Question 1090911
<br>Because the point (0,0) is on the graph, we know one of the x-intercepts is 0.<br>
That also tells us that, when the polynomial is in standard form<br>
{{{f(x)=ax^3+bx^2+cx+d}}}<br>
the constant term d is 0.<br>
We can then form three equations in the other coefficients a, b, and c by plugging the coordinates of the other three points in the standard equation.<br>
{{{-a+b-c =  15}}}
{{{a+b+c = -5}}}
{{{8a+4b+2c = 12}}}<br>
There are of course many paths to the solution of these equations.  Probably the easiest first step is to see that if we add the first two equations we immediately find 2b = 10, so b is 5.<br>
Substituting that value for b in the last two equations we get<br>
{{{a+c = -10}}}
{{{8a+2c = -8}}} or {{{4a+c = -4}}}<br>
Eliminating c gives us a = 2; and it follows that c = -12.  So the polynomial is<br>
{{{f(x) = 2x^3+5x^2-12x}}}<br>
Factoring gives us<br>
{{{f(x) = x(2x-3)(x+4)}}}<br>
So the three x-intercepts are 0, 3/2, and -4.<br>