Aus Online Mathematik Brückenkurs 2

If the equation has the root \displaystyle z=\text{1}, this means, according to the factor rule, that the equation mustcontain the \displaystyle z=\text{1}, i.e. the polynomial on the left-hand side can be written as

\displaystyle z^{3}-3z^{2}+4z-2=\left( z^{2}+Az+B \right)\left( z-1 \right)

for some constants \displaystyle A and \displaystyle B. We can determine the other unknown factor using polynomial division:

\displaystyle \begin{align} & z^{3}-3z^{2}+4z-2=\left( z^{2}+Az+B \right)\left( z-1 \right) \\ & z^{2}+Az+B=\frac{z^{3}-3z^{2}+4z-2}{z-1} \\ & =\frac{z^{3}-z^{2}+z^{2}-3z^{2}+4z-2}{z-1} \\ & =\frac{z^{2}\left( z-1 \right)-2z^{2}+4z-2}{z-1} \\ & =z^{2}+\frac{-2z^{2}+4z-2}{z-1} \\ & =z^{2}+\frac{-2z^{2}+2z-2z+4z-2}{z-1} \\ & =z^{2}+\frac{-2z\left( z-1 \right)+2z-2}{z-1} \\ & =z^{2}-2z+\frac{2z-2}{z-1} \\ & =z^{2}-2z+\frac{2\left( z-1 \right)}{z-1} \\ & =z^{2}-2z+2 \\ \end{align}

Thus, the equation can be written as

\displaystyle \left( z-1 \right)\left( z^{2}-2z+2 \right)=0

The advantage of writing the equation in this factorized form is that we can now conclude that the equation's two other roots must be zeros of the factor \displaystyle z^{2}-2z+2. This is because the left-hand side is zero only when at least one of the factors \displaystyle z-\text{1} or \displaystyle z^{2}-2z+2 is zero, and we see directly that \displaystyle z-\text{1} is zero only when \displaystyle z=\text{1}.

Hence, we determine the roots by solving the equation

\displaystyle z^{2}-2z+2=0

Completing the square gives

\displaystyle \begin{align} & \left( z-\text{1} \right)^{2}-1^{2}+2=0 \\ & \left( z-\text{1} \right)^{2}=-1 \\ \end{align}

and taking the root gives that \displaystyle z-\text{1}=\pm i i.e. \displaystyle z=1-i and \displaystyle z=1+i.

The equation's other roots are \displaystyle z=1-i and \displaystyle z=1+i.

As an extra check, we investigate whether \displaystyle z-\text{1}=\pm i really are roots of the equation.

\displaystyle \begin{align} & z=1+i:\quad z^{3}-3z^{2}+4z-2=\left( \left( z-3 \right)z+4 \right)z-2 \\ & =\left( \left( 1+i-3 \right)\left( 1+i \right)+4 \right)\left( 1+i \right)-2 \\ & =\left( \left( -2+i \right)\left( 1+i \right)+4 \right)\left( 1+i \right)-2 \\ & =\left( -2+i-2i-1+4 \right)\left( 1+i \right)-2 \\ & =\left( 1-i \right)\left( 1+i \right)-2 \\ & =1^{2}-i^{2}-2=1+1-2=0 \\ \end{align}

\displaystyle \begin{align} & z=1-i:\quad z^{3}-3z^{2}+4z-2=\left( \left( z-3 \right)z+4 \right)z-2 \\ & =\left( \left( 1-i-3 \right)\left( 1-i \right)+4 \right)\left( 1-i \right)-2 \\ & =\left( \left( -2-i \right)\left( 1-i \right)+4 \right)\left( 1-i \right)-2 \\ & =\left( -2-i+2i-1+4 \right)\left( 1-i \right)-2 \\ & =\left( 1+i \right)\left( 1-i \right)-2 \\ & =1^{2}-i^{2}-2=1+1-2=0 \\ \end{align}

NOTE: Writing

\displaystyle z^{3}-3z^{2}+4z-2=\left( \left( z-3 \right)z+4 \right)z-2

is known as the Horner scheme and is used to reduce the amount of the arithmetical work.