Question 1031404
{{{f(x)=a(x+5)(x-2)(x-3)}}}
{{{f(0)=a(0+5)(0-2)(0-3)=20}}}
{{{a(5)(-2)(-3)=20}}}
{{{a=10/3}}}
So then,
{{{f(x)=(2/3)(x+5)(x-2)(x-3)}}}
{{{f(x)=(2/3)(x+5)(x^2-5x+6)}}}
{{{f(x)=(2/3)(x^3-19x+30)}}}
{{{f(x)=(2/3)x^3-(38/3)x+20}}}
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*[illustration dq7.JPG].