Hello, I need some help with Part 2 question 6! Please show work as the instructions asked! If you want me to include other completed work from the assignment for extra information, please let me know. Thank you.


Problem N 6
we have the roots
3 and (4+i)
By the conjugate complex theorem
If (4+i) is a root
then
(4-i) is a root too
so
we have at least
Zeros
The polynomial function is given by
(x-3)(x-(4+i))(x-(4-i))
Multiply first
(x-(4+i))(x-(4-i))
x^2+(4+i)(4-i)-x(4-i)-x(4+i)
x^2+16-i^2-4x+xi-4x-xi
x^2+16-(-1)-8x
x^2-8x+17
so
(x-3)(x-(4+i))(x-(4-i))=(x-3)(x^2-8x+17)
Apply distributive property again
x^3-8x^2+17x-3x^2+24x-51
therefore