp''(x)+4p'(x)$$ where $\mathbb{R}_2[x]=\{ax^2+bx+c:a,b,c\in \mathbb{R}\}$ is the vector space of polynomials of degree $\leq 2$ with real coefficients.
How can we choose $\lambda$ so that $f$ is diagonisable?
If we write this in matrix form we have $$ f(p(x))=\begin{pmatrix}10 & 0 & 0 & 0 \\ 0 & 5 & 0 & 0 \\ 0& 0 & 3x\lambda & 0\\ 0 & 0 & 0 & 4\end{pmatrix}\begin{pmatrix}p(1) \\ p(x) \\ p''(x) \\ p'(x)\end{pmatrix}$$ Is this the correct matrix form?
Now we have to check when the matrix is diagonisable, or not?
The standard basis of $\mathbb R_2[x]$ is $\{1,x,x^2\}$ so that
$f(1)=15=15(1)+0(x)+0(x^2)$
$f(x)=14(1)+5(x)+0(x^2)$
$f(x^2)=10+(6\lambda+8) x+5x^2$
The matrix corresponding to $f$ is $T=\begin{pmatrix}15& 14& 10\\0& 5& 6\lambda+8\\ 0& 0& 5\end{pmatrix}$
For the repeated eigenvalue $t=5$, you need $AM=GM$ for which $Rank(T-5I)=1$ is required. This suggests $\lambda=-4/3$.