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Sagot :
bearing in mind that complex roots never come all by their lonesome, their sister is always with them, their conjugate, so if we have a root of 2i or 0 + 2i, its conjugate is also there or namely 0 - 2i, or just -2i.
[tex]\begin{cases} x=3\implies &x-3=0\\ x=4\implies &x-4=0\\ x=2i\implies &x-2i=0\\ x=-2i\implies &x+2i=0 \end{cases}\implies (x-3)(x-4)(x-2i)(x+2i)=\stackrel{y}{0} \\\\[-0.35em] ~\dotfill\\\\ \underset{\textit{difference of squares}}{(x-2i)(x+2i)}\implies [(x)^2-(2i)^2]\implies [x^2-(2^2i^2)]\implies [x^2-[4(-1)]] \\\\\\ x^2-(-4)\implies x^2+4 \\\\[-0.35em] ~\dotfill\\\\ (x-3)(x-4)(x^2+4)=0\implies \stackrel{F~O~I~L}{(x^2-7x+12)}(x^2+4)=0 \\\\\\ x^4-7x^3+12x^2+4x^2-28x+48=0\implies x^4-7x^3+16x^2-28x+48=y[/tex]
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