**– The degree of $ Q $ should be $ 3, space 0 $ and $ i $.**

The main objective of this question is to find the **polynomial** for the **given conditions**.

This question uses the concept of the **complex conjugate theorem.** According to the **conjugate root theorem**, if a **polynomial** for **one** **variable** has real coefficients and also the **complex number** which is $ a + bi $ is one of its** roots**, then its **complex conjugate**, a – bi, is also** one** of its **roots**.

## Expert Answer

We have to find the** polynomial** for the **given conditions**.

From the **complex conjugate theorem**, we know that if the** polynomial** $ Q (Â x ) $ has **real coefficients** and $ i $Â is a **zero**, it’s **conjugate** “-i” is also a **zero** ofÂ $ Q ( x ) $.

**Thus**:

- The e
**xpression**Â $Â (x – 0) $ is indeed a f**actor**of $ Q $ if $ 0 $ is indeed a**zero**of $ Q (x) $. - The
**expression**Â $Â (x – 0) $ is**indeed**a factor of $ Q $ if $ i $ is indeed a**zero**of $ Q (x) $. - The
**expression**Â $Â (x – 0) $ is indeed a**factor**of $ Q $ if $Â -i $ is**indeed**a zero of $ Q (x) $.

The **polynomial** is:

\[ \space Q ( x ) \space = \space ( x \space – \space 0 ) ( x \space – \space i) (x \space + \space 0) \]

We **know** that:

\[ \space a^2 \space – \space b^2 \space = \space ( a \space + \space b ) ( a \space – \space b ) \]

**Thus**:

\[ \space Q ( x ) \space = \space x ( x^2 \space – \space i^2 ) \]

\[ \space Q ( x ) \space = \space x ( x^2 \space + \space 1 ) \]

\[ \space Q ( x ) \space = \space x^3 \space + \space x \]

## Numerical Answer

The **polynomial** for the **given condition** is:

\[ \space Q ( x ) \space = \space x^3 \space + \space x \]

## Example

Find the **polynomial** which has a **degree** of $ 2 $ and **zeros** $ 1 \space +Â \space i $ with $ 1 \space –Â \space iÂ $.

We have to find the **polynomial** for the given **conditions**.

From the **complex conjugate theorem**, we know that if the **polynomial** $ Q (Â x ) $ has **real coefficients** and $ i $ is a** zero**, it’s **conjugate** “-i” is also a** zero** ofÂ $ Q ( x ) $.

**Thus**:

\[ \space ( x \space – \space (1 \space + i)) ( x \space – \space (1 \space –Â \space i )) \]

**Then**:

\[ \space (x \space –Â \space 1)^2 \space – \space (i)^2Â \]

\[ \space x^2 \space – \space 2 x \space + \space 1 \space – \space ( – 1 ) \]

\[ \space x^2 \space – \space 2 x \space + \space 2 \]

The **required polynomial** for the **given condition** is:

\[ \space x^2 \space – \space 2 x \space + \space 2 \]