# How do you write a polynomial in standard form given the zeros x=7, -13, and 3+2i?

Oct 2, 2016

$\text{The Reqd. Poly.} = k \left({x}^{4} - 114 {x}^{2} + 624 x - 1183\right) , k \in \mathbb{C} - \left\{0\right\} .$

#### Explanation:

Let $p \left(x\right)$ be the reqd. poly.

The Zeroes of $p \left(x\right) \text{ are } 7 , - 13 , \mathmr{and} , 3 + 2 i$.

We know that the complex zero of a poly. occurs in conjugate pairs.

So, the conjugate of $3 + 2 i , i . e , 3 - 2 i$ is also a zero of $p \left(x\right) .$

In all, thus, we have $4$ zeroes of $p \left(x\right)$, meaning that $p \left(x\right)$ must

be of degree $4$, having factors, derived using the zeroes,

(x-7), (x+13), (x-3-2i), & (x-3+2i).

We conclude that,

$p \left(x\right) = k \left(x - 7\right) \left(x + 13\right) \left(x - 3 - 2 i\right) \left(x - 3 + 2 i\right) , k \in \mathbb{C} - \left\{0\right\}$

$= k \left({x}^{2} + 6 x - 91\right) \left\{{\left(x - 3\right)}^{2} - {\left(2 i\right)}^{2}\right\}$

$= k \left({x}^{2} + 6 x - 91\right) \left\{{x}^{2} - 6 x + 9 - 4 {i}^{2}\right\}$

$= k \left({x}^{2} + 6 x - 91\right) \left\{{x}^{2} - 6 x + 9 - 4 \left(- 1\right)\right\}$

$= k \left({x}^{2} + 6 x - 91\right) \left({x}^{2} - 6 x + 13\right)$

$= - k \left\{6 x + \left({x}^{2} - 91\right)\right\} \left\{6 x - \left({x}^{2} + 13\right)\right\}$

$= - k \left\{{\left(6 x\right)}^{2} + 6 x \left({x}^{2} - 91 - {x}^{2} - 13\right) - \left({x}^{2} - 91\right) \left({x}^{2} + 13\right)\right\}$

$= - k \left\{36 {x}^{2} - 624 x - \left({x}^{4} - 78 {x}^{2} - 1183\right)\right\}$

$- k \left(114 {x}^{2} - 624 x - {x}^{4} + 1183\right)$

$\therefore p \left(x\right) = k \left({x}^{4} - 114 {x}^{2} + 624 x - 1183\right) , k \in \mathbb{C} - \left\{0\right\} .$