Find a polynomial function of degree 3 with the given numbers as zeros. Assume that the leading coefficient is 1 . \( -2 ; 6,8 \)
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To find a polynomial function of degree 3 with zeros at \( -2 \), \( 6 \), and \( 8 \), we can use the fact that if \( r_1, r_2, r_3 \) are the roots of a polynomial, then the polynomial can be expressed as: \[ f(x) = (x - r_1)(x - r_2)(x - r_3) \] Plugging in the values for the roots: \[ f(x) = (x + 2)(x - 6)(x - 8) \] Now, we can expand this product. First, let's multiply the first two factors: \[ (x + 2)(x - 6) = x^2 - 6x + 2x - 12 = x^2 - 4x - 12 \] Now, we multiply this result by the third factor \( (x - 8) \): \[ f(x) = (x^2 - 4x - 12)(x - 8) \] Expanding this: \[ = x^3 - 8x^2 - 4x^2 + 32x - 12x + 96 = x^3 - 12x^2 + 20x + 96 \] So, the polynomial function of degree 3 with the desired zeros and leading coefficient of 1 is: \[ f(x) = x^3 - 12x^2 + 20x + 96 \]