Given:
The given polynomial is
[tex]g(x)=x^4+4x^3+47x^2-110x+58[/tex]1 is a zero of multiplicity two.
Required:
We have to express g(x) as a product of linear factors.
Explanation:
Since 1 is a zero of multiplicity two,
[tex](x-1)^2[/tex]is a factor of g(x).
So we can divide g(x) by
[tex](x-1)^2=x^2-2x+1.[/tex][tex]g(x)=\text{ \_\_\_}(x^2-2x+1)+\text{ \_\_\_}(x^2-2x+1)+\text{ \_\_\_\_}[/tex]We will fill the blanks with suitable terms.
[tex]\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}[/tex]Final answer:
Hence the final answer is
[tex]g(x)=(x-2x^{2}-1)(x^{2}+6x+58)[/tex]