Explainer: Real and Complex Roots of Polynomials

In this explainer, we will learn how to understand the relationships between the degree of a polynomial, its coefficients, and its roots and how to apply this knowledge to solve problems.

We begin by considering two important theorems regarding polynomials.

The Fundamental Theorem of Algebra

A polynomial 𝑝(π‘₯) of degree 𝑛 with complex coefficients has, when counted with multiplicity, exactly 𝑛 roots.

The statement β€œwhen counted with multiplicity” means that we should count repeated roots by their multiplicity, that is, the number of times they are repeated. For example, in the equation (π‘₯βˆ’3)(π‘₯+3)=0, we have a polynomial of degree four. However, we only count two distinct real roots. This is because the root at π‘₯=3 is a multiple root with multiplicity three; therefore, the total number of roots, when counted with multiplicity, is four as the theorem states. Notice that this theorem applies to polynomials with real coefficients because real numbers are simply complex numbers with an imaginary part of zero.

The proof of this theorem is beyond the scope of this explainer and requires more advanced mathematical concepts such as completeness, whereas understanding this theorem and its implications is of great importance and will be the focus of this explainer.

Example 1: Number of Roots of a Polynomial

How many roots does the polynomial ο€Ή3π‘₯βˆ’1π‘₯+4π‘₯βˆ’2ο…οŠ¨οŠ© have?

Answer

Using the fundamental theorem of algebra, the number of roots is equal to the degree of the polynomial. In this case, we have been given a polynomial in factored form. To find the degree, we could expand the parentheses to find the highest-degree term. Alternatively, we could save ourselves some work by just looking for the highest-degree terms in each pair of parentheses; then, the degree of the polynomial will be the degree of their product. In the first set of parentheses, the highest-degree term is 3π‘₯ which has a degree of two. In the second set of parentheses, the highest-degree term is π‘₯ which has a degree of three. Therefore, the product of these two terms will have a degree of 5 which will be the degree of the polynomial. Hence, the polynomial has 5 roots.

We now consider the next important theorem regarding the nature of the roots of polynomials.

Conjugate Root Theorem

Let 𝑝 be a polynomial with real coefficients. If the complex number 𝑧=π‘Ž+𝑏𝑖 (where π‘Ž,π‘βˆˆβ„) is a root of 𝑝, then its conjugate 𝑧=π‘Žβˆ’π‘π‘–βˆ— is also a root.

By using the properties of the complex conjugate, we can prove this theorem as we will demonstrate. Let 𝑝 be a polynomial: 𝑝(π‘₯)=π‘Ž+π‘Žπ‘₯+π‘Žπ‘₯+β‹―+π‘Žπ‘₯, where π‘Ž,π‘Ž,…,π‘ŽοŠ¦οŠ§οŠ are real numbers. Furthermore, let 𝑧=π‘Ž+𝑏𝑖 be a root of 𝑝; that is, 𝑝(𝑧)=0. We consider the value of 𝑝(𝑧)βˆ—: 𝑝(𝑧)=π‘Ž+π‘Žπ‘§+π‘Ž(𝑧)+β‹―+π‘Ž(𝑧).βˆ—οŠ¦οŠ§βˆ—οŠ¨βˆ—οŠ¨οŠβˆ—οŠ

Using the property of complex conjugates that (𝑧)=(𝑧)βˆ—ο‰ο‰βˆ—, we can rewrite this as 𝑝(𝑧)=π‘Ž+π‘Žπ‘§+π‘Žο€Ήπ‘§ο…+β‹―+π‘Ž(𝑧).βˆ—οŠ¦οŠ§βˆ—οŠ¨οŠ¨βˆ—οŠοŠβˆ—

Since π‘Žο‡ for π‘˜=0,1,…,𝑛 are real numbers, we know that π‘Ž=π‘Žβˆ—ο‡ο‡; hence, 𝑝(𝑧)=π‘Ž+π‘Žπ‘§+π‘Žο€Ήπ‘§ο…+β‹―+π‘Ž(𝑧).βˆ—βˆ—οŠ¦βˆ—οŠ§βˆ—βˆ—οŠ¨οŠ¨βˆ—βˆ—οŠοŠβˆ—

Using the multiplicative property of complex conjugates, 𝑀𝑧=(𝑧𝑀)βˆ—βˆ—βˆ—, we can rewrite this as 𝑝(𝑧)=π‘Ž+(π‘Žπ‘§)+ο€Ήπ‘Žπ‘§ο…+β‹―+(π‘Žπ‘§).βˆ—βˆ—οŠ¦οŠ§βˆ—οŠ¨οŠ¨βˆ—οŠοŠβˆ—

Using the additive properties of complex conjugates 𝑧+𝑀=(𝑧+𝑀)βˆ—βˆ—βˆ—, we have 𝑝(𝑧)=ο€Ήπ‘Ž+π‘Žπ‘§+π‘Žπ‘§+β‹―+π‘Žπ‘§ο…=(𝑝(𝑧)).βˆ—οŠ¦οŠ§οŠ¨οŠ¨οŠοŠβˆ—βˆ—

Since 𝑧 is a root of 𝑝, we know that 𝑝(𝑧)=0; hence, 𝑝(𝑧)=0=0,βˆ—βˆ— which demonstrates that π‘§βˆ— is also a root of 𝑝.

Example 2: The Nature of the Roots of Polynomials

Is it possible for a polynomial with real coefficients to have exactly 3 nonreal roots?

Answer

The conjugate root theorem tells us that for every nonreal root 𝑧=π‘Ž+𝑏𝑖 of a polynomial with real coefficients, its conjugate is also a root. Therefore, if a polynomial 𝑝 had exactly 3 nonreal roots, 𝛼, 𝛽, and 𝛾, then for alpha we know that π›Όβˆ— is also a nonreal root. Therefore, π›Όβˆ— is equal to either 𝛽 or 𝛾. Without loss of generality, we let 𝛼=π›½βˆ—. Now, we also know that since 𝛾 is a nonreal root, π›Ύβˆ— will also be a nonreal root. Hence, it will be equal to one of 𝛼 or 𝛽. If 𝛾=π›Όβˆ—, then, taking complex conjugates, we have

which implies 𝛾=𝛽. Similar reasoning shows that if 𝛾=π›½βˆ—, 𝛾=𝛼. Hence, we have a contradiction to the idea that we have three distinct nonreal roots which implies that it is impossible to have exactly 3 nonreal roots.

We will now consider the implications of these theorems for quadratics, cubics, and quartics with real coefficients. Beginning with quadratics, from the fundamental theorem of algebra, we know that any quadratic will have two roots. From the conjugate root theorem, we know that if the polynomial has real coefficients, then if it has any nonreal root, its roots will be a complex conjugate pair. If it has real roots, it could either have two distinct real roots or a single repeated root. To distinguish between these three different cases, we have the concept of the discriminant of the equation.

Discriminant

The discriminant of a quadratic equation π‘Žπ‘₯+𝑏π‘₯+𝑐=0 is defined as π‘βˆ’4π‘Žπ‘οŠ¨. Often Ξ” is used to denote the discriminant.

Using the discriminant, we identify the three different cases of quadratic equations as follows:

  1. Positive discriminant: π‘βˆ’4π‘Žπ‘>0, two real roots;
  2. Zero discriminant: π‘βˆ’4π‘Žπ‘=0, one repeated real root;
  3. Negative discriminant: π‘βˆ’4π‘Žπ‘<0, complex conjugate roots.

The graphs below depict each case.

Using the discriminant, we can determine the nature of the roots of a quadratic equation.

Example 3: Nature of the Roots of Quadratics

Determine the type of the roots of the equation π‘₯+4π‘₯+1=3.

Answer

We start by rearranging the equation into the standard form of a quadratic. Given π‘₯+4π‘₯+1=3, we multiply through by π‘₯+1, which gives π‘₯(π‘₯+1)+4=3(π‘₯+1).

Expanding the parentheses, we have π‘₯+π‘₯+4=3π‘₯+3.

Subtracting 3π‘₯+3 from both sides gives π‘₯βˆ’2π‘₯+1=0.

By simple inspection, we can factor this equation to (π‘₯βˆ’1)=0, from which we can see there is one repeated real root. It is not always possible to simply factor the equation, so a more general method is for us to calculate the discriminant π‘βˆ’4π‘Žπ‘οŠ¨. For this equation we have π‘Ž=1, 𝑏=βˆ’2, and 𝑐=1. Hence, π‘βˆ’4π‘Žπ‘=(βˆ’2)βˆ’4Γ—1Γ—1=4βˆ’4=0.

Since the discriminant is zero, we know the equation has one repeated real root.

Turning our attention to cubic equations, the fundamental theorem of algebra tells us that each cubic will have three roots. If it also has real coefficients, we know that for any nonreal roots their complex conjugate is also a root. Therefore, we essentially have two possible cases:

  1. one real root and a complex conjugate pair of nonreal roots,
  2. three real roots.

In the case of three real roots, it is possible to have repeated roots. We could have one repeated real root with a multiplicity of three; for example, (π‘₯βˆ’5)=0. Alternatively, we could have two real roots, one of which is repeated with a multiplicity of two; for example, (π‘₯+4)(π‘₯+10)=0.

One of the interesting results of the fundamental theorem of algebra and the conjugate root theorem is that a cubic with real coefficients always has at least one real root. In fact, these theorems show that any polynomial of odd degree has at least one real root.

The idea of the discriminant can be extended to general polynomials. However, it quickly becomes computationally complex and is usually best evaluated using a computer algebra system.

The next example will demonstrate how the conjugate root theorem can help us solve cubic equations.

Example 4: Roots of Cubics

Given that 𝑖 is one of the roots of the equation π‘₯βˆ’5π‘₯+π‘₯βˆ’5=0, find the other two roots.

Answer

Since 𝑖 is a root of π‘₯βˆ’5π‘₯+π‘₯βˆ’5=0, we know that π‘₯βˆ’π‘– is a divisor of π‘₯βˆ’5π‘₯+π‘₯βˆ’5. We could use this to factor out π‘₯βˆ’π‘– and then solve the resulting quadratic equation with complex coefficients. However, it will be far simpler to first apply the conjugate root theorem which tells us that βˆ’π‘– will also be a root. Hence, (π‘₯βˆ’π‘–)(π‘₯+𝑖)=π‘₯+1 is a factor of π‘₯βˆ’5π‘₯+π‘₯βˆ’5. Therefore, we can write π‘₯βˆ’5π‘₯+π‘₯βˆ’5=ο€Ήπ‘₯+1(π‘₯+𝛼).

Expanding the parentheses, we have π‘₯βˆ’5π‘₯+π‘₯βˆ’5=π‘₯+𝛼π‘₯+π‘₯+𝛼.

Equating coefficients, we see that 𝛼=βˆ’5. Therefore, π‘₯=5 is also a root of the equation. Hence, the roots of the equation are 5,βˆ’π‘–, and 𝑖.

Finally we turn our attention to quartic equations. Recall that a quartic equation is an equation of the form π‘Žπ‘₯+𝑏π‘₯+𝑐π‘₯+𝑑π‘₯+𝑒=0.οŠͺ

Once again, by applying the fundamental theorem of algebra, we know that a quartic equation will have four roots. Similarly, the conjugate root theorem implies that, for a quartic with real coefficients, we essentially have three possible cases:

  1. all four of the roots are real,
  2. two of the roots are real and the other two form a nonreal complex conjugate pair,
  3. all of the roots are nonreal and formed of two complex conjugate pairs.

In the case of quartics, it is also possible to have repeated roots; we can have repeated real roots or even repeated complex roots.

Using the conjugate root theorem, we can solve quartic equations with real coefficients if we are given a single nonreal root as the final example will demonstrate.

Example 5: Roots of Quartics

Given that 2+π‘–βˆš3 is a root of π‘₯βˆ’12π‘₯+55π‘₯βˆ’120π‘₯+112=0οŠͺ, find all the roots.

Answer

We begin by applying the conjugate root theorem to find the second root of the equation. Since π‘₯βˆ’12π‘₯+55π‘₯βˆ’120π‘₯+112οŠͺ is a polynomial with real coefficients and 2+π‘–βˆš3 is a nonreal root, by the conjugate root theorem, 2βˆ’π‘–βˆš3 will be a root of the equation. Therefore, ο€»π‘₯βˆ’ο€»2βˆ’π‘–βˆš3π‘₯βˆ’ο€»2+π‘–βˆš3=π‘₯βˆ’4π‘₯+7 is a factor of π‘₯βˆ’12π‘₯+55π‘₯βˆ’120π‘₯+112οŠͺ. Hence, π‘₯βˆ’12π‘₯+55π‘₯βˆ’120π‘₯+112=ο€Ήπ‘₯βˆ’4π‘₯+7π‘₯+𝑏π‘₯+𝑐.οŠͺ

Expanding the parentheses of the right-hand side, we have π‘₯βˆ’12π‘₯+55π‘₯βˆ’120π‘₯+112=π‘₯+𝑏π‘₯+𝑐π‘₯βˆ’4π‘₯βˆ’4𝑏π‘₯βˆ’4𝑐π‘₯+7π‘₯+7𝑏π‘₯+7𝑐.οŠͺοŠͺ

Gathering like terms, we get π‘₯βˆ’12π‘₯+55π‘₯βˆ’120π‘₯+112=π‘₯+(π‘βˆ’4)π‘₯+(π‘βˆ’4𝑏+7)π‘₯+(7π‘βˆ’4𝑐)π‘₯+7𝑐.οŠͺοŠͺ

We can solve for 𝑏 and 𝑐 by equating coefficients. From the constant term, we have 7𝑐=112, which gives 𝑐=16. From the coefficient of π‘₯, we have π‘βˆ’4=βˆ’12 which gives 𝑏=βˆ’8. Using these values of 𝑏 and 𝑐, we can check the equality of our other coefficients. This will ensure we have done our calculations correctly. Substituting 𝑏 and 𝑐 back into our equation, we have 0=π‘₯βˆ’12π‘₯+55π‘₯βˆ’120π‘₯+112=ο€Ήπ‘₯βˆ’4π‘₯+7π‘₯βˆ’8π‘₯+16.οŠͺ

By inspection, we see that π‘₯βˆ’8π‘₯+16=(π‘₯βˆ’4). Hence, π‘₯=4 is a repeated root. Therefore, the roots of the equation are π‘₯=4, 2+π‘–βˆš3, and 2βˆ’π‘–βˆš3.

Key Points

  1. The fundamental theorem of algebra tells us that a polynomial of degree 𝑛 will have, when counted with multiplicity, 𝑛 roots.
  2. The conjugate root theorem tells us that nonreal roots of polynomials with real coefficients occur in complex conjugate pairs.
  3. As a result of these two theorems, we can categorize the nature of the roots of polynomials.
  4. We can use the conjugate root to help us solve cubic and quartic equations with real coefficients.
  5. All polynomials of odd degree with real coefficients have at least one real root.

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