Since x = 3, -2, 1/3, and -1/2 are zeroes, by the Factor Theorem, we have that:
x - 3, x - (-2) = x + 2, x - 1/3, and x - (-1/2) = x + 1/2 are factors.
So, since the polynomial has degree 4, one such polynomial function is just the product of these factors:
(x - 3)(x + 2)(x - 1/3)(x + 1/2).
Just a note here: any other polynomial of degree 4 with the given roots must be a constant multiple of the above polynomial. If you want to get rid of the fractions, we can multiply the above by 6 to get:
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Since x = 3, -2, 1/3, and -1/2 are zeroes, by the Factor Theorem, we have that:
x - 3, x - (-2) = x + 2, x - 1/3, and x - (-1/2) = x + 1/2 are factors.
So, since the polynomial has degree 4, one such polynomial function is just the product of these factors:
(x - 3)(x + 2)(x - 1/3)(x + 1/2).
Just a note here: any other polynomial of degree 4 with the given roots must be a constant multiple of the above polynomial. If you want to get rid of the fractions, we can multiply the above by 6 to get:
6(x - 3)(x + 2)(x - 1/3)(x + 1/2) = (x - 3)(x + 2)[3(x - 1/3)][2(x + 1/2)]
= (x - 3)(x + 2)(3x - 1)(2x + 1),
which will expand to yield a polynomial with solely integer coefficients.
I hope this helps!
(x - 3)(x + 2)(3x -1)(2x + 1)
Multiply out, and you have your function.