# Easy way to factorize $z^4 + 3z^2 - 6z + 10 = 0$

We are given that $$(1 + i)$$ ia a root of the equation $$z^4 + 3z^2 - 6z + 10 = 0$$

What is a straightforward way to find the other factors?

Here's my approach:

$$(z-1-i)(z^3 + az^2 + bz + c ) \equiv z^4 + 3z^2 - 6z + 10 = 0$$

The values that I find for the unknowns through matching the coefficients are:
a = 1 + i
b = 3 + 2i
c = -5 + 5i

But these values for a, b, and c must be wrong because when I plug them into $$(z-1-i)(z^3 + az^2 + bz + c )$$ and expand, I don't get the original equation.

## So my questions really are:

a) How does one factorize $$z^4 + 3z^2 - 6z + 10 = 0$$ given the root $$(1+ i)$$
b) Am I using a false approach, and if so, what is it that is wrong?
c) Using my approach, it gets very complex very soon. So in case it is correct, is there not an easier method?

P.S. I checked my work several times over, and I still don't see the mistake.

• – lab bhattacharjee Jan 6 '17 at 16:13
• Your values for $a,b,c$ are correct, according to WolframAlpha. You must have made a mistake with your multiplication. But as others have already explained, this is not the way to go. – TonyK Jan 6 '17 at 16:28

Remember that complex solutions come in pairs when the coefficients of the polynomial are real, so $z-1+i$ is also a factor. Since $$(z-1-i)(z-1+i)=z^2-2z+2,$$ you can divide $z^4+3z^2-6z+10$ by $z^2-2z+2$ to get a second degree polynomial. Then you can use the usual formula to solve the remaining second degree equation.

• Complex solutions come in pairs when the coefficients of the polynomial are real, to be more precise. (Sorry so nitpicky). Nice solution, +1! – cool.coolcoolcool Jan 6 '17 at 16:48

I'm adding a second answer to this post not because I think it is different in a fundamental way, but because it looks very different and feels very different in the execution. The title in the OP asks for an "easy way" to factorize, given that you know one of the factors already. One handy method is synthetic division, which works even with complex roots: one divides $z^4+3z^2-6z+10$ by $(z-(1+i))$ as follows:

$$\begin{array}{r r r r r r} \begin{array}{r |} 1+i\\ \hline \end{array} & 1 & 0 & 3 & -6 & 10 \\ %\hline & & 1+i & 2i & 1+5i & -10\\ \hline & 1 & 1+i& 3+2i & -5+5i &\begin{array}{| r}0 \\ \hline \end{array} \\ \end{array}$$

If you aren't familiar with the synthetic division algorithm: We work left-to-right, writing the sum of each column below the horizontal line, then multiplying that sum by the number in the top-left box and placing it one cell up and to the right.

The $0$ in the bottom-right cell confirms that $1+i$ is indeed a root of the polynomial, and the rest of the entries in the bottom row tell us that the quotient is $z^3 + (1+i)z^2 + (3+2i)z + (-5+5i)$ (the same as you found in your original solution). Now we have to factor this cubic.

At this point, we invoke the knowledge that complex roots of a polynomial with real coefficients come in conjugate pairs, so $1-i$ must be a root as well. We again use synthetic division on the quotient:

$$\begin{array}{r r r r r } \begin{array}{r |} 1-i\\ \hline \end{array} & 1 & 1+i & 3+2i & -5+5i \\ %\hline & & 1-i & 2-2i & 5-5i \\ \hline & 1 & 2& 5 &\begin{array}{| r}0 \\ \hline \end{array} \\ \end{array}$$

Again the $0$ in the bottom right corner confirms that $1-i$ is indeed a root, and the remaining coefficients in the bottom row tell us that the quotient is $z^2+2z+5$. Now use the quadratic formula to finish it.

• I like this solution. I would always reach for synthetic division by linear factors when it's an option. – Deepak Jan 9 '17 at 1:56

The other answerers have told you the most efficient way to solve the problem, so I will instead focus on the second question:

b) Am I using a false approach, and if so, what is it that is wrong?

In principle there is nothing incorrect with your method, insofar as if $1+i$ is a root of a polynomial $p(z)$, then $(z-1-i)$ is a factor of $p(z)$; so writing $$(z-(1+i))(z^3+az^2+bz+c) = z^4+3z^2-6z+10$$ does indeed lead to a system of equations that can be solved for $a,b,$ and $c$. Specifically one gets $$a-1-i=0$$ $$b -a(1+i)=3$$ $$c -b(1+i) =-6$$ $$-c(1+i) = 10$$ which can be solved to get $$a=1+i$$ $$b=3+(1+i)a=3+(1+i)(1+i)=3+2i$$ $$c=-6 + b(1+i) = -6 +(3+2i)(1+i) = -5 + 5i$$ as you said.

If you plug those coefficients back in and expand you should find that $$(z-(1+i))(z^3 + (1+i)z^2 + (3+2i)z + (-5+5i)) = z^4+3z^2-6z+10$$ as intended. Essentially, you have done synthetic division with the root $1+i$.

However, take a step back. Yes, everything above is correct -- but it is also almost useless, because now you have a third-degree polynomial with complex coefficients, and what are you going to do with that? How will you factor it further?

If you happen to remember the fact that complex roots come in conjugate pairs, you could repeat the method above to divide your complex polynomial by $(z-(1-i))$. You would then find (fortunately) that the quotient has real coefficients, so you can proceed by simpler methods. But all of this is much more complicated than it needs to be, because if you know that both $1+i$ and $1-i$ are roots, then $(z-(1+i))(z-(1-i)) = (z-1)^2 + 1 = z^2 - 2z + 2$ is a factor of your original 4th degree polynomial, and you can proceed from there.