Precalculus Complex Zeros V. J. Motto

Introduction We have already seen that an nth-degree polynomial can have at most n real zeros. In the complex number system, an nth-degree polynomial has exactly n zeros. Thus, it can be factored into exactly n linear factors. This fact is a consequence of the Fundamental Theorem of Algebra, which was proved by the German mathematician C. F. Gauss in 1799.

Fundamental Theorem of Algebra The following theorem is the basis for much of our work in: Factoring polynomials. Solving polynomial equations.

The Fundamental Theorem of Algebra Every polynomial P(x) = a n x n + a n-1 x n a 1 x + a 0 (n ≥ 0, a n ≠ 0) with complex coefficients has at least one complex zero. As any real number is also a complex number, it applies to polynomials with real coefficients too.

Fundamental Theorem of Algebra The Fundamental Theorem of Algebra and the Factor Theorem together show that a polynomial can be factored completely into linear factors—as we now prove.

Complete Factorization Theorem If P(x) is a polynomial of degree n ≥ 1, then there exist complex numbers a, c 1, c 2,..., c n (with a ≠ 0) such that P(x) = a(x – c 1 ) (x – c 2 )... (x – c n )

Complete Factorization Theorem—Proof By the Fundamental Theorem of Algebra, P has at least one zero—which we will call c 1. By the Factor Theorem, P(x) can be factored as: P(x) = (x – c 1 ) · Q 1 (x) where Q 1 (x) is of degree n – 1.

Complete Factorization Theorem—Proof Applying the Fundamental Theorem to the quotient Q 1 (x) gives us the factorization P(x) = (x – c 1 ) · (x – c 2 ) · Q 2 (x) where: Q 2 (x) is of degree n – 2. c 2 is a zero of Q 1 (x).

Complete Factorization Theorem—Proof Continuing this process for n steps, we get a final quotient Q n (x) of degree 0— a nonzero constant that we will call a. This means that P has been factored as: P(x) = a(x – c 1 )(x – c 2 ) ··· (x – c n )

Complex Zeros To actually find the complex zeros of an nth-degree polynomial, we usually: First, factor as much as possible. Then, use the quadratic formula on parts that we can’t factor further.

E.g. 1—Factoring a Polynomial Completely Let P(x) = x 3 – 3x 2 + x – 3 (a)Find all the zeros of P. (b)Find the complete factorization of P.

E.g. 1—Factoring Completely We first factor P as follows. Example (a)

E.g. 1—Factoring Completely We find the zeros of P by setting each factor equal to 0: P(x) = (x – 3)(x 2 + 1) Setting x – 3 = 0, we see that x = 3 is a zero. Setting x = 0, we get x 2 = –1; so, x = ±i. Thus, the zeros of P are 3, i, and –i. Example (a)

E.g. 1—Factoring Completely Since the zeros are 3, i, and i, by the Complete Factorization Theorem, P factors as: Example (b)

E.g. 2—Factoring a Polynomial Completely Let P(x) = x 3 – 2x + 4. (a)Find all the zeros of P. (b)Find the complete factorization of P.

E.g. 2—Factoring Completely The possible rational zeros are the factors of 4: ±1, ±2, and ±4. Using synthetic division, we find that –2 is a zero, and the polynomial factors as: P(x) = (x + 2) (x 2 – 2x + 2) Example (a)

E.g. 2—Factoring Completely To find the zeros, we set each factor equal to 0. Of course, x + 2 = 0 means x = –2. We use the quadratic formula to find when the other factor is 0. Example (a)

E.g. 2—Factoring Completely So, the zeros of P are –2, 1 + i, and 1 – i. Example (a)

E.g. 2—Factoring Completely Since the zeros are – 2, 1 + i, and 1 – i, by the Complete Factorization Theorem, P factors as: Example (b)

Zeros and Their Multiplicities

In the Complete Factorization Theorem, the numbers c 1, c 2,..., c n are the zeros of P. These zeros need not all be different. If the factor x – c appears k times in the complete factorization of P(x), we say that c is a zero of multiplicity k.

Zeros and Their Multiplicities For example, the polynomial P(x) = (x – 1) 3 (x + 2) 2 (x + 3) 5 has the following zeros: 1(multiplicity 3) –2(multiplicity 2) –3(multiplicity 5)

Zeros and Their Multiplicities The polynomial P has the same number of zeros as its degree. It has degree 10 and has 10 zeros—provided we count multiplicities. This is true for all polynomials—as we prove in the following theorem.

Zeros Theorem Every polynomial of degree n ≥ 1 has exactly n zeros—provided a zero of multiplicity k is counted k times.

Zeros Theorem—Proof Let P be a polynomial of degree n. By the Complete Factorization Theorem, P(x) = a(x – c 1 )(x – c 2 ) ··· (x – c n )

Zeros Theorem—Proof Now, suppose that c is a zero of P other than c 1, c 2,..., c n. Then, P(c) = a(c – c 1 )(c – c 2 ) ··· (c – c n ) = 0

Zeros Theorem—Proof Thus, by the Zero-Product Property, one of the factors c – c i must be 0. So, c = c i for some i. It follows that P has exactly the n zeros c 1, c 2,..., c n.

E.g. 3—Factoring a Polynomial with Complex Zeros Find the complete factorization and all five zeros of the polynomial P(x) = 3x x x

E.g. 3—Factoring a Polynomial with Complex Zeros Since 3x is a common factor, we have: To factor x 2 + 4, note that 2i and –2i are zeros of this polynomial.

E.g. 3—Factoring a Polynomial with Complex Zeros Thus, x = (x – 2i )(x + 2i ). Therefore, The zeros of P are 0, 2i, and –2i. Since the factors x – 2i and x + 2i each occur twice in the complete factorization, the zeros 2i and –2i are of multiplicity 2 (or double zeros). Thus, we have found all five zeros.

Factoring a Polynomial with Complex Zeros The table gives further examples of polynomials with their complete factorizations and zeros.

E.g. 4—Finding Polynomials with Specified Zeros (a)Find a polynomial P(x) of degree 4, with zeros i, –i, 2, and –2 and with P(3) = 25. (b)Find a polynomial of degree 4, with zeros –2 and 0, where –2 is a zero of multiplicity 3.

E.g. 4—Specified Zeros The required polynomial has the form We know that P(3) = a(3 4 – 3 · 3 2 – 4) = 50a = 25. Thus, a = ½. So, P(x) = ½x 4 – 3/2x 2 – 2 Example (a)

E.g. 4—Specified Zeros We require: Example (b)

E.g. 4—Specified Zeros We are given no information about Q other than its zeros and their multiplicity. So, we can choose any number for a. If we use a = 1, we get: Q(x) = x 4 + 6x x 2 + 8x Example (b)

E.g. 5—Finding All the Zeros of a Polynomial Find all four zeros of P(x) = 3x 4 – 2x 3 – x 2 – 12x – 4 Using the Rational Zeros Theorem from Section 3-3, we obtain this list of possible rational zeros: ±1, ±2, ±4, ±1/3, ±2/3, ±4/3

E.g. 5—Finding All the Zeros of a Polynomial Checking them using synthetic division, we find that 2 and -1/3 are zeros, and we get the following factorization.

E.g. 5—Finding All the Zeros of a Polynomial The zeros of the quadratic factor are: So, the zeros of P(x) are:

Finding All the Zeros of a Polynomial The figure shows the graph of the polynomial P in Example 5. The x-intercepts correspond to the real zeros of P. The imaginary zeros cannot be determined from the graph.

Complex Zeros Come in Conjugate Pairs

As you may have noticed from the examples so far, the complex zeros of polynomials with real coefficients come in pairs. Whenever a + bi is a zero, its complex conjugate a – bi is also a zero.

Conjugate Zeros Theorem If the polynomial P has real coefficients, and if the complex number z is a zero of P, then its complex conjugate is also a zero of P.

Conjugate Zeros Theorem—Proof Let P(x) = a n x n + a n-1 x n a 1 x + a 0 where each coefficient is real. Suppose that P(z) = 0. We must prove that.

Conjugate Zeros Theorem—Proof We use the facts that: The complex conjugate of a sum of two complex numbers is the sum of the conjugates. The conjugate of a product is the product of the conjugates.

Conjugate Zeros Theorem—Proof This shows that is also a zero of P(x), which proves the theorem.

E.g. 6—A Polynomial with a Specified Complex Zero Find a polynomial P(x) of degree 3 that has integer coefficients and zeros ½ and 3 – i. Since 3 – i is a zero, then so is 3 + i by the Conjugate Zeros Theorem.

E.g. 6—A Polynomial with a Specified Complex Zero That means P(x) has the form

E.g. 6—A Polynomial with a Specified Complex Zero To make all coefficients integers, we set a = 2 and get: P(x) = 2x 3 – 13x x – 10 Any other polynomial that satisfies the given requirements must be an integer multiple of this one.

E.g. 7—Counting Real and Imaginary Zeros Without actually factoring, determine how many positive real zeros, negative real zeros, and imaginary zeros this polynomial could have: P(x) = x 4 + 6x 3 – 12x 2 – 14x – 24

E.g. 7—Counting Real and Imaginary Zeros There is one change of sign. So, by Descartes’ Rule of Signs, P has one positive real zero. Also, P(–x) = x 4 – 6x 3 – 12x x – 24 has three changes of sign. So, there are either three or one negative real zero(s).

E.g. 7—Counting Real and Imaginary Zeros So, P has a total of either four or two real zeros. Since P is of degree 4, it has four zeros in all, which gives the following possibilities. Positive RealNegative RealImaginary

Linear and Quadratic Factors

We have seen that a polynomial factors completely into linear factors if we use complex numbers. If we don’t use complex numbers, a polynomial with real coefficients can always be factored into linear and quadratic factors.

Linear and Quadratic Factors A quadratic polynomial with no real zeros is called irreducible over the real numbers. Such a polynomial cannot be factored without using complex numbers.

Linear and Quadratic Factors Theorem Every polynomial with real coefficients can be factored into a product of linear and irreducible quadratic factors with real coefficients.

Linear and Quadratic Factors Theorem—Proof We first observe that, if c = a + bi is a complex number, then The last expression is a quadratic with real coefficients.

Linear and Quadratic Factors Theorem—Proof If P is a polynomial with real coefficients, by the Complete Factorization Theorem, P(x) = a(x – c 1 )(x – c 2 ) ··· (x – c n ) The complex roots occur in conjugate pairs. So, we can multiply the factors corresponding to each such pair to get a quadratic factor with real coefficients. This results in P being factored into linear and irreducible quadratic factors.

E.g. 8—Factoring into Linear and Quadratic Factors Let P(x) = x 4 + 2x 2 – 8. (a)Factor P into linear and irreducible quadratic factors with real coefficients. (b)Factor P completely into linear factors with complex coefficients

E.g. 8—Linear & Quadratic Factors The factor x is irreducible since it has only the imaginary zeros ±2i. Example (a)

E.g. 8—Linear & Quadratic Factors To get the complete factorization, we factor the remaining quadratic factor. Example (b)