Question 1038159
ax^3+bx^2+cx+d is the polynomial
-a+b-c+d=15
d=0, so one intercept is (0,0).  The reason d=0 is that f(0)=0.  When f(0) is applied to the function, all the x terms disappear, since they are all zero.  The only term left is d, the constant.  But d is 0, because f(0)=0.
a+b+c=-5
8a+4b+2c=12
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-a+b+c=15.  This comes from the first equation, removing d, since it is 0.  I can then add that and the one below it.
a+b+c=-5
2b+2c=10
b+c=5, dividing by 2.
a=-10
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-80+10+2b=12, because 2b+2c+10
2b=82
b=41
c=-36
-10x^3+41x^2-36x=0
{{{graph(300,200,-10,10,-20,20,-10x^3+41x^2-36x)}}}
factor out an -x
-x(10x^2-41x+36)=0
x=0
x=(1/20)(41 +/-sqrt (1681-1440); sqrt 241=15.52
x=56.52/20, or 2.83
x=25.48/20, or 1.274