# Question 2, Exercise 1.3

Solutions of Question 2 of Exercise 1.3 of Unit 01: Complex Numbers. This is unit of A Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB) Peshawar, Pakistan.

Factorize the polynomial $P(z)$ into linear factors. $$P\left( z \right)={{z}^{3}}+6z+20$$

Given: $$p\left( z \right)={{z}^{3}}+6z+20$$ By factor theorem, $(z-a)$ is a factor of $P(z)$ iff $P(a)=0$. Put $z=-2$ \begin{align} P(-2)&=(-2)^3+6(-2)+20\\ &=-8-12+20\\ &=0\end{align} Thus $z+2$ is a factor of ${{z}^{3}}+6z+20$.
By using synthetic division, we have $$\begin{array}{c|cccc} -2 & 1 & 0 & 6 & 20 \\ & \downarrow & -2 & 4 & -20 \\ \hline & 1 & -2 & 10 & 0 \\ \end{array}$$ This gives \begin{align} P(z)&=(z+2)(z^2-2z+10)\\ &=(z+2)\left(z^2-2z+1+9\right)\\ &=(z+2)\left[(z-1)^2-(3i)^2\right]\\ &=(z+2)(z-1+3i)(z-1-3i)\end{align}

Factorize the polynomial $P(z)$ into linear factors. $$P(z)=3z^2+7.$$

\begin{align} P(z)&=3z^2+7\\ &=\left(\sqrt{3}z\right)^2-\left(\sqrt{7}i \right)^2\\ &=\left( \sqrt{3}z+\sqrt{7}i \right)\left( \sqrt{3}z-\sqrt{7}i \right)\end{align}

Factorize the polynomial $P\left( z \right)$ into linear factors. $$P\left( z \right)={{z}^{2}}+4$$

\begin{align}P(z)&={{z}^{2}}+4\\ &={{\left( z \right)}^{2}}-{{\left( 2i \right)}^{2}}\\ &=\left( z+2i \right)\left( z-2i \right).\end{align}

Factorize the polynomial $P(z)$ into linear factors. $$P(z)={{z}^{3}}-2{{z}^{2}}+z-2.$$

Given: $$P\left( z \right)={{z}^{3}}-2{{z}^{2}}+z-2$$
By factor theorem $(z-a)$ is a factor of $()$ iff $P(a)=0$.
Put $z=2$
\begin{align}P(-2)&=(-2)^3+6(-2)+20\\ &=-8-12+20=0\end{align} Thus $z-2$ is a factor of ${{z}^{3}}-2{{z}^{2}}+z-2$.
By using synthetic division, we have $$\begin{array}{c|cccc} 2 & 1 & -2 & 1 & -2 \\ & \downarrow & 2 & 0 & 2 \\ \hline & 1 & 0 & 1 & 0 \\ \end{array}$$ This implies \begin{align}P(z)&=(z-2)\left( {{z}^{2}}+1 \right)\\ &=(z-2)\left(z^2-i^2\right)\\ &=(z-2)(z-i)(z+i).\end{align}