Question

Find the polynomial function $f\left(x\right)$ of least degree having only real coefficients and zeros as given.

Assume multiplicity $1$.

$3+2i,-1\text{and}2$..

Answer 1

Find the polynomial function $f\left(x\right)$ of least degree with real coefficients and and given zeros:

$\begin{array}{rcl}\text{If}x& =& a,b,c...\text{are the zeros of a function}f\left(x\right)\\ & \Rightarrow & f\left(x\right)=\left(x-a\right)\left(x-b\right)\left(x-c\right).....\end{array}$

The given zeros are $3+2i,-1\text{and}2$.

$\because$The complex roots are exists in conjugate pair.

So the zeros are $x=3+2i,3-2i,-1\text{and}2$.

$\Rightarrow f\left(x\right)=\left[\left(x-3-2i\right)\left(x-3+2i\right)\right]\left(x+1\right)\left(x-2\right)$

Simplify the equation $\Rightarrow f\left(x\right)=\left[\left(x-3-2i\right)\left(x-3+2i\right)\right]\left(x+1\right)\left(x-2\right)$:

$\begin{array}{rcl}f\left(x\right)& =& \left[\left(x-3-2i\right)\left(x-3+2i\right)\right]\left(x+1\right)\left(x-2\right)\\ & =& \left[{\left(x-3\right)}^2-{\left(2i\right)}^2\right]\left[x^2-2x+x-2\right]\\ & =& \left[x^2-6x+9-\left(-4\right)\right]\left(x^2-x-2\right)\\ & =& \left(x^2-6x+13\right)\left(x^2-x-2\right)\\ & =& x^4-x^3-2x^2-6x^3+6x^2+12x+13x^2-13x-26\\ & =& x^4-7x^3+17x^2-x-26\end{array}$

Hence, the required polynomial function is $f\left(x\right)=x^4-7x^3+17x^2-x-26$.