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.

Factorize the polynomial into linear functions: $z^{2}+169$.

Solution.

$$\begin{align} & z^{2} + 169 \\ = & z^{2} - (13i)^2 \\ = &(z + 13i)(z - 13i). \end{align}$$

Factorize the polynomial into linear functions: $2 z^{2}+18$.

Solution.

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

Factorize the polynomial into linear functions: $3 z^{2}+363$.

Solution.

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

Factorize the polynomial into linear functions: $z^{2}+\dfrac{3}{25}$.

Solution.

$$\begin{align} & z^2 + \dfrac{3}{25} \\ = & z^2 - \left(\dfrac{\sqrt{3}}{5} i \right)^2 \\ = & \left(z + \dfrac{\sqrt{3}}{5}i\right)\left(z - \dfrac{\sqrt{3}}{5}i\right) \end{align}$$

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 $$\begin{align}& 2z^3 + 3z^2 - 10z - 15 \\ = & \left(z + \dfrac{3}{2}\right)(2 z^2 - 10) \\ = & \frac{1}{2}(2z + 3)\cdot 2(z^2-5) \\ = & (2z + 3)(z^2-(\sqrt{5})^2)\\ = & (2z + 3)(z+\sqrt{5})(z - \sqrt{5})\end{align}$$

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 $$\begin{align} & z^3 - 7z + 6 \\ =& (z-1)(z^2+z-6) \\ =& (z-1)(z^2+3z-2z-6) \\ =& (z-1)(z(z+3)-2(z+3)) \\ =& (z-1)(z-2)(z+3). \end{align}$$

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 $$\begin{align} & z^3 + 2z^2 - 23z - 60 \\ & = (z + 3)(z^2 - z - 20)\\ & = (z + 3)(z^2 +4z-5z - 20)\\ & = (z + 3)(z(z + 4)-5(z + 4))\\ &=(z + 3)(z + 4)(z - 5). \end{align}$$

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 $$\begin{align} & 2z^3 + 9z^2 - 11z - 30 \\ = & (z - 2)(2z^2 + 13z + 15)\\ =& (z - 2)(2z^2 + 10z + 3z + 15)\\ =&(z - 2) (2z(z + 5) + 3(z + 5)) \\ =&(z - 2) (z + 5) (2z + 3). \end{align}$$

Factorize the polynomial into linear functions: $z^{2}-7 z-8$.

Solution.

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

Factorize the polynomial into linear functions: $4 z^{2}-7 z-11$.

Solution.

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

Question 2>