====== 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.

=====Question 2(i)=====
Factorize the polynomial $P(z)$ into linear factors.
$$P\left( z \right)={{z}^{3}}+6z+20$$
====Solution==== 
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}

=====Question 2(ii)=====
Factorize the polynomial $P(z)$ into linear factors.
$$P(z)=3z^2+7.$$ 
====Solution====
\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}


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

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

====Go To====
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