# How do you find the zeros of 5x^8+4x^7+20x^5+42x^4+20x^3+4x+5 algebraically?

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#### Explanation:

We can factor this as a product of fourth degree polynomials

$5 {x}^{8} + 4 {x}^{7} + 20 {x}^{5} + 42 {x}^{4} + 20 {x}^{3} + 4 x + 5 = \left({x}^{4} + 4 x + 5\right) \left(5 {x}^{4} + 4 {x}^{3} + 1\right)$

The first ${x}^{4} + 4 x + 5$ can be factored as

(-x+sqrt(1-i)-i) (-x+sqrt(1+i)+i) (x+sqrt(1-i)+i) (x+sqrt(1+i)-i)

The second $5 {x}^{4} + 4 {x}^{3} + 1$ can be factored as

1/125 (-5 x+sqrt(7-i)-(1+2 i)) (-5 x+sqrt(7+i)-(1-2 i)) (5 x+sqrt(7-i)+(1+2 i)) (5 x+sqrt(7+i)+(1-2 i))

Now we can easily describe the zeros of the original equation.

Footnote

For general solutions of fourh degree equations the following article may be useful

Jun 11, 2016

Part one: Factoring into quartics

$5 {x}^{8} + 4 {x}^{7} + 20 {x}^{5} + 42 {x}^{4} + 20 {x}^{3} + 4 x + 5$

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

#### Explanation:

$f \left(x\right) = 5 {x}^{8} + 4 {x}^{7} + 20 {x}^{5} + 42 {x}^{4} + 20 {x}^{3} + 4 x + 5$

Part One - Factoring into quartics

Note that since the coefficients are symmetric, this can be factored as a product of quartics with coefficients in mirror image to one another.

So there is a factorisation of the form:

$5 {x}^{8} + 4 {x}^{7} + 20 {x}^{5} + 42 {x}^{4} + 20 {x}^{3} + 4 x + 5$

$= \left({x}^{4} + a {x}^{3} + b {x}^{2} + c x + 5\right) \left(5 {x}^{4} + c {x}^{3} + b {x}^{2} + a x + 1\right)$

$= 5 {x}^{8} + \left(5 a + c\right) {x}^{7} + \left(a c + 6 b\right) {x}^{6} + \left(a + a b + b c + 5 c\right) {x}^{5} + \left({a}^{2} + {b}^{2} + {c}^{2} + 26\right) {x}^{4} + \left(a + a b + b c + 5 c\right) {x}^{3} + \left(a c + 6 b\right) {x}^{2} + \left(5 a + c\right) x + 5$

Equating coefficients, we find:

$\left\{\begin{matrix}5 a + c = 4 \\ 5 b + a c + b = 0 \\ a + a b + b c + 5 c = 20 \\ {a}^{2} + {b}^{2} + {c}^{2} + 26 = 42\end{matrix}\right.$

From the last equation, we find:

${a}^{2} + {b}^{2} + {c}^{2} = 16$

and hence:

$\left\{\begin{matrix}\left\mid a \right\mid \le 4 \\ \left\mid b \right\mid \le 4 \\ \left\mid c \right\mid \le 4\end{matrix}\right.$

If there are integer solutions, then the first equation implies $c \ne 0$, hence either $\left(a , c\right) = \left(1 , - 1\right)$ or $\left(a , c\right) = \left(0 , 4\right)$

If $\left(a , c\right) = \left(1 , - 1\right)$ then we find ${b}^{2} = 14$, which would make $b$ not an integer.

If $\left(a , c\right) = \left(0 , 4\right)$ then we find $b = 0$.

The values $a = b = 0$, $c = 4$ satisfy all of the equations, so we have found our factorisation:

$5 {x}^{8} + 4 {x}^{7} + 20 {x}^{5} + 42 {x}^{4} + 20 {x}^{3} + 4 x + 5$

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

The zeros of the second quartic will be the reciprocals of the first one.

Jun 11, 2016

Part Two - Solving the first quartic

Find zeros of ${x}^{4} + 4 x + 5$ by an algebraic method...

#### Explanation:

Part Two - Solving the first quartic

$g \left(x\right) = {x}^{4} + 4 x + 5$

$= \left({x}^{2} - A x + B\right) \left({x}^{2} + A x + C\right)$

$= {x}^{4} + \left(B + C - {A}^{2}\right) {x}^{2} + A \left(B - C\right) x + B C$

Equating coefficients and rearranging a little:

$\left\{\begin{matrix}B + C = {A}^{2} \\ B - C = \frac{4}{A} \\ B C = 5\end{matrix}\right.$

So:

${\left({A}^{2}\right)}^{2} = {\left(B + C\right)}^{2} = {\left(B - C\right)}^{2} + 4 B C = \frac{16}{A} ^ 2 + 20$

Hence:

${\left({A}^{2}\right)}^{3} - 20 \left({A}^{2}\right) - 16 = 0$

Let $t = {A}^{2}$

We want to solve:

${t}^{3} - 20 t - 16 = 0$

By the rational root theorem, any rational roots are expressible in the form $\frac{p}{q}$ for integers $p , q$ with $p$ a divisor of the constant term $- 16$ and $q$ a divisor of the coefficient $1$ of the leading term.

That means that the only possible rational roots are:

$\pm 1$, $\pm 2$, $\pm 4$, $\pm 8$, $\pm 16$

We find $t = - 4$ is a zero, so $\left(t + 4\right)$ is a factor:

${t}^{3} - 20 t - 16 = \left(t + 4\right) \left({t}^{2} - 4 t - 4\right)$

$= \left(t + 4\right) \left({\left(t - 2\right)}^{2} - {\left(2 \sqrt{2}\right)}^{2}\right)$

$= \left(t + 4\right) \left(t - 2 - 2 \sqrt{2}\right) \left(t - 2 + 2 \sqrt{2}\right)$

We could use the positive root $t = 2 + 2 \sqrt{2}$ to get a factorisation into quadratics with real coefficients, but the sums may be a little easier by choosing ${A}^{2} = t = - 4$.

Let $A = 2 i$

Then:

$\left\{\begin{matrix}B + C = - 4 \\ B - C = \frac{4}{2 i} = - 2 i\end{matrix}\right.$

$2 B = - 4 - 2 i$

Hence $B = - 2 - i$ and $C = - 2 + i$

So we have found:

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

So using the quadratic formula on the first of these quadratic factors we find:

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

The second quadratic is the Complex conjugate, so we can deduce zeros:

$x = - i \pm \sqrt{1 + i}$

Next use the formula I derived in https://socratic.org/s/avemtNYy , which simplified for square roots in Q1 becomes:

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

So:

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

$= \left(\sqrt{\frac{\sqrt{2} + 1}{2}}\right) + \left(\sqrt{\frac{\sqrt{2} - 1}{2}}\right) i$

Hence (using the normal definition of $A r g \left(z\right) \in \left(- \pi , \pi\right]$):

$\sqrt{1 - i} = \left(\sqrt{\frac{\sqrt{2} + 1}{2}}\right) - \left(\sqrt{\frac{\sqrt{2} - 1}{2}}\right) i$

So:

$i \pm \sqrt{1 - i} = i \pm \left(\left(\sqrt{\frac{\sqrt{2} + 1}{2}}\right) - \left(\sqrt{\frac{\sqrt{2} - 1}{2}}\right) i\right)$

$- i \pm \sqrt{1 + i} = - i \pm \left(\left(\sqrt{\frac{\sqrt{2} + 1}{2}}\right) + \left(\sqrt{\frac{\sqrt{2} - 1}{2}}\right) i\right)$

So the four zeros in $a + b i$ form are:

${x}_{1} = \left(\sqrt{\frac{\sqrt{2} + 1}{2}}\right) - \left(1 + \sqrt{\frac{\sqrt{2} - 1}{2}}\right) i$

${x}_{2} = - \left(\sqrt{\frac{\sqrt{2} + 1}{2}}\right) - \left(1 - \sqrt{\frac{\sqrt{2} - 1}{2}}\right) i$

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

${x}_{4} = - \left(\sqrt{\frac{\sqrt{2} + 1}{2}}\right) - \left(1 + \sqrt{\frac{\sqrt{2} - 1}{2}}\right) i$