If a polynomial can be factored as the product of k linear factors with exponents n₁, n₂, ..., n_k, as follows:
![P(x)=C\cdot(x-a_1)^(n_1)*(x-a_2)^(n_2)*\ldots*(x-a_k)^(n_k)](https://img.qammunity.org/2023/formulas/mathematics/college/863ulym4j1v8xiycjn7n1t04h3sdgzw9ky.png)
Then, the zeros of P(x) are a₁, a₂, ...,a_k and their respective multiplicities are n₁, n₂, ..., n_k.
For the given polynomial, notice that 3x^2 is a common factor for all three terms. Then, factor it out:
![\begin{gathered} G(x)=3x^4+6x^3+3x^2 \\ \Rightarrow G(x)=3x^2(x^2+2x+1) \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/iaed5s0gcgsi16vvqlzzney56w6z4b5z99.png)
The factor (x^2+2x+1) is a perfect square binomial that can be expressed as a binomial squared:
![x^2+2x+1=(x+1)^2](https://img.qammunity.org/2023/formulas/mathematics/college/9uv6hakmdayblr24pddv3rkq0jfplur7x8.png)
Then, G(x) can be written as:
![\begin{gathered} G(x)=3x^2(x+1)^2 \\ \Rightarrow G(x)=3(x-0)^2(x-\lbrack-1\rbrack)^2 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/7sp3pb609tw6n0agu6kk3xbh1975egax2x.png)
As we can see from the expression, the roots of G(x) from the smallest to largest are -1 and 0, and the multiplicities are 2 in both cases.
Therefore, the first zero is -1 with a multiplicity of 2, and the second zero is 0 with a multiplicity of 2.