Question

Can someone help please

Answer 1

let's bear in mind that complex roots never come alone, their conjugate sister is always with her, so if we have the complex root of "i" or namely "0 + i", her conjugate is also coming along, or "0 - i", so we really have four roots, so

`\begin{cases} x = 0+i &\implies x -i=0\\ x = 0-i &\implies x +i=0\\ x = √(2) &\implies x -√(2)=0\\ x = 3 &\implies x -3=0\\ \end{cases} \\\\[-0.35em] ~\dotfill\\\\ \stackrel{original~polynomial}{a ( x -i )( x +i )( x -√(2) )( x -3 ) = \stackrel{0}{y}}\hspace{5em}\stackrel{\textit{we are assuming that}}{a=1} \\\\\\ 1( x -i )( x +i )( x -√(2) )( x -3 ) = y\implies ( x -i )( x +i )( x -√(2) )( x -3 ) = y \\\\[-0.35em] ~\dotfill`

`\stackrel{ \textit{difference of squares} }{( x -i )( x +i )}\implies x^2 - i^2\implies x^2-(-1)\implies x^2+1 \\\\[-0.35em] ~\dotfill\\\\ (x^2+1)( x -√(2) )( x -3 )\implies (x^2+1)(x^2-3x-x√(2)+3√(2)) \\\\\\ (x^2+1)[x^2-x(3+√(2))+3√(2)] \\\\\\ x^4-x^3(3+√(2))+3x^2√(2)+x^2-x(3+√(2))+3√(2) \\\\\\ \boxed{x^4-x^3(3+√(2))+x^2(3√(2)+1)-x(3+√(2))+3√(2)~~ = ~~y}`