# Find the polynomial g(x) with integer coefficients of least degree (and positive leading coefficients as small as possible) for which 1-2i is a double root?

Oct 5, 2017

$g \left(x\right) = {x}^{4} - 4 {x}^{3} + 14 {x}^{2} - 20 x + 25$

#### Explanation:

As the polynomial $g \left(x\right)$ has integer coefficients and $1 - 2 i$ is a double root of the polynomial, it has its factor ${\left(x - \left(1 - 2 i\right)\right)}^{2}$.

But it will give us complex coefficients and hence for integers as coefficients we must have another double root which is complex conjugate of $1 - 2 i$ i.e. $1 + 2 i$. Hence ${\left(x - \left(1 + 2 i\right)\right)}^{2}$ too is a factor.

Hence the polynomial with least degree would be

$g \left(x\right) = {\left(x - \left(1 - 2 i\right)\right)}^{2} {\left(x - \left(1 + 2 i\right)\right)}^{2}$

= ${\left(\left(x - 1 + 2 i\right) \left(x - 1 - 2 i\right)\right)}^{2}$

= ${\left({x}^{2} - x \textcolor{red}{- 2 i x} - x + 1 \textcolor{red}{+ 2 i + 2 i x - 2 i} - 4 {i}^{2}\right)}^{2}$

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

= ${\left({x}^{2} - 2 x + 5\right)}^{2}$

= ${x}^{4} + 4 {x}^{2} + 25 - 4 {x}^{3} + 10 {x}^{2} - 20 x$

(using ${\left(a + b + c\right)}^{2} = {a}^{2} + {b}^{2} + {c}^{2} + 2 a b + 2 a c + 2 a b$)

= ${x}^{4} - 4 {x}^{3} + 14 {x}^{2} - 20 x + 25$