# What are the roots of x^4-6x^3+14x^2-14x+5=0 with their multiplicities?

##### 1 Answer
Mar 19, 2017

The roots are:

$x = 1$ with multiplicity $2$

$x = 2 \pm i$ each with multiplicity $1$

#### Explanation:

Given:

${x}^{4} - 6 {x}^{3} + 14 {x}^{2} - 14 x + 5 = 0$

Note that the sum of the coefficients is $0$, that is:

$1 - 6 + 14 - 14 + 5 = 0$

Hence $x = 1$ is a root and $\left(x - 1\right)$ a factor:

${x}^{4} - 6 {x}^{3} + 14 {x}^{2} - 14 x + 5 = \left(x - 1\right) \left({x}^{3} - 5 {x}^{2} + 9 x - 5\right)$

Note that the sum of the coefficients of the remaining cubic is also $0$, that is:

$1 - 5 + 9 - 5 = 0$

Hence $x = 1$ is a root again and $\left(x - 1\right)$ a factor again:

${x}^{3} - 5 {x}^{2} + 9 x - 5 = \left(x - 1\right) \left({x}^{2} - 4 x + 5\right)$

We can factor the remaining quadratic by completing the square and using the difference of squares identity:

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

with $a = \left(x - 2\right)$ and $b = i$ as follows:

${x}^{2} - 4 x + 5 = {x}^{2} - 4 x + 4 + 1$

$\textcolor{w h i t e}{{x}^{2} - 4 x + 5} = {\left(x - 2\right)}^{2} - {i}^{2}$

$\textcolor{w h i t e}{{x}^{2} - 4 x + 5} = \left(\left(x - 2\right) - i\right) \left(\left(x - 2\right) + i\right)$

$\textcolor{w h i t e}{{x}^{2} - 4 x + 5} = \left(x - 2 - i\right) \left(x - 2 + i\right)$

So the remaining two zeros are:

$x = 2 \pm i$