Chapter 2.5, Problem 128E

To determine

To find: The equation for $f$ with leading coefficient $a=1$ and $f\left(i\right)=f\left(2i\right)=0$ .

Answer to Problem 128E

The required equation is $f\left(x\right)=x^4+5x^2+4$ .

Explanation of Solution

Given information:

The leading coefficient is $a=-1$ and $f\left(2\right)=f\left(i\right)=0$ .

The standard cubic equation is,

  $f\left(x\right)=a{x}^3+b{x}^2+cx+d$

Here, $a,b,c$ and $d$ are real constant.

Substitute -1 for $a$ .

  $f\left(x\right)=-x^3+b{x}^2+cx+d$

The given condition is,

  $\begin{array}{c}f\left(2\right)=f\left(i\right)\\ =0\end{array}$

We know that the complex solution is always in conjugate pairs. Therefore the actual condition we have

  $\begin{array}{c}f\left(2\right)=f\left(i\right)\\ =f\left(-i\right)\\ =0\end{array}$

Consider on $f\left(2\right)=0$

  $\begin{array}{l}f\left(2\right)=0\\ -8+4b+2c+d=\mathrm{0...........}\left(1\right)\end{array}$

Now consider

  $\begin{array}{l}f\left(i\right)=0\\ i-b+ci+d=\mathrm{0...........}\left(2\right)\end{array}$

Also take

  $\begin{array}{l}f\left(-i\right)=0\\ -i-b-ci+d=\mathrm{0...........}\left(3\right)\end{array}$

On solving above two equations we get the values $b=2,c=-1,d=2$ .

Substitute $2$ for $b$ , $-1$ for $c$ and $2$ for $d$ in the standard aquatic equation.

  $f\left(x\right)=-x^3+2x^2-x+2$

Therefore, the required equation is $f\left(x\right)=-x^3+2x^2-x+2$ .