# Solving $z^4 + 2z^3 + 6z - 9 = 0$

I'm trying to solve $z^4 + 2z^3 + 6z - 9 = 0$.

$z$ is a complex number.

I usually can solve those equations when they are of second degree.

I don't know what to do, breaking out $z$ doesn't help...

EDIT: Sorry I forgot to mention that $z$ has a solution where $\Re(z) = 0$.

• There is a formula for equations of (up to) fourth degree, but you don't want to use it unless you have to. If you can guess a solution $a$ by trying (like $z=1$ in this case :-) you can divide the polynomial by $(z-a)$ (polynomial division), resulting in a product $(z-a) p(z)$. The degree of $p$ is then smaller than the degree of the original polynomial, and with some luck you can continue like this till you end up with $(z-a)(z-b) q(z)$ with $q$ of degree 2....
– user20266
Commented Dec 19, 2011 at 17:21

As mentioned in comments, $z=1$ is one of the roots (always check whether 1 or -1 is one of the roots, it's worth it). So, to calculate other factors, you should calculate: $$\frac{z^4+2z^3+6z-9}{z-1}$$

Which would be $z^3+3z^2+3z+9$ by polynomial division. Now, it's easy to factorize $(z+3)$. $$z^2(z+3)+3(z+3)=(z+3)(z^2+3)$$

To satisfy the condition $\Re(z) = 0$, you should find the values of $z$ in $z^2+3$ ($z=\pm i \sqrt3$).

• Did you mean $(z+3)$ instead of $(z-3)$?
– user6701
Commented Dec 25, 2011 at 17:55
• @Tim: Right, I made a small mistake there. Thank you. Commented Dec 25, 2011 at 19:11

The complex number $z = c$ is a solution to the equation $f(z) = 0$ (also called a zero or root of $f(z)$) if and only if $(z-c)$ is a factor of $f(z)$. The proof of this statement follows easily by adapting the Euclidean algorithm which concerns finding the greatest common divisor of integers to finding the greatest common divisor of functions. A proof of this statement would be easy enough that I could add it to the post if you like.

At any rate, use this idea and try to find a root to the equation $z^4 +2z^3 + 6z -9 = 0$. Finding a root then gives a factor, and reduces the degree of the polynomial you want to factor by $1$.

Now there are many ways to find the zeroes of $f(z)$. In the simplest cases, you can simply guess and check for solutions. The obvious candidates to check for zeroes are $z = 1, z = -1, z = 2, z= -2$, etc. Note that $z = 0$ is a zero if and only if $f(z)$ has no constant term. One of the more useful tests for finding roots of polynomials is the Rational Root Test. Other known methods for finding zeroes of polynomials include completing the square (for quadratic functions), using the cubic formula, or using the quartic formula, though the cubic and quartic formulae are a bit unwieldy in practice.

• Edited my question. Commented Dec 19, 2011 at 17:39
• @Dimme, did you forget to say "thank you"? Commented Dec 19, 2011 at 21:23

Hint: are there any rational roots? If $z=\frac{n}{m}$ is a rational root in reduced form, then $n$ must be a divisor of $\pm 9$, and $m=1$.

• Edited my question. Commented Dec 19, 2011 at 17:39
• Your edit does not change my answer. As Javaman points out, if $a$ is a root, then $z-a$ is a factor of your polynomial. In this case, you can use my hint to find two rational roots $a$ and $b\in\mathbb{Q}$, and then factor your polynomial $p(z)$ as $(z-a)(z-b)q(z)$, where $q(z)$ is quadratic (and therefore easy to factor). Commented Dec 19, 2011 at 18:56

The given information means there is a solution of the form $it$ with $t$ real. Then $-it$ is also a solution (since all coefficients are real). Plugging these two solutions in and using a little algebra you should be able to solve for $t$ and therefore find a quadratic factor $z^2 + t^2$.

I guess the highroad here is to recognise the factor $z^2+3$:

$(z^4-9) + (2z^3 + 6z) = (z^2+3)(z^2-3) + (z^2+3)(2z) = (z^2+3)(z^2+2z-3)$ $= (z+i\sqrt{3})(z-i\sqrt{3})(z-1)(z+3)$

• And is there some align* or eqnarray environment possible with mathjax? My attempts were futile. Commented Dec 19, 2011 at 20:45
• assuming all the algebra before is correct, you factored $z^2 + 2z - 3$ incorrectly. It should be $(z-1)(z+3)$. Commented Dec 19, 2011 at 21:19
• Would you really have to assume that? :P But I editted the signs, thanks. Commented Dec 19, 2011 at 21:25

The Quartic Formula can be very useful to solve this. But there is an alternate way, which my lecturer suggested me : Summing and Subtracting some powers of $z$ to make expression factorable. For example $z^4+2z^3+6z-9=0$ can be factorized as:

$z^4+(3z^3-z^3)+(3z^2-3z^2)+(9z-3z)-9=0$ $(z^4+3z^3+3z^2+9z)-(z^3+3z^2+3z+9)=0$ $z(z^3+3z^2+3z+9)-(z^3+3z^2+3z+9)=0$ $z(z^2(z+3)+3(z+3))-(z^2(z+3)+3(z+3))=0$ $(z-1)(z+3)(z^2+3)=0$.

Thus, $z^2+3=0$ or $z=\pm i \sqrt{3}$ are the required solutions for your given conditions.

• This method seems very interesting but I don't quite understand how you pick your numbers. Do you want to explain? Commented Dec 19, 2011 at 18:15
• I think, I am not a good person to explain it. This is matter of practice. Point that is kept in mind: > You have to break whole expression in two or more parts each having 'some terms' common. Commented Dec 19, 2011 at 18:24