Question 1090083
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Let  f(x) = ax^3 + bx^2 + cx + d  with unknown coefficients  a, b, c and d.


Since f(0) = 0, it implies d = 0   (to see it, simply substitute x= 0 into the polynomial).


So, you need to determine a, b and c.


f(-1) = 15  ====> -a +  b -  c = 15,     (1)

f(1)  = -5  ====>  a +  b +  c = -5,     (2)

f(2)  = 12  ====> 8a + 4b + 2c = 12.     (3)


Add equations (1) and (2). You will get   2b = 15 + (-5) = 10  ====>  b = 5.


Then equations (2) and (3) take the form

 a +   5 +  c = -5     (4)    (instead of (2))
8a + 4*5 + 2c = 12     (5)    (instead of (3))

or, which is the same

 a + c = -10,          (4')
4a + c =  -4,          (5')


Subtract (4') from (5') to get 3a = 6,   a = 2.

Then  from (4')  c = -12.


Thus the polynomial is  f(x) = {{{2*x^3 + 5x^2 - 12*x}}} = {{{x*(2*x^2 + 5x - 12)}}} =  (factor ! ) = {{{2x*(x-3/2)*(x+4)}}} = {{{x*(2x-3)*(x+4)}}}.


Thus the x-intercepts are {{{3/2}}}, 0 (given !)  and  -4.
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Solved.