Step 1 :The real zeros of a polynomial function are the x-values for which the function equals zero. In this case, the function will be zero when either \((x-9)\) or \((x+2)^{2}\) equals zero. So, we need to solve these two equations to find the real zeros.
Step 2 :The solutions to the equations are x = 9 and x = -2. However, since \((x+2)^{2}\) is squared, the zero x = -2 has a multiplicity of 2, meaning it is a root of the polynomial twice. The zero x = 9 has a multiplicity of 1, as it is not raised to any power.
Step 3 :Final Answer: The real zeros of the function \(f(x)=7(x-9)(x+2)^{2}\) are \(x = -2\) with multiplicity 2 and \(x = 9\) with multiplicity 1. In Latex format, the answer is \(\boxed{x = -2 \text{ with multiplicity 2 and } x = 9 \text{ with multiplicity 1}}\).