Given:
The given polynomial is
![g(x)=x^4+4x^3+47x^2-110x+58](https://img.qammunity.org/2023/formulas/mathematics/college/k135xrh37yoj31fkg1g3ecephpxddvdtxj.png)
1 is a zero of multiplicity two.
Required:
We have to express g(x) as a product of linear factors.
Step-by-step explanation:
Since 1 is a zero of multiplicity two,
![(x-1)^2](https://img.qammunity.org/2023/formulas/mathematics/college/d0wql51sqw3r661px9agtf0qc0djq4puup.png)
is a factor of g(x).
So we can divide g(x) by
![(x-1)^2=x^2-2x+1.](https://img.qammunity.org/2023/formulas/mathematics/college/nxzq30o2r7odnksyt3uwiaa9qoszf5pq9b.png)
![g(x)=\text{ \_\_\_}(x^2-2x+1)+\text{ \_\_\_}(x^2-2x+1)+\text{ \_\_\_\_}](https://img.qammunity.org/2023/formulas/mathematics/college/mm6scmhr01ccgw11hg6wf61po7m5q3w2kh.png)
We will fill the blanks with suitable terms.
![\begin{gathered} g(x)=x^2(x-2x^2-1)+6x(x-2x^2-1)+58(x-2x^2-1) \\ g(x)=(x-2x^2-1)(x^2+6x+58) \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/y4rhb7emnw826dd53nl59klsj6gw3iv6g3.png)
Final answer:
Hence the final answer is
![g(x)=(x-2x^(2)-1)(x^(2)+6x+58)](https://img.qammunity.org/2023/formulas/mathematics/college/b8zihiyrfuptmklilizamrpie8p34ou0ne.png)