Find an expression for a cubic function f if f(3) = 108 and f(-3) = f(0) = f(4) = 0.
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1) As, f(-3) = f(0) = f(4) = 0, {-3, 0, 4} are the roots of the required cubic polynomial. Hence, the cubic polynomial is given by: (x+3)(x-0)(x-4) = 0
2) Expanding and simplifying the above, x³ - x² - 12x = 0
3) So, in general the polynomial, f(x) = a(x³ - x² - 12x)
As f(3) = 108,
==> 108 = a(27 - 9 - 36); solving a = -6
Thus the required polynomial is f(x) = -6(x³ - x² - 12x)
or in another form the same is: f(x) = 6x(12x + x - x²)
2) Expanding and simplifying the above, x³ - x² - 12x = 0
3) So, in general the polynomial, f(x) = a(x³ - x² - 12x)
As f(3) = 108,
==> 108 = a(27 - 9 - 36); solving a = -6
Thus the required polynomial is f(x) = -6(x³ - x² - 12x)
or in another form the same is: f(x) = 6x(12x + x - x²)