# How do you find the zeroes for f(x)=x^5+3x^3-x+6?

Dec 6, 2015

Use a graphing calculator.

#### Explanation:

Use a graphing calculator.

graph{x^5+3x^3-x+6 [-15.09, 16.95, -5.13, 10.89]}

There's only one zero: $\left(- 1.178 , 0\right)$

Dec 6, 2015

Use Newton's method to find Real root:

$x \approx - 1.178259848$

and Complex roots:

$x \approx - 0.20255440359096 \pm 1.893126314652 i$

$x \approx 0.79168432752598 \pm 0.882050595027 i$

#### Explanation:

$f \left(x\right) = {x}^{5} + 3 {x}^{3} - x + 6$

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

Starting with a first approximation ${a}_{0}$, iterate using the formula:

${a}_{i + 1} = {a}_{i} - \frac{f \left({a}_{i}\right)}{f ' \left({a}_{i}\right)}$

Putting this into a spreadsheet, choosing ${a}_{0} = - 1$, we find:

${a}_{0} = - 1$

${a}_{1} = - 1.230769231$

${a}_{2} = - 1.181552182$

${a}_{3} = - 1.178273619$

${a}_{4} = - 1.178259848$

${a}_{5} = - 1.178259848$

The same method can be applied to find the $4$ Complex roots, using Complex arithmetic and Complex initial approximations.

Since my spreadsheet application does not support Complex arithmetic directly, I created columns for $R e \left(z\right)$, $I m \left(z\right)$, $R e \left({z}^{2}\right)$, $I m \left({z}^{2}\right)$,.., $R e \left({z}^{5}\right)$, $I m \left({z}^{5}\right)$, $R e \left(f \left(z\right)\right)$, $I m \left(f \left(z\right)\right)$, $R e \left(f ' \left(z\right)\right)$, $I m \left(f ' \left(z\right)\right)$, $R e \left(f \left(z\right) \overline{f ' \left(z\right)}\right)$, $I m \left(f \left(z\right) \overline{f ' \left(z\right)}\right)$, $R e \left(f ' \left(z\right) \overline{f ' \left(z\right)}\right)$, $I m \left(f ' \left(z\right) \overline{f ' \left(z\right)}\right)$.

With an initial approximation of $0 + 2 i$ I saw the following approximations:

${a}_{0} = 0 + 2 i$

${a}_{1} = - 0.13953488372093 + 1.860465116279 i$

${a}_{2} = - 0.21027440054555 + 1.889317276079 i$

${a}_{3} = - 0.20248503148821 + 1.893034069471 i$

${a}_{4} = - 0.20255442138413 + 1.893126325251 i$

${a}_{5} = - 0.20255440359096 + 1.893126314652 i$

${a}_{6} = - 0.20255440359096 + 1.893126314652 i$

With an initial approximation of $1 + i$ I saw the following approximations:

${a}_{0} = 1 + i$

${a}_{1} = 0.83921568627451 + 0.909803921569 i$

${a}_{2} = 0.79458495418768 + 0.883839764146 i$

${a}_{3} = 0.79169561531938 + 0.882058297006 i$

${a}_{4} = 0.79168432769581 + 0.882050595168 i$

${a}_{5} = 0.79168432752598 + 0.882050595027 i$

${a}_{6} = 0.79168432752598 + 0.882050595027 i$

Note that Complex roots occur in conjugate pairs so the $4$ Complex roots are approximately:

$- 0.20255440359096 \pm 1.893126314652 i$

$0.79168432752598 \pm 0.882050595027 i$