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The question that I have trouble solving is the following:

Factor the polynomial $z^5 + 32$ in real factors. The answer should not use trigonometric functions. (Hint: you are allowed to use the fact that: $cos(\pi/5) = \frac{1+\sqrt{5}}{4}$ and $cos(3\pi/5) = \frac{1-\sqrt{5}}{4}$.

In the previous question you were supposed to find the complex roots to the equation $z^5 = -32$ in polar form, resulting in the roots:

$$z_1 = 2e^{i\frac{\pi}{5}}$$$$z_2 = 2e^{i\frac{3\pi}{5}}$$ $$z_3 = 2e^{i\pi} = -2$$ $$z_4 = 2e^{i\frac{7\pi}{5}}$$$$ z_5 = 2e^{i\frac{9\pi}{5}} $$

My attempt at a solution:

First we know that there were only one real root to the equation $z^5 = -32$, namely $-2$.

So if we write the polynomial as the equation: $z^5 + 32 = 0$ we know that $(z+2)$ must be a factor and we have four potential factors left to find. Since we know that the polynomial $z^5 +32$ only has real coefficients, the non real roots must come in pairs.

We should then find two roots that are composed of conjugates of the roots with imaginary components to produce the other two real roots... But I am stuck here.

Lukas Arvidsson
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  • The n complex nth roots of a number all have the same magnitude and are arranged regularly around the zero-centred circle in the complex plane with that magnitude. – Joffan Jan 09 '15 at 16:24
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    For a method that only involves quadratic equations and algebraic substitutions (no trigonometry, no complex exponentials), see my answer to The roots of $t^5+1$. (Added moments later) Well, actually I find the complex roots this way. But you might be able to use the ideas (and certainly the roots, if all else fails) to get the complete factorization over the reals. And indeed, it appears egreg has done this. – Dave L. Renfro Jan 09 '15 at 16:53

4 Answers4

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Set $t=2z$, for the moment. Then \begin{align} z^5+32=32(t^5+1) &=32(t+1)(t^4-t^3+t^2-t+1)\\ &=32(t+1)t^2\left(t^2+\frac{1}{t^2}-\left(t+\frac{1}{t}\right)+1\right)\\ &=32(t+1)t^2\left(\left(t+\frac{1}{t}\right)^{\!2}-\left(t+\frac{1}{t}\right)-1\right) \end{align} Consider $u^2-u-1=(u-\alpha)(u-\beta)$, where $$ \alpha=\frac{1+\sqrt{5}}{2},\qquad\beta=\frac{1-\sqrt{5}}{2} $$ and so $$ z^5+32=32(t+1)t^2\left(t+\frac{1}{t}-\alpha\right)\left(t+\frac{1}{t}-\beta\right) $$ which is to say $$ z^5+32=32(t+1)(t^2-\alpha t+1)(t^2-\beta t+1)= (z+2)(z^2-2\alpha z+4)(z^2-2\beta z+4) $$

egreg
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You found the roots so $z^5+32=(z+2)(z-z_1)(z-z_2)(z-z_4)(z-z_5)$.

Notice that $cos(x)=\dfrac{e^{ix}+e^{-ix}}{2}$ and that $z_1z_5\in\mathbb{R}$ and $z_2z_4\in\mathbb{R}$

Scientifica
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  • Thank you for your answer! Just a quick question: How do you see that the development will be $z^5 + 32 = (z+2)(x - \alpha)(z - \overline {\alpha})(z - \beta)(z- \overline {\beta})$ and not $z^5 + 32 = (z+2)(z + \alpha)(z + \overline {\alpha})(z + \beta)(z + \overline {\beta})$? – Lukas Arvidsson Jan 09 '15 at 20:00
  • You're welcome. If $\alpha$ is a root of a polynomial $P$ then $(z-\alpha)|P$. $-2$ is a root that's why we have $(z-(-2))=(z+2)$. – Scientifica Jan 10 '15 at 10:19
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HINT:

$$a^5+b^5=(a+b)(a^4-a^3b+a^2b^2-ab^3+b^4)$$

$$a^4-a^3b+a^2b^2-ab^3+b^4=(a^2+b^2)^2-ab(a^2+b^2)-a^2b^2$$

If $ab\ne0,$

$$(a^2+b^2)^2-ab(a^2+b^2)-a^2b^2=a^2b^2\left[\left(\frac{a^2+b^2}{ab}\right)^2-\frac{a^2+b^2}{ab}-1\right]$$ So, we can express $a^2+b^2$ in terms of $ab$

lab bhattacharjee
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$z^2 + 32 = (z+2)(x - \alpha)(z - \overline {\alpha})(z - \beta)(z- \overline {\beta})$ where $\overline {\alpha}, \overline {\beta}$ are complex conjugate of $\alpha, \beta$ respectively. Also $(z - \alpha)(z - \overline {\alpha})$ and $(z - \beta)(z- \overline {\beta})$ are real polynomials (why?). Now choose $\alpha = z_1, \beta = z_2.$ Then $\overline {\alpha} = z_5, \overline {\beta} = z_4.$

Krish
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  • Thank you for your answer! Shouldn't the equality be: $z^5 + 32 = (z+2)(z + \alpha)(z + \overline {\alpha})(z + \beta)(z + \overline {\beta})$?

    Now lets start with $(z + \alpha)(z + \overline {\alpha})$. This could be written as: $(z + 2e^{i\frac{\pi}{5}})(z + 2e^{i\frac{-\pi}{5}})$.

    If I try to develop this I get:

    $z^2 + 2z(e^{i\frac{\pi}{5}} + e^{i\frac{-\pi}{5}}) + 4(e^{i\frac{\pi}{5}})(e^{i\frac{-\pi}{5}})$ How do I continue from there?

     – Lukas Arvidsson Jan 09 '15 at 18:34
  • the complex roots occurs in conjugate pair. so if $\alpha$ is a complex root then $\overline{\alpha}$ is also a root. now $\alpha$ is root means $z - \alpha$ is factor. similarly $z - \overline{\alpha}$ is also a root. (there was a mistake. the first factor should be $z+2,$ not $z-2$ as I wrote previously. now fixed.) – Krish Jan 09 '15 at 18:42
  • Thank you for your comment, if you would have time to do the actual calculations I would be very happy (I am not sure that I understand this correctly yet...) – Lukas Arvidsson Jan 09 '15 at 18:51
  • $(z - \alpha)(z - \overline {\alpha}) = z^2 -z(\alpha + \overline {\alpha}) + \alpha \overline{\alpha}.$ Now $\alpha + \overline {\alpha} =2 \cos(\frac{\pi}{5}) = 2 \cdot \frac{1 + \sqrt 5}{4}, \alpha \overline {\alpha} = 4.$ similarly for $(z - \beta)(z - \overline {\beta}).$ Also the expression you wrote: $z^5 + 32 = (z+2)(z + \alpha)(z + \overline {\alpha})(z + \beta)(z + \overline {\beta})$ is not correct. If you simplify right hand side, you will get something else. there will be a $z^3$ term. – Krish Jan 09 '15 at 19:02
  • BTW, how do you see that the development will be $z^5 + 32 = (z+2)(x - \alpha)(z - \overline {\alpha})(z - \beta)(z- \overline {\beta})$ and not $z^5 + 32 = (z+2)(z + \alpha)(z + \overline {\alpha})(z + \beta)(z + \overline {\beta})$? – Lukas Arvidsson Jan 09 '15 at 19:59
  • it's a general fact: if $p(x)$ is a complex polynomial of degree $n \geq 1,$ then it has exactly $n$ roots (counting with multiplicity) and if $\alpha_1, \alpha_2, \cdots , \alpha_n$ are all the roots (not necessarily distinct) then $p(x) =a \prod_{1 \leq i \leq n}(x - \alpha_i),$ where $a$ is the coefficient of the highest degree term of $p(x).$ This is called Fundamental Theorem of Algebra. See http://en.wikipedia.org/wiki/Fundamental_theorem_of_algebra – Krish Jan 10 '15 at 04:23
  • Thank you very much for your time and effort!!! This has really helped my understanding! – Lukas Arvidsson Jan 10 '15 at 08:21
  • @LukasArvidsson: You are welcome. :) – Krish Jan 10 '15 at 12:54