Question 896428
One root of p is zero, so you can begin with {{{p(x)=x(ax^2+bx+c)}}}, to try to find a CUBIC polynomial.


The other three points let you form a system of three equations.


{{{p(x)=ax^3+bx^2+cx=p(x)}}}
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{{{-a+b-c=2}}}, using (-1,2).


{{{a+b+c=1}}}, using (1,1).


{{{a*4^3+b*4^2+c*4=58}}}


{{{64a+16b+4c=58}}}, using point (4,58).


Solve for the coefficients a, b, c in the system
{{{highlight_green(system(-a+b-c=2,a+b+c=1,64a+16b+4c=58))}}}


{{{(matrix(3,4,
-1,1,-1,2,
1,1,1,1,
32,8,2,29))}}}


R2+R1, and R3+32*R1,


{{{(matrix(3,4,
-1,1,-1,2,
0,2,0,3,
0,40,-30,93))}}}


R3-20*R2
and one small division of R3 by constant, scaler


{{{(matrix(3,4,
1,-1,1,-2,
0,2,0,3,
0,0,1,-11))}}}


Which indicates {{{highlight(c=-11)}}}.
Using that continuing without further matrix operations, 
{{{highlight(b=3/2)}}} and {{{highlight(a=29/2)}}}.


Putting this all into expected cubic p(x),
{{{highlight(p(x)=(29/2)x^3+(3/2)x^2-11x)}}}.
This was not checked but it should be checked.