Question 4.5.3: Finding the Complex Zeros of a Polynomial Function Given tha...

Because P(x) has real coefficients, the conjugate $\overline{2-i}=2+i$ is also a zero. By the Factorization Theorem, the linear factors $[x-(2-i)] \text { and }[x-(2+i)]$ appear in the factorization of P(x). Consequently, their product

$\begin{aligned}{[x-(2-i)][x-(2+i)] } &=(x-2+i)(x-2-i) \\&=[(x-2)+i][(x-2)-i] \quad \text { Regroup. } \\&=(x-2)^{2}-i^{2} \quad(A+B)(A-B)=A^{2}-B^{2} \\&=x^{2}-4 x+4+1 \quad \text { Expand }(x-2)^{2}, i^{2}=-1 \\&=x^{2}-4 x+5 \quad \text { Simplify. }\end{aligned}$

is also a factor of P(x). We divide P(x) by x²-4 x+5 to find the other factor.

$\text{Divisor} \rightarrow x^{2}-4 x+5 ) \overset{x^{2}-2 x+1} {\overline{x^{4}-6 x^{3}+14 x^{2}-14 x+5} \overset{\leftarrow \text{Quotient}}{\leftarrow \text{Dividend}} }$

$\begin {aligned} x^{4}-4 x^{3}+5 x^{2} \\ \hline \\ \quad -2 x^{3}+9 x^{2}-14 x \\ -2 x^{3}+8 x^{2}-10 x \\ \hline \\ x^{2}-4 x+5 \\ x^{2}-4 x+5 \\ \hline \\ o \leftarrow \text{Remainder} \end {aligned}$

$\begin{aligned}P(x) &=\left(x^{2}-2 x+1\right)\left(x^{2}-4 x+5\right) & & P(x)=\text { Quotient } \\&=(x-1)(x-1)\left(x^{2}-4 x+5\right) & & \text { Factor } x^{2}-2 x+1 \\&=(x-1)(x-1)[x-(2-i)][x-(2+i)] & & \text { Factor } x^{2}-4 x+5\end{aligned}$

The zeros of P(x) are 1 (of multiplicity 2), 2 – i, and 2 + i.