Synthetic Division: Calculations with Complex Numbers
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Why do complex numbers show up in synthetic division?
When using synthetic division to factor a polynomial, you will sometimes be given an initial root that is a complex number. These roots do not "just show up"; instead, the author of the exercise constructed a quadratic factor for the polynomial which itself had complex-valued roots.
In other words, and in practical terms, you'll never come up with a complex-valued test zero; the Rational Roots Test only provides real-valued test zeroes. You'll only ever work with complex zeroes if they give you one and tell you to start with that. You'll only ever (on your own) find zeroes containing the imaginary i after you've found all the regular roots, and find yourself left with that quadratic factor that the author had used for his complex factors.
What do you do, after dividing out the complex factor?
When you have been provided with a complex number for one of the zeroes of a polynomial, and after you've divided out that factor, your next step will be to divide out the conjugate. That is, when they've given you a + bi as one zero, then your next step will be to divide out a − bi. (Recall that the conjugate is the exact same two terms, but the sign in the middle has been flipped.)
- Given that 2 − i is a zero of x5 −6x4 + 11x3 − x2 − 14x + 5, fully solve the equation x5 − 6x4 + 11x3 − x2 − 14x + 5 = 0.
I will do my work carefully, keeping in mind the properties of complex numbers. In particular, I need to remember that i2 = −1.
I do the first couple of steps:
Now I need to multiply to get the entry that goes below the 11:
I add down:
Again, I need to multiply to get the entry that goes below the −1:
I add down again:
I multiply to get the entry that goes below the −14:
I add down:
I multiply to get the entry that goes below the 5:
I add down:
And now I'm ready for the next division, which works out like this:
Note that, since since this second synthetic division handled the conjugate complex root, all the complex coefficients disappeared. You should expect this to happen. Whenever you have roots that are conjugates, dividing out one of those roots will make things very messy, but then dividing out the other root will clean things back up.
If you want to get the right answers, do not try to do the messier parts in your head or in the margins; take out a sheet of scratch paper and do your work properly.
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