Answer:
Step-by-step explanation:
each zero of the function will have a factor of (x - x₀)
h(x) = a(x + 3)(x + 2)(x - 1)
h(x) = a(x + 3)(x² + x - 2)
h(x) = a(x³ + 4x² + x - 6)
or the third option works if a = 1
however this equation gives us the points (0, -6) and (-1. -4), so "a" must be -2
h(x) = -2x³ - 8x² - 2x + 12
to fit ALL of the given points as it fits the three zeros and also h(0) and h(-1) so I guess that is why the given group is a <u><em>partial</em></u> set of solution sets
