How to express $2x^3-x^2-3x+2$ as a linear combination of Legendre polynomials I have used the formula
\begin{align}p_0(x)&=1\\
p_1(x)&=x\\
p_2(x)&=\frac12(3x^2−1)\\
p_3(x)&=\frac12(5x^3−3x)
\end{align}
$$2x^3-x^2-3x+2=Ap_3(x)+Bp_2(x)+Cp_1(x)+Dp_0(x)$$
EDIT-
\begin{align}&=\frac A2(5x^3−3x) + \frac B2(3x^2−1) + Cx +D \end{align}
but don't know how to continue after this.

Continuing from your steps, we compare the coefficients of each term on the L.H.S and R.H.S.
$$2x^3-x^2-3x+2=\dfrac A2(5x^3−3x) + \dfrac B2(3x^2−1) + Cx +D$$
So, $$ 2x^3-x^2-3x+2=\dfrac{5}{2}Ax^3+\dfrac{3}{2}Bx^2+(C-\dfrac{3}{2}A)x+(D-\dfrac{B}{2})$$


*

*For the $x^3$ term, we have
$\dfrac{5}{2}A=2 \Rightarrow A=\dfrac{4}{5}$

*For the $x^2$ term, we have 
$\dfrac{3}{2}B=-1 \Rightarrow B=\dfrac{-2}{3}$

*For the $x$ term, we have 
$C-\dfrac{3}{2}A=-3 \Rightarrow C-\dfrac{3}{2}\cdot\dfrac{4}{5}=-3 \Rightarrow C-\dfrac{6}{5}=-3 \Rightarrow C=\dfrac{6}{5}-3\Rightarrow C=\dfrac{-9}{5}$

*For the constant term, we have
$D-\dfrac{1}{2}B=2 \Rightarrow D-\dfrac{1}{2} \cdot \dfrac{(-2)}{3}=2 \Rightarrow D=2-\dfrac{1}{3} \Rightarrow D=\dfrac{5}{3}$
