====== Question 1, Exercise 1.3 ======
Solutions of Question 1 of Exercise 1.3 of Unit 01: Complex Numbers. This is unit of Model Textbook of Mathematics for Class XI published by National Book Foundation (NBF) as Federal Textbook Board, Islamabad, Pakistan. 

====Question 1(i)====
Factorize the polynomial into linear functions: $z^{2}+169$.

**Solution.**

\begin{align} 
& z^{2} + 169 \\
= & z^{2} - (13i)^2 \\
= &(z + 13i)(z - 13i).
\end{align}
====Question 1(ii)====
Factorize the polynomial into linear functions: $2 z^{2}+18$.

**Solution.**

\begin{align}
& 2z^2 + 18 \\
= &2(z^2 - (3i)^2)\\ 
= &2(z + 3i)(z - 3i)
\end{align}

====Question 1(iii)====
Factorize the polynomial into linear functions: $3 z^{2}+363$.
  
**Solution.**

\begin{align}
& 3z^2 + 363 \\ 
= & 3(z^2 - (11i)^2)\\
= & 3(z + 11i)(z - 11i)
\end{align}

====Question 1(iv)====
Factorize the polynomial into linear functions: $z^{2}+\dfrac{3}{25}$.

**Solution.**

\begin{align}
& z^2 + \dfrac{3}{25} \\
= & z^2 - \left(\dfrac{\sqrt{3}}{5} i \right)^2 \\
= & \left(z + \dfrac{\sqrt{3}}{5}i\right)\left(z - \dfrac{\sqrt{3}}{5}i\right)
\end{align}
====Question 1(v)====
Factorize the polynomial into linear functions: $2 z^{3}+3 z^{2}-10 z-15$.

**Solution.**

Suppose 
$$ P(z) = 2 z^{3}+3 z^{2}-10 z-15.$$
Since
\begin{align}P\left(-\dfrac{3}{2} \right) &= 2\left(-\dfrac{3}{2}\right)^3 + 3\left(-\dfrac{3}{2}\right)^2 - 10\left(-\dfrac{3}{2}\right) - 15\\ &= -\dfrac{27}{4} + \dfrac{27}{4} + 15 - 15 \\
&= 0\end{align}
So $z-(-\dfrac{3}{2})=z+\dfrac{3}{2}$ is the factor of polynomial.
Then by using synthetic division:
\begin{align}
\begin{array}{r|rrrr}
-\dfrac{3}{2} & 2 & 3 & -10 & -15 \\
&   & -3 & 0 & 15 \\
\hline
& 2 & 0 & -10 & 0 \\
\end{array}\end{align}
Thus
\begin{align}& 2z^3 + 3z^2 - 10z - 15 \\ 
= & \left(z + \dfrac{3}{2}\right)(2 z^2 - 10) \\
= & \frac{1}{2}(2z + 3)\cdot 2(z^2-5) \\
= & (2z + 3)(z^2-(\sqrt{5})^2)\\
= & (2z + 3)(z+\sqrt{5})(z - \sqrt{5})\end{align}
	

====Question 1(vi)====
Factorize the polynomial into linear functions: $z^{3}-7 z+6$.

**Solution.**

Suppose $$P(z) = z^3 - 7z + 6.$$
Since
$$ P(1) = 1^3 - 7 \cdot 1 + 6 = 0 $$
So $z-(1)=z-1$ is the factor of polynomial.
Then by using synthetic division:
\begin{align}
\begin{array}{c|ccc}
1 & 1 & 0 & -7 & 6 \\
&   & 1 & 1 & -6 \\
\hline
& 1 & 1 & -6 & 0 \\
\end{array}
\end{align}

Thus
\begin{align}
& z^3 - 7z + 6 \\
=& (z-1)(z^2+z-6) \\
=& (z-1)(z^2+3z-2z-6) \\
=& (z-1)(z(z+3)-2(z+3)) \\
=& (z-1)(z-2)(z+3).
\end{align}
====Question 1(vii)====
Factorize the polynomial into linear functions: $z^{3}+2 z^{2}-23 z-60$.

**Solution.**

Given: $$z^{3}+2 z^{2}-23 z-60$$

Putting $z = -3$:
\begin{align}
(-3)^3 + 2(-3)^2 - 23(-3) - 60 &= -27 + 18 + 69 - 60\\
&= 0\end{align}
So $z-(-3)=z+3$ is the factor of polynomial.
Then by using synthetic division:
Now, by synthetic division:
\begin{align}
\begin{array}{r|rrrr}
-3 & 1 & 2 & -23 & -60 \\
&   & -3 & 3  & 60 \\
\hline
& 1 & -1 & -20 & 0 \\
\end{array}
\end{align}
This gives
\begin{align}
& z^3 + 2z^2 - 23z - 60 \\
& = (z + 3)(z^2 - z - 20)\\
& = (z + 3)(z^2 +4z-5z - 20)\\
& = (z + 3)(z(z + 4)-5(z + 4))\\
&=(z + 3)(z + 4)(z - 5).
\end{align}
GOOD


====Question 1(viii)====
Factorize the polynomial into linear functions: $2 z^{3}+9 z^{2}-11 z-30$.

**Solution.**

Suppose $$P(z)=2z^3 + 9z^2 - 11z - 30.$$

Since
\begin{align} 
P(2) & = 2(2)^3 + 9(2)^2 - 11(2) - 30 \\
& = 16 + 36 - 22 - 30 = 0.
\end{align}
So $z-2$ is the factor of polynomial.\\
Then by using synthetic division:
\begin{align}
\begin{array}{r|rrrr}
2 & 2 & 9 & -11 & -30 \\
&   & 4 & 26 & 30 \\
\hline
& 2 & 13 & 15 & 0 \\
\end{array}
\end{align}
Thus
\begin{align} 
& 2z^3 + 9z^2 - 11z - 30 \\
= & (z - 2)(2z^2 + 13z + 15)\\
=& (z - 2)(2z^2 + 10z + 3z + 15)\\
=&(z - 2) (2z(z + 5) + 3(z + 5)) \\
=&(z - 2) (z + 5) (2z + 3).
\end{align}
GOOD

====Question 1(ix)====
Factorize the polynomial into linear functions: $z^{2}-7 z-8$.

**Solution.**

\begin{align}
& z^2 - 7z - 8 \\
= & z^2 -8z +z -8 \\
= & z(z-8)+1(z-8) \\
= & (z - 8)(z + 1).
\end{align}

====Question 1(x)====
Factorize the polynomial into linear functions: $4 z^{2}-7 z-11$.

**Solution.**

\begin{align}
& 4z^2 - 7z - 11 \\
= & 4z^2 - 11z + 4z -11 \\
= & z(4z-11)+1(4z-1)\\
= &(4z - 11)(z + 1).
\end{align}

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