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I have a simple question here but I don't know how to solve it.

Given that $i$ is a root of: $P(x)=x^4 + 2x^3+ 3x^2 + 2x+2$ find all the other roots.

Help is much appreciated. Thanks.

amWhy
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Jason
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    Recall that if a complex number is a root of a polynomial, its complex conjugate must also be a root. Furthermore, as this polynomial is quartic, it has at most four distinct complex roots. You have two roots. Can you take it from here? – Lost Jul 08 '14 at 19:43
  • We know that $i$ and $-i$ are roots and we have $P(x) = (x^2+1)\cdot (ax^2+bx+c)$. It is immediate that $a=1$ and $c=2$. We also know that $b=2$. To factor $x^2+2x+2$, we know that $z$ and its conjugate are solutions and that there sum is $-2$. So the real part is $-1$. Then their product (or the square of their modulus) is $2$ and so the imaginary part is $\sqrt{2-1} = +-1$. Then the other roots are $1+i$ and $1-i$. – Christopher K Jul 08 '14 at 19:58
  • @Lost: to be pedantic, if a complex number is a root of a polynomial with real coefficients, then […] – Ben Millwood Jul 08 '14 at 22:18

6 Answers6

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Since $i$ is in fact a root of

$P(x)=x^4 + 2x^3+ 3x^2 + 2x+2 \tag{1}$

(which indeed may easily be checked), and $P(x) \in \Bbb R[x]$ has real coefficients, $-i$ is also a root; thus $(x + i)(x - i) = x^2 + 1$ must divide $P(x)$. We see, by synthetic division of polynomials, that

$P(x) = x^4 + 2x^3+ 3x^2 + 2x+2 = (x^2 + 1)(x^2 + 2x + 2), \tag{2}$

so the remaining roots must satisfy

$x^2 + 2x + 2 = 0; \tag{3}$

Now the quadratic formula yields

$r_\pm = \dfrac{1}{2}(-2 \pm \sqrt{2^2 - 4*2}) = -1 \pm i; \tag{4}$

the roots are thus $\pm i$, $-1 \pm i$.

NOTE added Tuesday 8 July 2014 1:14 PM PST: The division process mimics to a great degree ordinary long division. We start by dividing the leading term of $x^2 + 1$ into $x^4$, the leading term of $P(x)$, obtaining $x^2$; then we form

$P_1(x) = P(x) - x^2(x^2 + 1) = 2x^3 + 2x^2 + 2x + 2; \tag{5}$

repeating the process with $P_1(x)$ in place of $P(x)$ we find that we get $2x$, and

$P_2(x) = P_1(x) - 2x(x^2 + 1) = 2x^2 + 2, \tag{6}$

which is exactly divisible by $x^2 + 1$, yielding $2$; adding up all the intermediate quotients gives $x^2 + 2x + 2$; a lot more can be found here. End of Note.

Hope this helps. Cheers,

and as always,

Fiat Lux!!!

Robert Lewis
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  • @Thanks for the detailed answer. I seem to have trouble coming up with the division you did in step (2). Is there any trick I can use for that? – Jason Jul 08 '14 at 19:58
  • Nice and very exact! – Mohammad Khosravi Jul 08 '14 at 20:00
  • @Jason et al: Thanks gentlemen! I'm adding a description of the division process to my post! – Robert Lewis Jul 08 '14 at 20:02
  • @Jason: There is a famous trick: compare similar monomial on the sides! :) Some of them are very helpful! – Mohammad Khosravi Jul 08 '14 at 20:02
  • @RobertLewis: Thanks a lot. Now I got it. – Jason Jul 08 '14 at 20:24
  • @MohammadKhosravi: I did not understand your trick :) English is not my first language (despite my name ;p) what are monomial? – Jason Jul 08 '14 at 20:25
  • @Jason: glad to help out! And thanks for the "acceptance"! – Robert Lewis Jul 08 '14 at 20:25
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    Note that you can do long division of polynomials very quickly using standard long division techniques—you just have to do it in "base x" instead of base 10. (this is basically the same as what Robert is saying here, it's just another way to look at it) – Andrew Dudzik Jul 08 '14 at 20:40
  • @you-sir-33433: yea thanks, I have been reading up a bit on this, found out about synthetic division and how to use it with quadratic expression. This seems like the quickest way to do it. – Jason Jul 08 '14 at 20:56
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Since the coefficients are real valued, the conjugate root theorem tells you that $-i$ is a root. Now use long division and divide $P(x)$ by $(x+i)(x-i)=x^2+1$. This will give you a quadratic polynomial that you can factor using the quadratic formula or some other technique to find the other roots.

Joel
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As $i$ is a root and all coefficients are real so $-i$ is again a root and $P(x)$ is dividable by $x^2+1$. Therefore $$ P(x) = x^4+2x^3+3x^2+2x+2 = (x^2+1)(x^2+2x^2+2) $$ Now compute the roots of $Q(x) = (x^2+2x^2+2)$.

Mohammad Khosravi
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You have two roots $x =\pm i$ (because conjugate will also be root). Thus $(x-i)(x+i)=x^2+1$ divides your polynomial. Now divide your polynomial by $x^2+1$ and you will get a second degree polynomial which you can factor.

Anurag A
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Note that:

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

agha
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notice that $+i$ must also be a root since the poly has real coefficients. now do polynomial (long) division by $(x+i)(x-i)$ leaving you with a second degree polynomial. use the quadratic formula

Zlatan der Zechpreller
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