In this video, we’re going to learn about the real and complex roots of polynomials. We’ll see that we can easily tell how many roots a given polynomial has. And we’ll see that in the case of polynomials with real coefficients that if we have one complex root of the polynomial, we can easily find another. We’ll start with the fundamental theorem of algebra. Let’s work up to the statement of this theorem.
Recall that the polynomial 𝑝 of 𝑥 equals 𝑥 squared plus one has no real roots. But we defined 𝑖 to be a nonreal root of this polynomial. 𝑝 of 𝑖 is 𝑖 squared plus one and 𝑖 squared is negative one. So 𝑝 of 𝑖 is zero. We can also see this negative 𝑖 is a root. 𝑝 of negative 𝑖 is negative 𝑖 squared plus one. And writing negative 𝑖 as negative one times 𝑖, we can rearrange the factors to get negative one squared times 𝑖 squared plus one. Negative one squared is one and 𝑖 squared is negative one and their product is negative one. So negative 𝑖 is also a root.
And having found the two roots of the quadratic expression 𝑥 squared plus one, we can factor this quadratic. It’s some constant 𝐴 times 𝑥 minus 𝑖 times 𝑥 minus negative 𝑖. 𝑥 minus negative 𝑖 is just 𝑥 plus 𝑖. And as the coefficient of 𝑥 squared in the definition of 𝑝 of 𝑥 is one, the value of 𝐴 is also one. And so, we don’t need to write this explicitly. Writing 𝑝 of 𝑥 as 𝑥 squared plus one, we see that using complex numbers we can factor the quadratic polynomial 𝑥 squared plus one into a product of linear factors. We couldn’t do this when we were working with just real numbers.
You might know that by completing the square, you can find roots to any quadratic with real coefficients. And so, we can factor any quadratic writing it as the product of linear factors with perhaps a constant multiply as well. The fundamental theorem of algebra vastly generalizes this fact. It says that every polynomial of degree 𝑛 — so we’re not just dealing here with quadratics which have degree two — with complex coefficients — so the coefficients don’t have to be real; they can be, but they don’t have to for the theorem to apply — can be factored into linear factors.
It might be helpful to use some symbols here. A polynomial of degree 𝑛 that’s something of the form 𝑎 subscript 𝑛 times 𝑥 to the power of 𝑛 plus 𝑎 subscript 𝑛 minus one times 𝑥 to the power of 𝑛 minus one and so on all the way down to 𝑎 one 𝑥 plus 𝑎 zero, where we require that the coefficient of 𝑥 to the power of 𝑛 — that’s 𝑎 subscript 𝑛 — is not equal to zero; otherwise, its polynomial isn’t really of degree 𝑛 with complex coefficients. So 𝑎 zero, 𝑎 one, and so on all the way up to 𝑎 𝑛 are complex. They could be real numbers. Any real number is also a complex number. But they don’t have to be. The theorem states that such a polynomial can be factored into linear factors.
We generally like to take a factor of 𝑎 𝑛 out first. So we only have to consider the case when the coefficient of 𝑥 to the power of 𝑛 is one. And then, the coefficient of 𝑥 in each of these linear factors can be one. How many of these linear factors are there? I’ve let this number be 𝑚. But when multiplying all the 𝑥s from the parentheses together, we expect to get an 𝑥 to the 𝑛 term. And so, there should be 𝑛 linear factors. The number of linear factors we get is the degree of the polynomial 𝑛. A quadratic has degree two. And so, we expect two linear factors, which is what we got. A cubic would have three, a quartic four, and so on.
Now, we hopefully understand better what the theorem tells us. Let me just remark that we couldn’t really expect any better result. We couldn’t expect to factor any further. Does this theorem mean that 𝑝 of 𝑥 has 𝑛 roots? Well, in a certain sense, no. We’re not told that the factors are unique. So 𝑥 minus 𝑟 one could be the same as 𝑥 minus 𝑟 two. In other words, 𝑟 one could equal 𝑟 two.
For example, consider the quartic polynomial 𝑥 minus two times 𝑥 minus two times 𝑥 minus two times 𝑥 plus three. This factor of 𝑥 minus two appears three times. And so, 𝑝 of 𝑥 doesn’t have four distinct roots as we might expect, but only two. The root two corresponds to the 𝑥 minus two factors and root negative three corresponds to the factor 𝑥 plus three. However, we say that the root two has multiplicity three because the factor 𝑥 minus two occurs three times in the factorization of 𝑝 of 𝑥. When we count the roots with multiplicity, we count two three times, one for each of those factors. And we say that 𝑝 of 𝑥 has four roots when those roots are counted with multiplicity. Three of those roots are two and the other root is negative three.
This leads to another way of stating the fundamental theorem of algebra. A polynomial 𝑝 of 𝑥 of degree 𝑛 with complex coefficients has, when counted with multiplicity, exactly 𝑛 roots. These 𝑛 roots come from the 𝑛 factors of 𝑝 of 𝑥. And counting with multiplicity, we don’t mind that they might not all be distinct. There are 𝑛 of them in total including repeats. The proof of this theorem is quite difficult and so outside the scope of this video. So we’ll have to take this theorem on faith as we apply it to an example.
How many roots does the polynomial three 𝑥 squared minus one times 𝑥 cubed plus four 𝑥 minus two have?
We don’t have to find the roots to count them. We can use the fundamental theorem of algebra instead. This theorem states that a polynomial 𝑝 of 𝑥 of degree 𝑛 with complex coefficients has, when counted with multiplicity, exactly 𝑛 roots. So assuming that we’re accounting the roots with multiplicity as we’re not asked how many distinct roots the polynomial has, we just need to find the degree of our polynomial. What is this degree? Distributing, we find that the highest pair of 𝑥 is the pair 𝑥 to the power of five. And so, the degree of this polynomial is five.
We could have saved some work here by noticing the highest pair of 𝑥 would have come from the product of the highest power of 𝑥 in the first set of parentheses and the highest pair of 𝑥 in the second set as we only needed the one term to determine the degree of the polynomial. And the fundamental theorem of algebra tells us that there are therefore five roots counted with multiplicity. The number of roots is the same as the degree of the polynomial. Let’s now learn about the conjugate root theorem.
For this theorem, we need 𝑝 to be a polynomial with real coefficients. If the complex number 𝑧 equals 𝑎 plus 𝑏𝑖, where 𝑎 and 𝑏 are real, is a root of 𝑝, then its conjugate 𝑧-star equals 𝑎 minus 𝑏𝑖 is also a root. So if we have a polynomial with real coefficients and a complex root of that polynomial, then we can immediately find another root. It’s a given root, complex conjugate. Another way of saying this is that if 𝑝 of 𝑧 is zero, where 𝑝 is a polynomial with real coefficients, then 𝑝 of 𝑧 star is zero. Let’s prove this.
We first prove a more general result. We prove that 𝑝 evaluated at the conjugate of 𝑧 is a conjugate of 𝑝 of 𝑧. We take an arbitrary polynomial with real coefficients and call it 𝑝. We let 𝑛 be the degree of 𝑝 and 𝑎 zero up to a 𝑛 be its real coefficient. We evaluate this polynomial at 𝑧 star, the complex conjugate of some complex number 𝑧. And now, our task is to rearrange this somehow to get the complex conjugate of 𝑝 of 𝑧. Let’s be hopeful and write this at the end of our derivation in the belief that we’ll be able to prove this. Using the definition of the polynomial 𝑝, we can evaluate it at 𝑧 and then take the complex conjugate. We write this on the line above and hope to meet in the middle.
The complex conjugate of a sum is the sum of the complex conjugate and the complex conjugate of a product is the product of the complex conjugate. Both of these facts can be proved. Now, we might like to compare what we have working upward with what we have working downward. One difference is that we have 𝑎 zero instead of 𝑎 zero star and 𝑎 one instead of 𝑎 one star and so on. But these coefficients are all real numbers. And the complex conjugate of any real number is just itself. We can therefore replace each real coefficient with its complex conjugate without changing anything.
Are we done? Have we met in the middle? Well, almost, we have the square of the conjugate of 𝑧 here and the conjugate of the square of 𝑧 here and similarly the 𝑛th power of the conjugate of 𝑧 here and the conjugate of the 𝑛th power of 𝑧 here. We want to swap the order in which we take the power and the conjugate. And we can in fact do this, although this is a little more tricky to prove than the other fact that we’ve used. We can, for example, prove this fact using de Moivre’s theorem. Applying this fact, we get exactly the same expression on the right-hand side that we got working upward. And so, we’ve proved what we wanted to.
Now, how does this help us prove the conjugate root theorem? Well, in the special case that 𝑧 is a root of 𝑝, we can apply the conjugate on both sides to show that the conjugate of 𝑝 of 𝑧 is a conjugate of zero. And the conjugate of zero is just zero. And we’ve just proved that the conjugate of 𝑝 of 𝑧 is 𝑝 of the conjugate of 𝑧. And so, the conjugate of 𝑧 is also a root. This proves the conjugate root theorem. Let’s now apply this theorem.
Is it possible for a polynomial with real coefficients to have exactly three nonreal roots?
Let 𝑧 be one of the nonreal roots of 𝑝. The conjugate root theorem tells us that the conjugate of 𝑧, 𝑧 star, will be a root of 𝑝 also. And as 𝑧 is nonreal, 𝑧 star is also nonreal. So we’ve got two nonreal roots of 𝑝. What about the third? Let’s call this 𝑤. But 𝑤 star must also be a nonreal root. So if 𝑤 is distinct from 𝑧 and 𝑧 star, then there are four nonreal roots. So if a polynomial with real coefficients has three distinct nonreal roots, then it must also have a fourth. It can’t have exactly three distinct nonreal roots.
But what about if 𝑤 is 𝑧 or 𝑧 star? If 𝑤 is 𝑧 or 𝑧 star, there’s no such problem. What we can do is write our polynomial 𝑝 of 𝑥 as the product of its factors 𝑥 minus 𝑧 and 𝑥 minus 𝑧 star and the polynomial 𝑞 of 𝑧. Distributing, we get a quadratic factor with real coefficients as the coefficient negative 𝑧 plus 𝑧 star and 𝑧 times 𝑧 star are both real numbers. This means that 𝑞 of 𝑥 must also have real coefficients as it is a quotient of two polynomials with real coefficients. If 𝑝 of 𝑥 has exactly three nonreal roots, then 𝑞 of 𝑥 has exactly one nonreal root. And this is impossible due to the conjugate root theorem. The answer to our question is therefore no.
Recall that the discriminant of 𝑎𝑥 squared plus 𝑏𝑥 plus 𝑐 is 𝑏 squared minus four 𝑎𝑐, often denoted by capital Δ. When working with just real numbers, we found that a quadratic with discriminant greater than zero has two distinct real roots, a quadratic with discriminant zero has one repeated real root, and a quadratic with discriminant less than zero has no real roots. But working in the complex numbers, the fundamental theorem of algebra tells us to always expect two roots.
We have to count with multiplicity to make this a case when the discriminant is zero. And when the discriminant is less than zero as there are no real roots, there must be two complex roots. And the conjugate root theorem tells us these two complex roots are complex conjugates. Be careful here. The conjugate root theorem is only true for polynomials with real coefficients. This classification only holds true for quadratics with real coefficients.
What we say about the roots of cubics with real coefficients? The fundamental theorem of algebra tells us that there must be three roots counting with multiplicity. It could be the case that these three roots are real, they could be distinct, one could be repeated, or in fact all three roots could be the same. If we have one nominal root, then we must have a second, the complex conjugate of that root. We can’t have exactly one nonreal root and the other root must be real. We can’t have exactly three nonreal roots of a polynomial with real coefficients.
Here are a couple of factored cubics. You can check that after distributing, they have real coefficients. These are the only two possibilities that they’re either three real roots or a complex conjugate pair of roots and one real root. From this, we can see that a cubic with real coefficients has at least one real root. And we can show in much the same way that any polynomial with odd degree has at least one real root. The concept of a discriminant can be generalized from quadratics to cubics and other higher-degree polynomials. But the expressions involved are quite complicated and beyond the scope of this video. Let’s now see how we can use this classification to solve cubic equations.
Given that 𝑖 is one of the roots of the equation 𝑥 cubed minus five 𝑥 squared plus 𝑥 minus five equals zero, find the other two roots.
The factor theorem tells us that 𝑥 minus 𝑖 is one of the factors of this polynomial. And so, we could divide the left-hand side by this factor to get the quadratic 𝑎𝑥 squared plus 𝑏𝑥 plus 𝑐, which we could then solve. But we can make things easier for ourselves by using the conjugate root theorem which tells us that the complex conjugate of 𝑖 must also be a root as we’re dealing with a polynomial with real coefficients. And so, 𝑥 plus 𝑖 again by the factor theorem must be a factor of this polynomial.
Multiplying the two known factors together, we get 𝑥 squared plus one. And we can distribute again. And comparing coefficients, we can see that 𝑚 is one and 𝑛 is negative five. We can substitute these values then, factoring our cubic as 𝑥 minus one times 𝑥 plus one times 𝑥 minus five. Remember that we’re looking for the roots of this equation. And we can just read them off from the factored form. We find that they are five, negative 𝑖, and 𝑖. For this problem, the conjugate root theorem saved us some working, but wasn’t essential. Let’s now turn to quartic equations, where the conjugate root theorem can be essential.
Let’s classify the roots of quartics with real coefficients. A quartic equation has the form 𝑎𝑥 to the four plus 𝑏𝑥 cubed plus 𝑐𝑥 squared plus 𝑑𝑥 plus 𝑒 equals zero. And as it has real coefficients, 𝑎, 𝑏, 𝑐, 𝑑, and 𝑒 must be real numbers. By the fundamental theorem of algebra, a quartic has four roots count in multiplicity. All four could be real. If there’s a nonreal root, there must by the conjugate root theorem be a complex conjugate pair of them. The other two roots might be real or they might themselves form a complex conjugate pair, not necessarily distinct from the first pair.
Note that while a cubic with real coefficients must have at least one real root, the same is not true of quartics. A polynomial with real coefficients is only guaranteed to have at least one real root if its degree is odd. Let’s now see an example of solving a quartic equation given one of its roots.
Given that two plus 𝑖 root three is a root of 𝑥 to the power of four minus 12𝑥 cubed plus 55𝑥 squared minus 120𝑥 plus 112 equals zero, find all the roots.
We have a complex root of a polynomial with real coefficients. And so, by the conjugate root theorem, its complex conjugate two minus 𝑖 root three is also a root of this equation. We found two roots then. But how do we find the other two? The factor theorem gives us two linear factors of our quartic from these two roots. What can we say about the remaining factor? Well, it must be quadratic so that distributing our right-hand side gives a quartic as on the left. And the roots of this unknown quadratic factor are the remaining roots of our quartic equation. Let’s set about finding this quadratic factor then.
We multiply the two linear factors together to get a known quadratic factor. There are a couple of identities which reduce the amount of calculation required. Now, we can just divide our quartic by this known quadratic factor to find the unknown quadratic factor. Alternatively, we can distribute on the right-hand side multiplying every term in the first set of parentheses by every term in the second set and then collecting like terms. We can then compare coefficients.
For example, the coefficient of 𝑥 to the four tells us that 𝑎 is one. We make the substitution and then compare the coefficients of 𝑥 cubed, finding that 𝑏 minus four is negative 12. And so, 𝑏 is negative eight. Again, making the substitution, we then compare the coefficients of 𝑥 squared, finding that 𝑐 is 16. And making the substitution really does make the right-hand side equal to the left. We found the unknown quadratic factor then. It’s one 𝑥 squared minus eight 𝑥 plus 16. And we can factor this as 𝑥 minus four squared. It has repeated root of four. And so, the original quartic must do too.
Having factored our quartic, we read off the roots. The roots are four, repeated root with multiplicity two; two plus 𝑖 root three, the complex root that we were given; and its complex conjugate two minus 𝑖 root three.
Here are the key points we’ve covered in this video. The fundamental theorem of algebra tells us that a polynomial of degree 𝑛 will have, when counted with multiplicity, 𝑛 roots. The conjugate root theorem tells us that nonreal roots of polynomials with real coefficients appear in complex conjugate pairs. As a result of these two theorems, we can categorize the nature of the roots of polynomials. We can use the conjugate root theorem to help us solve cubic and quartic equations with real coefficients. And all polynomials of odd degree with real coefficients have at least one real root.