# How do you find all the zeros of g(x) = x^5 – 8x^4 + 28x^3 – 56x^2 + 64x – 32 ?

Feb 28, 2016

Use the rational root theorem, polynomial division and completing the square to find roots:

$x = 2$ (with multiplicity $3$)

$x = 1 \pm \sqrt{3} i$

#### Explanation:

$g \left(x\right) = {x}^{5} - 8 {x}^{4} + 28 {x}^{3} - 56 {x}^{2} + 64 x - 32$

By the rational root theorem, any rational zeros of $g \left(x\right)$ must be expressible in the form $\frac{p}{q}$ for integers $p$ and $q$ with $p$ a divisor of the constant term $32$ and $q$ a divisor of the coefficient $1$ of the leading term.

In addition note that $g \left(- x\right) = - {x}^{5} - 8 {x}^{4} - 28 {x}^{3} - 56 {x}^{2} - 64 x - 32$ has no changes of sign in its coefficients, so there are no negative Real zeros.

Hence the only possible rational zeros are:

$1 , 2 , 4 , 8 , 16 , 32$

We find:

$g \left(2\right) = 32 - 128 + 224 - 224 + 128 - 32 = 0$

So $x = 2$ is a zero and $\left(x - 2\right)$ a factor:

${x}^{5} - 8 {x}^{4} + 28 {x}^{3} - 56 {x}^{2} + 64 x - 32$

$= \left(x - 2\right) \left({x}^{4} - 6 {x}^{3} + 16 {x}^{2} - 24 x + 16\right)$

Let $h \left(x\right) = {x}^{4} - 6 {x}^{3} + 16 {x}^{2} - 24 x + 16$

$h \left(2\right) = 16 - 48 + 64 - 48 + 16 = 0$

So $x = 2$ is a zero of $h \left(x\right)$ and $\left(x - 2\right)$ is a factor:

${x}^{4} - 6 {x}^{3} + 16 {x}^{2} - 24 x + 16 = \left(x - 2\right) \left({x}^{3} - 4 {x}^{2} + 8 x - 8\right)$

Let $k \left(x\right) = {x}^{3} - 4 {x}^{2} + 8 x - 8$

$k \left(2\right) = 8 - 16 + 16 - 8 = 0$

So $x = 2$ is a zero of $k \left(x\right)$ and $\left(x - 2\right)$ is a factor:

${x}^{3} - 4 {x}^{2} + 8 x - 8 = \left(x - 2\right) \left({x}^{2} - 2 x + 4\right)$

The remaining quadratic factor can be factored by completing the square and using the difference of squares identity:

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

with $a = x - 1$ and $b = \sqrt{3} i$

${x}^{2} - 2 x + 4$

$= {x}^{2} - 2 x + 1 + 3$

$= {\left(x - 1\right)}^{2} - {\left(\sqrt{3} i\right)}^{2}$

$= \left(\left(x - 1\right) - \sqrt{3} i\right) \left(\left(x - 1\right) + \sqrt{3} i\right)$

$= \left(x - 1 - \sqrt{3} i\right) \left(x - 1 + \sqrt{3} i\right)$

Hence the last two zeros are:

$x = 1 \pm \sqrt{3} i$