Question

Form a fifth-degree polynomial function with real coefficients such that i, 1-2i, and 5 are zeros and f(0) = -75.

f(x)=
(Simplify your answer. Type an expression using x as the variable.)

Answer 1

well, hmmm keeping in mind that complex roots never come all by their lonesome, their sister always comes along, namely their conjugate, so if we have the complex roots of "i" or namely "0 + i", we also have her sister "0 - i", and if we have "1 - 2i", she also came with "1 + 2i", and we also have the root of 5, and that'd give us the fifth degree polynomial, so

`\begin{cases} x = 0+i &\implies x -i=0\\ x = 0-i &\implies x +i=0\\ x = 1-2i &\implies x -1+2i=0\\ x = 1+2i &\implies x -1-2i=0\\ x = 5 &\implies x -5=0\\ \end{cases} \\\\[-0.35em] ~\dotfill\\\\ \stackrel{original~polynomial}{a ( x -i )( x +i )( x -1+2i )( x -1-2i )( x -5 ) = \stackrel{0}{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 -1+2i )( x -1-2i )\implies \stackrel{ \textit{difference of squares} }{( [x -1]+2i )( [x -1]-2i )} \\\\\\ (x-1)^2 -(2i)^2\implies (x^2-2x+1)-4i^2\implies (x^2-2x+1)-4(-1) \\\\\\ x^2-2x+1+4\implies x^2-2x+5 \\\\[-0.35em] ~\dotfill\\\\ \stackrel{\textit{so we can say}}{a(x^2+1)(x^2-2x+5)(x-5)=y}\hspace{5em}\textit{we also know that } \begin{cases} x=0\\ y=-75 \end{cases} \\\\\\ a(0^2+1)(0^2-2(0)+5)(0-5)=-75\implies -25a=-75 \\\\\\ a=\cfrac{-75}{-25}\implies a=3 \\\\[-0.35em] ~\dotfill`

`3(x^2+1)(x^2-2x+5)(x-5)=y\implies 3(x^3-5x^2+x-5)(x^2-2x+5)=y \\\\\\ 3(x^5-7x^4+16x^3-32x^2+15x-25)=y \\\\[-0.35em] ~\dotfill\\\\ ~\hfill {\Large \begin{array}{llll} 3x^5-21x^4+48x^3-96x^2+45x-75=y \end{array}}~\hfill`

Check the picture below.