The given polynomial is
[tex]x^4-6x^3+25x^2-96x+144[/tex]
Using Desmos online graphing calculator, the graph of the function is shown below
From the graph, a zero of the polynomial occurs at the point (3,0)
Hence x = 3 is a zero of the polynomial
Hence x = 3
This implies that x - 3 is a factor of the polynomial
Using synthetic division to verify the zeros
Perform the division
[tex]\frac{x^4-6x^3+25x^2-96x+144}{x-3}[/tex]
The division is shown below
Since the remainder of the polynomial is 0, hence 3 is a zero of the polynomial
Factoring the polynomial gives
Since x - 3 is a factor then
[tex]x^4-6x^3+25x^2-96x+144=\mleft(x-3\mright)\frac{x^4-6x^3+25x^2-96x+144}{x-3}[/tex]
From the synthetic division
[tex]\frac{x^4-6x^3+25x^2-96x+144}{x-3}=x^3-3x^2+16x-48[/tex]
It follows
[tex]x^4-6x^3+25x^2-96x+144=(x-3)(x^3-3x^2+16x-48)[/tex]
Factoring
[tex]x^3-3x^2+16x-48[/tex]
gives
[tex]\mleft(x-3\mright)\mleft(x^2+16\mright)[/tex]
It follows
[tex]x^4-6x^3+25x^2-96x+144=(x-3)(x-3)(x^2+16)[/tex]
Hence the imaginary zeros occur at
[tex](x^2+16)[/tex]
Equate to zero and solve
[tex]\begin{gathered} x^2+16=0 \\ \Rightarrow x^2=-16 \\ \Rightarrow x=\pm\sqrt[]{-16}_{} \\ x=\pm4i \end{gathered}[/tex]
Therefore, the imaginary zeros are 4i and -4i
