Guided Example #2 - Polynomial Equations

If the polynomial has real coefficients, then its imaginary roots occur in conjugate pairs. So a polynomial with these two roots must actually have at least three roots: ${\displaystyle 2,1-i}$,and ${\displaystyle 1+i}$. Since each root of the equation corresponds to a factor of the polynomial, we can write the following equation.
${\displaystyle (x-2)[(x-(1-i)][x-(1+i)]=0}$

Regroup
${\displaystyle (x-2)[((x-1)+i][(x-1)-i]=0}$

Recall that ${\displaystyle (a+b)(a-b)=a^2-b^2}$
${\displaystyle (x-2)[{(x-1)}^2-i^2=0}$

But ${\displaystyle i^2=-1}$
${\displaystyle (x-2)[(x^2-2x+1)+1]=0}$

Simplify
${\displaystyle (x-2)(x^2-2x+2)=0}$

So
${\displaystyle x^3-4x^2+6x-4=0}$
Note that any multiple of this equation would also have the required roots but would not be as simple.

If the polynomial has real coefficients, then its imaginary roots occur in conjugate pairs. So a polynomial with these two roots must actually have at least three roots: ${\displaystyle 2,1-i}$,and ${\displaystyle 1+i}$. Since each root of the equation corresponds to a factor of the polynomial, we can write the following equation.
${\displaystyle (x-2)[(x-(1-i)][x-(1+i)]=0}$

Regroup
${\displaystyle (x-2)[((x-1)+i][(x-1)-i]=0}$

Recall that ${\displaystyle (a+b)(a-b)=a^2-b^2}$
${\displaystyle (x-2)[{(x-1)}^2-i^2=0}$

But ${\displaystyle i^2=-1}$
${\displaystyle (x-2)[(x^2-2x+1)+1]=0}$

Simplify
${\displaystyle (x-2)(x^2-2x+2)=0}$

So
${\displaystyle x^3-4x^2+6x-4=0}$
Note that any multiple of this equation would also have the required roots but would not be as simple.