Roots of a quintic function

Question

I need some pointers in the right direction for this question:

Three of the roots of the equation

$ax^5+bx^4+cx^3+dx^2+ex+f=0$

are $-2$, $2i$ and $1+i$. Find $a$, $b$, $c$, $d$, $e$ and $f$.

I know that if $-2$, $2i$, and $1+i$ are the roots of the equation, then the factors of the equation must be $(x+2)$, $(x-2i)$ and $(x-(1+i))$.

I am assuming that another root of the equation must be $1-i$ (therefore another factor must be $(x-(1-i))$), to make a complex conjugate pair.

I have tried to expand out

$(x+2)(x-2i)(x-(1+i))(x-(1-i))$ = $x^4-2x^3i+4ix^2+8ix+2x^2-8i$

but I am unsure of what to do next (and I'm unsure if that's the correct expansion!). Can I set this equal to my quintic polynomial above? Is there a quicker way to work this out?

@gammatester: Don't they have to be real since this is a fifth degree equation and we know there are 4 complex roots (2 conjugates of the two stated) and a real one? — Sheheryar Zaidi, Sep 16 '14 at 08:40
@Jack: It has to specify! Otherwise the question is impossible. — TonyK, Sep 16 '14 at 08:41
@Sheheryar Zaidi: This is about complex numbers, why should they be real? If there are no restrictions, you can add just two other real or complex numbers as roots. — gammatester, Sep 16 '14 at 08:42
@TonyK That is the question straight from the textbook. It does not say if the coefficients are real. — Jack G, Sep 16 '14 at 08:42
@gammatester: Of course, I see what you mean. Then, shouldn't we assume that they are real, because as TonyK said, it's impossible to solve otherwise. — Sheheryar Zaidi, Sep 16 '14 at 08:43
@JackGriffiths: What is the source? Which book are you using? — Sheheryar Zaidi, Sep 16 '14 at 08:44
[http://www.amazon.co.uk/gp/product/0435519239/ref=pd_lpo_sbs_dp_ss_1?pf_rd_p=479289247&pf_rd_s=lpo-top-stripe&pf_rd_t=201&pf_rd_i=0199117837&pf_rd_m=A3P5ROKL5A1OLE&pf_rd_r=00C358VHRCNN3WVEJ1BT](It is this book) (you can click on the cover and search for 'Three of the roots of'. Question 11, Page 29. — Jack G, Sep 16 '14 at 08:47
I checked this link: book chapter pdf and clearly, they are dealing only with real roots. — Sheheryar Zaidi, Sep 16 '14 at 08:54
@SheheryarZaidi: I think you mean "Clearly, they are dealing only with real coefficients." The roots can be complex. — TonyK, Sep 19 '14 at 11:53
Yes, that's what I meant, wasn't able to edit the comment for some reason! — Sheheryar Zaidi, Sep 19 '14 at 12:02
In my opinion, that is not a super-well written book. On p. 24, it does state for skill 1.8: "You can solve some types of polynomial equations with real coefficients", and then I guess you're supposed to take that for granted throughout the rest of the section and exercises without further comment. — Daniel R. Collins, Dec 16 '15 at 19:57

score 3 · Accepted Answer · answered Sep 16 '14 at 08:37

3

Hint: expand this $$(x+2)(x-2i)(x+2i)(x-(1+i))(x-(1-i))$$

answered Sep 16 '14 at 08:37

Sheheryar Zaidi

2,714

1

Yes, this is correct. Expanding this will give the values of a=1, b=0, c=2, d=4, e=-8 and f=16 (assuming a,b,c,d,e and f are real). – Jack G Sep 16 '14 at 08:52

score 0 · Answer 2 · answered Dec 14 '15 at 14:05

Substituting the three roots in the polynomial expression, you get the system

$$\begin{align}2\to&&-32a+16b-8c+4d-2e+f&=0\\ 2i\to&&32ia+16b-8ic-4d+f+2ei&=0\\ 1+i\to&&(-4-4i)a-4b+(-2+2i)c+2id+f+e(1+i)&=0\end{align}\\ $$ or, separating the real and imaginary parts,

$$\begin{align}-32a+16b-8c+4d-2e+f=0\\ 16b-4d+f=0\\ 32a-8c+2e=0\\ -4a-4b-2c+e+f=0\\ -4a+2c+2d+e=0\end{align}\\ $$

A solution of this linear system is

$$b=0, c=2a, d=4a, e=-8a, f=16a.$$

This solution is much less efficient than direct multiplication. — , Dec 14 '15 at 14:07

Roots of a quintic function

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