A third-degree polynomial with zeros of x=2 and x=-3 (multiplicity of 2) is found by multiplying the factors (x-2) and (x+3)^2, which results in the polynomial f(x) = x^3 + 4x^2 - 3x - 18.
To write a third-degree polynomial with zeros of x=2 and x=-3 (with a multiplicity of 2), we first understand that zeros of a polynomial are the solutions to the equation when we set the polynomial equal to zero. For a zero at x=2, we have a factor of (x-2). For a zero at x=-3 with a multiplicity of 2, we have a factor of (x+3)². Multiplying these factors together gives us the polynomial.
f(x) = (x-2)(x+3)²
Expanding the equation, f(x) becomes:
f(x) = (x-2)(x+3)(x+3)
f(x) = (x-2)(x²+6x+9)
f(x) = x³+6x²+9x-2x²-12x-18
f(x) = x³+4x²-3x-18
This is the third-degree polynomial with the given zeros.