ANSWER
The factored polynomial is (x - 4 + 3i)(x - 4 - 3i)(x + 9)(x - 3)
EXPLANATION
The conjugate roots theorem tells us that if a+bi is a zero for the polynomial, then its conjugate a - bi is a zero too.
In this problem, it is given that 4 - 3i is a zero, then 4 + 3i is a zero too.
Therefore, the given polynimal can be divided by:
[tex]\begin{gathered} (x-4+3i)(x-4-3i)=((x-4)+3i)((x-4)-3i) \\ \text{ by the difference of two squares rule:} \\ (x-4)^2-(3i)^2=(x^2-8x+16)-(-9)=x^2-8x+25 \end{gathered}[/tex]The division gives as a result this polynomial:
[tex]x^2+6x-27[/tex]Which is a second degree polynomial and we can use this rule to find its two roots:
[tex]\begin{gathered} ax^2+bx+c \\ x=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a} \end{gathered}[/tex]In this polynomial a = 1, b = 6 and c = -27:
[tex]\begin{gathered} x=\frac{-6\pm\sqrt[]{6^2-4\cdot1(-27)}}{2\cdot1} \\ x=\frac{-6\pm\sqrt[]{36+108}}{2} \\ x=\frac{-6\pm\sqrt[]{144}}{2}=\frac{-6\pm12}{2} \\ x_1=\frac{-6-12}{2}=\frac{-18}{2}=-9 \\ x_2=\frac{-6+12}{2}=\frac{6}{2}=3 \end{gathered}[/tex]The other two roots (or zeros) of our polynomial are 3 and -9.
To write a factored polynomial, if x1, x2, x3 and x4 are roots, we have to write:
[tex](x-x_1)(x-x_2)(x-x_3)(x-x_4)[/tex]For our polynomial:
[tex]\begin{gathered} (x-(4-3i))(x-(4+3i))(x-(-9))(x-3) \\ \text{ simplifying (with less parenthesis)} \\ (x-4+3i)(x-4-3i)(x+9)(x-3) \end{gathered}[/tex]
