Question 1036832
{{{f(x)=-2x^3+ax^2+bx+c}}}
So,
{{{f(2)=0}}}
{{{-2(2)^3+a(2)^2+2b+c=0}}}
{{{-16+4a+2b+c=0}}}
1.{{{4a+2b+c=16}}}
Similarly,
{{{f(1)=1}}}
{{{-2(1)^3+a(1)^2+b+c=1}}}
{{{-2+a+b+c=1}}}
2.{{{a+b+c=3}}}
and
{{{f(-2)=-2}}}
{{{-2(-2)^3+a(-2)^2-2b+c=-2}}}
{{{16+4a-2b+c=-2}}}
3.{{{4a-2b+c=-18}}}
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Subtract 2 from 1,
{{{4a+2b+c-a-b-c=16-3}}}
4.{{{3a+b=13}}}
Subtracting 2 from 3,
{{{4a-2b+c-a-b-c=-18-2}}}
{{{3a-3b=-21}}}
5.{{{a-b=-7}}}
Adding 4 and 5,
{{{3a+b+a-b=13-7}}}
{{{4a=6}}}
{{{a=3/2}}}
Then,
{{{3(3/2)+b=13}}}
{{{9/2+b=26/2}}}
{{{b=17/2}}}
And,
{{{3/2+17/2+c=3}}}
{{{10+c=3}}}
{{{c=3-10}}}
{{{c=-7}}}
So,
{{{f(x)=-2x^3+(3/2)x^2+(17/2)x-7}}}
and 
{{{f(-1)=-2(-1)^3+(3/2)-(17/2)-7}}}
{{{f(-1)=-12}}}