# Factoring over a set of complex numbers

Ahem. I can only now laugh at how blind I am to not see such a simple answer, LMAO. Thanks to Isaac ♦, Paul VanKoughnett, and Bill Dubuque!

Case closed!

How would you factor $6x^2 + 18x - 60$ over a set of complex numbers?

When I did it myself, I had $6(x - 1.5 + 3.25i)(x - 1.5 - 3.25i)$ .

I am not 100% sure about my answer, so I am asking anyone who knows how to do this to help me confirm the answer! (:

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Well, first thing that I did was to factor out 6. $6(x^2 + 3x - 10)$ , then I changed the expression into a perfect square, so : $6(x^2 - 3x + 2.25) - 12.25$ .

After that: $6(x - 1.5)^2 + (-12.25)$ .

Then I took the square roots of that, so I got:

$6(1 - 1.5)^2 + 3.5i)$ .

Thus my answer: $6(x - 1.5 + 3.25i)(x - 1.5 - 3.25i)$

And I see that it is wrongggggg~

My teacher did not go over it much; she was rushing throughout the whole lesson and her accent does not make it any easier. The textbook does not have anything on it. I looked around the index and there is nothing. -_____-

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The answer is wrong since $\rm\ 3.25^2 \ne 10\:\$ It appears that you made the error of computing the discriminant as $\rm\ 9^2 - 24\cdot 60\ = 39^2\$ instead of $\rm\ 18^2 -24\cdot 60 = 42^2\:$.

Simpler: apply the quadratic formula to $\rm\ p(x)/6 = x^2 + 3x -10\:$, or apply the Rational Root test.

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Polynomials over the complex numbers still have the unique factorization property, so if a polynomial has a real factorization, that will also be its complex factorization. This also means that you can go about factorizing in the same way you would over the reals. The ways you already know for quadratics (quadratic formula, completing the square, etc.) still work fine, except now you keep complex answers instead of throwing them out.

To check your answer, just multiply the factors back together, using complex multiplication. In this case, your answer isn't right, since $(1.5-3.25i)(1.5+3.25i)=1.25^2+3.25^2=12.8125$. In fact, this polynomial has real roots, which I'm sure you can find if you look at $6(x^2+3x-10)$ for a while.

(As a reminder, a quadratic has real roots if and only if its discriminant is nonnegative.)

EDIT: appears somebody else got to the answer first.

EDIT 2 in light of your most recent edit:

I don't think you really understand completing the square. That's actually okay, since you can do everything with the quadratic formula, but at your level in school you'll probably get tested on that specific technique, so you might as well learn it. Also, it's how you derive the quadratic formula in the first place.

As far as I can see, you made three mistakes:

First, you had $x^2-3x-10$ instead of $x^2+3x-10$.

Second, you forgot some parentheses when you completed the square. $$6(x^2-3x-10)=6((x^2-3x+2.25)-12.25)=6(x-1.5)^2-73.5.$$

Third, "taking the square root" didn't really work out. Here, it's better to work with the equation $f(x)=0$ than just the expression $f(x)$. Then, you can move the $12.25$ to the other side of the equation and take the square root, giving $\pm 3.5$, not $\pm 3.5i$. (Also, you changed $3.5i$ to $3.25i$ on the way back.) When you add it back in, this gives you the expected answer, $6(x-5)(x+2)$.

That isn't the answer to the equation the first way you wrote it, but I can't tell which is the right equation.

I guess the moral is that you can check your work at nearly any step of these types of problems. If you do this frequently, you can avoid little mistakes like these. Good luck!

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I've fixed the errors, thanks! – Paul VanKoughnett Oct 5 '10 at 23:42
LOL, I do not have those mistakes in my notes.... Typos, sorry! xP – Danielle Oct 6 '10 at 1:01

\begin{align} 6x^2+18x-60&=6(x^2+3x-10) \\ &=6(x+5)(x-2) \end{align}

When I expand yours: $$6(x - 1.5 + 3.25i)(x - 1.5 - 3.25i)=6x^2-18x+76.875$$

So, I'm not entirely sure what went wrong. How did you arrive at your answer?

edit: Given your work, it's worth having another look. You factored out the 6, as I did. Then, you went to complete the square, which is reasonable, but you somehow went from $+3x$ to $-3x$ and there are some missing/misplaced parentheses: \begin{align} 6(x^2+3x-10)&=6(x^2+3x+2.25-12.25) \\ &=6((x+1.5)^2-12.25) \\ &=6((x+1.5)^2-(3.5)^2) \\ &=6(x+1.5+3.5)(x+1.5-3.5) \\ &=6(x+5)(x-2) \end{align}

So, to summarize, the ideas you tried to use would have worked, but there were one or more instances where something went wrong with +/-, and you should be careful with parentheses and the factored-out leading coefficient (the 6).

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OOPS.... I went a bit too far then... Thanks for pointing it out! I should have seen that I could have just factored that out easily, but I overlooked it, LMAO. Ty~ – Danielle Oct 5 '10 at 23:43
@Danielle: I've edited my answer to address the work you added to your question. Please let me know if there are any steps that do not make sense to you. – Isaac Oct 6 '10 at 0:13
What you did makes perfect sense. Again, thank you! – Danielle Oct 6 '10 at 0:54