Lesson: Roots of Polynomial Equations

Question 2

1 point


Another use of the complex conjugate is seen in the following theorem. Let p(x) = an(x-to-the-n) + through to a1x + a0, where our aj are all real numbers for j from 1 to n. Then if z the complex number is a root of p(x), then the conjugate of z is also a root of p(x). So note that, in this theorem, p(x) is a polynomial, but its coefficients are from the real numbers, not the complex numbers. This theorem is most certainly not true of polynomials with complex coefficients.

The proof of this theorem is actually quite simple, using properties of the complex conjugate. We’ll simply start by supposing that z is a root of p(x). Well, then we know that by plugging in z to p(x), we will get 0. So this means that an(z-to-the-n) + through to a1z + a0 = 0. Now, we had to restrict our aj to be real numbers, but of course, real numbers are also complex numbers, so we can still use properties of the complex conjugate on them as well. And so, if we plug z-conjugate into p, then simply we get an((conjugate of z)-to-the-n) + through to a1(conjugate of z) + a0. Now, we have a property of complex conjugation that says that (the conjugate of z)-to-the-n is (z-to-the-n)-conjugate, so we’ll make use of that for our next step. Then there’s also a property that says that (z-conjugate)(w-conjugate) would equal (zw)-conjugate, so that we can distribute the conjugate over our product. Keeping in mind that, with our real numbers, they are their own complex conjugate, we see that (an(z-to-the-n))-conjugate + through to (a1z)-conjugate + a0-conjugate is also still p(z-conjugate). But we also have a property that allows us to distribute over a plus sign. That is to say that w-conjugate + z-conjugate = (w + z)-conjugate. So now we can take all of our individual conjugates and instead replace it with one big conjugate over the value (an(z-to-the-n) + through to a1z + a0). But of course, our assumption was that z is a root of p(x), and so we know that this value under the conjugate symbol is a 0. And of course, being a real number, the conjugate of 0 is still 0. And so we’ve seen that the conjugate of z is a root of p(x).

So we can use this fact to help us factor polynomials with real coefficients, and find all of their roots in the complex numbers.

So for example, let’s factor the polynomial p(x) = x-squared + 9, given that 3i is one of the roots. Well, from our theorem, we know that if 3i is a root, then the conjugate of 3i, which is -3i, is also a root. Well, that means that both (x – 3i) and (x + 3i) are factors of x-squared + 9. And since (x – 3i)(x + 3i) = x-squared + 3ix – 3ix – 9(i-squared), which equals x-squared + 9, we see that x-squared + 9 factors as (x – 3i)(x + 3i).

Let’s look at another example. Let’s factor p(x) = x-cubed – 9(x-squared) + 36x – 54, given that (3 + 3i) is one of the roots. Well, from our theorem, we know that if (3 + 3i) is one of the roots, then the conjugate of (3 + 3i), which equals 3 – 3i, is also a root. This means that both (x – 3 – 3i) and (x – 3 + 3i) are factors of p, so we know that their product is also a factor of p. We see that (x – 3 – 3i)(x – 3 + 3i) = x-squared – 3x + 3ix – 3x + 9 – 9i – 3ix + 9i – 9(i-squared), which is equal to x-squared – 6x + 18. If we divide p(x) by x-squared – 6x + 18, we get (x – 3). And so we have that p(x) = (x – 3 – 3i)(x – 3 + 3i)(x – 3).

Now, could we have started by simply dividing p(x) by (x – 3 – 3i)? Well, yes, and we would have found that p(x) = (x – 3 – 3i)(x-squared + (-6 + 3i)x + (9 – 9i)). We can even use the quadratic formula to factor the remaining degree-2 polynomial. The work would look something like this. This, of course, confirms our earlier result.

Here’s another example that combines both techniques. Let’s factor p(x) = x-to-the-fourth + 4(x-cubed) + 6(x-squared) + 4x + 5, given that i is one of the roots. Well, since i is one of the roots, -i is also one of the roots, and so we know that both (x – i) and (x + i) are factors of this polynomial. As such, their product (x – i)(x + i), which equals x-squared + ix – ix – i-squared, which is x-squared + 1, is a factor of p. If we divide p(x) by x-squared + 1, we get x-squared + 4x + 5. Now let’s plug this into the quadratic formula to find its roots. Recall that the quadratic formula says the roots will be (-b plus-or-minus the square root of (b-squared – 4ac))/(2a). In this case, our b is 4, our a is 1, and our c is 5. Under the square root symbol, we’ll see that we get a (16 – 20), which is a -4. Well, we can think of the square root of -4 as the square root of ((-1)(4)). Of course, the square root of -1 is an i, and the square root of 4 is a 2. Now dividing by the 2 on the bottom, we see that the roots are (-2 + i) and (-2 – i). And you’ll note that, of course, these two complex roots are conjugates of each other, as they would have to be. And so we have found that p(x) factors into (x – i)(x + i)(x + 2 – i)(x + 2 + i).

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