Step-by-step explanation:
Use this algebra 2 theorem:
If r and q are roots of a polynomial function then
the polynomial function can be expressed as
[tex](x - r)(x - q)[/tex]
Here the roots are
1- root of 6, 1+ root 6, and. 7-i so our. function can be expressed as
[tex](x - (1 - \sqrt{6} ))(x - (1 + \sqrt{6} ))(x - (7 - i))[/tex]
The first two binomials are difference off squares so the we have
[tex] {x}^{2} + x( - 1 + \sqrt{6} ) + x( - 1 - \sqrt{6} ) + ( - 5)[/tex]
[tex] {x}^{2} - 2x - 5[/tex]
The other root is (7-i).
Also note since (7-i) is a root, then (7+I) is also a root.
So
[tex](x - (7 - i)(x - (7 + i)( {x}^{2} - 2x - 5)[/tex]
[tex]( {x}^{2} - 14x + 50)( {x}^{2} - 2x - 5)[/tex]
Simplify and it gives us
[tex] {x}^{4} - 16 {x}^{3} + 73 {x}^{2} - 30x - 250[/tex]
