Question 1043703
Use the general equation,
{{{y=ax^3+bx^2+cx+d}}}
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{{{15=a(-1)^3+b(-1)^2+c(-1)+d}}}
1.{{{-a+b-c+d=15}}}
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{{{0=a(0)^3+b(0)^2+c(0)+d}}}
2.{{{d=0}}}
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{{{-5=a(1)^3+b(1)^2+c(1)}}}
3.{{{a+b+c=-5}}}
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{{{12=a(2)^3+b(2)^2+c(2)}}}
{{{8a+4b+2c=12}}}
4.{{{4a+2b+c=6}}}
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Updating eq. 1,
1.{{{-a+b-c=15}}}
Adding eq. 1 to eq. 3,
{{{a+b+c-a+b-c=-5+15}}}
{{{2b=10}}}
{{{b=5}}}
Adding eq. 1 to eq. 4, 
{{{4a+2(5)+c-a+5-c=15+6}}}
{{{3a+15=21}}}
{{{3a=6}}}
{{{a=2}}}
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So then using eq. 3,
{{{2+5+c=-5}}}
{{{c=-12}}}
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{{{y=2x^3+5x^2-12x}}}
Now solve for 
{{{2x^3+5x^2-12x=0}}}
{{{(2x^2+5x-12)x=0}}}
{{{(2x-3)(x+4)x=0}}}
Solve for the three zeros or look at the graph where the func tion crosses the x-axis.
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*[illustration gr3.JPG].