keeping in mind that -3i, is a complex solution, or zero, now, complex zeros do not come all by their lonesome, their sister is always with them, the conjugate, so if -3i, namely 0 - 3i is there, her sister the conjugate 0 + 3i is also there, so there are three zeros, let's check them out,
[tex]\bf \textit{difference of squares}
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(a-b)(a+b) = a^2-b^2
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\textit{also recall that }i^2=-1\\\\
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\begin{cases}
x=5\implies &x-5=0\\
x=0-3i\implies &x+3i=0\\
x=0+3i\implies &x-3i=0
\end{cases}
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(x-5)(x+3i)(x-3i)=\stackrel{original~polynomial}{0}
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(x-5)~~~[(x)^2~~-~~(3i)^2] = y\implies (x-5)~~~[(x)^2~~-~~(3^2i^2)] = y
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(x-5)~~~[(x)^2~~-~~(9\cdot -1)] = y\implies (x-5)(x^2+9)=y
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x^3+9x-5x^2-45=y\implies x^3-5x^2+9x-45=y[/tex]
