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Section 3.5 The Complex Zeros of a Polynomial Function

Before Starting this Section, Review

1 Long division (Section 3.3 , page 353)
2 Synthetic division (Section 3.3 , page 355)
3 Rational zeros theorem (Section 3.4 , page 365)
4 Complex numbers (Section 1.4 , page 120)
5 Conjugate of a complex number (Section 1.4 , page 123)

Objectives

1 Learn basic facts about the complex zeros of polynomials.
2 Use the Conjugate Pairs Theorem to find zeros of polynomials.

Cardano Accused of Stealing Formula

In sixteenth-century Italy, getting and keeping a teaching job at a university was difficult. University appointments were, for the most part, temporary. One way a professor could convince the administration that he was worthy of continuing in his position was by winning public challenges. Two contenders for a position would present each other with a list of problems, and later in a public forum, each person would present his solutions to the other’s problems. As a result, scholars often kept secret their new methods of solving problems.

In 1535, in a public contest primarily about solving the cubic equation, Nicolo Tartaglia (1499–1557) won over his opponent Antonio Maria Fiore. Gerolamo Cardano (1501–1576), in the hope of learning the secret of solving a cubic equation, invited Tartaglia to visit him. After many promises and much flattery, Tartaglia revealed his method on the promise, probably given under oath, that Cardano would keep it confidential. When Cardano’s book Ars Magna (The Great Art) appeared in 1545, the formula and the method of proof were fully disclosed. Angered at this apparent breach of a solemn oath and feeling cheated out of the rewards of his monumental work, Tartaglia accused Cardano of being a liar and a thief for stealing his formula. Thus began the bitterest feud in the history of mathematics, carried on with name-calling and mudslinging of the lowest order. As Tartaglia feared, the formula (see Exercise 54) has forever since been known as Cardano’s formula for the solution of the cubic equation.

1 Learn basic facts about the complex zeros of polynomials.

The Factor Theorem connects the concept of factors and zeros of a polynomial, and we continue to investigate this connection.

Because the equation x2+1=0 $x^{2} + 1 = 0$ has no real roots, the polynomial function P(x)=x2+1 $P (x) = x^{2} + 1$ has no real zeros. However, if we replace x with the complex number i, we have

P (i) = i 2 + 1 = (- 1) + 1 = 0 .

$P (i) = i^{2} + 1 = (- 1) + 1 = 0 .$

Consequently, i is a complex number for which P(i)=0; $P (i) = 0;$ that is, i is a complex zero of P(x). In Section 3.2, we stated that a polynomial of degree n can have, at most, n real zeros. We now extend our number system to allow the variables and coefficients of polynomials to represent complex numbers. When we want to emphasize that complex numbers may be used in this way, we call the polynomial P(x) a complex polynomial. If P(z)=0 $P (z) = 0$ for a complex number z, we say that z is a zero or a complex zero of P(x). In the complex number system, every nth-degree polynomial equation has exactly n roots and every nth-degree polynomial can be factored into exactly n linear factors. This fact follows from the Fundamental Theorem of Algebra.

This theorem was proved by the mathematician Karl Friedrich Gauss at age 20. The proof is beyond the scope of this book, but if we are allowed to use complex numbers, we can use the theorem to prove that every polynomial has a complete factorization.

To prove the Factorization Theorem for Polynomials, we notice that if P(x) is a polynomial of degree 1 or higher with leading coefficient a, then the Fundamental Theorem of Algebra states that there is a zero r1 $r_{1}$ for which P(r1)=0. $P (r_{1}) = 0 .$ By the Factor Theorem, (x−r1) $(x - r_{1})$ is a factor of P(x) and

P (x) = (x - r 1) Q 1 (x),

$P (x) = (x - r_{1}) Q_{1} (x),$

Side Note

The Factor Theorem in the real number setting was proved in Section 3.3. It can be verified for the complex numbers by the same reasonings.

where the degree of Q1(x) $Q_{1} (x)$ is n−1 $n - 1$ and its leading coefficient is a. If n−1 $n - 1$ is positive, then we apply the Fundamental Theorem of Algebra to Q1(x); $Q_{1} (x);$ this gives us a zero, say, r2, $r_{2},$ of Q1(x). $Q_{1} (x) .$ By the Factor Theorem, (x−r2) $(x - r_{2})$ is a factor of Q1(x) $Q_{1} (x)$ and

Q 1 (x) = (x - r 2) Q 2 (x),

$Q_{1} (x) = (x - r_{2}) Q_{2} (x),$

where the degree of Q2(x) $Q_{2} (x)$ is n−2 $n - 2$ and its leading coefficient is a. Then

P (x) = (x - r 1) (x - r 2) Q 2 (x) .

$P (x) = (x - r_{1}) (x - r_{2}) Q_{2} (x) .$

This process can be continued until P(x) is completely factored as

P (x) = a (x - r 1) (x - r 2) \dots (x - r n),

$P (x) = a (x - r_{1}) (x - r_{2}) \dots (x - r_{n}),$

where a is the leading coefficient and r1,r2,…,rn $r_{1}, r_{2}, \dots, r_{n}$ are zeros of P(x). The proof is now complete.

In general, the numbers r1,r2,…,rn $r_{1}, r_{2}, \dots, r_{n}$ may not be distinct. As with the polynomials we encountered previously, if the factor (x−r) $(x - r)$ appears m times in the complete factorization of P(x), then we say that r is a zero of multiplicity m. The polynomial P(x)=(x−2)3(x−i)2(x+i)2 $P (x) = {(x - 2)}^{3} {(x - i)}^{2} {(x + i)}^{2}$ has zeros:

2 (of multiplicity 3), i (of multiplicity 2), and - i (of multiplicity 2) .

$2 (of multiplicity 3), i (of multiplicity 2), and - i (of multiplicity 2) .$

We have the following theorem:

Example 1 Constructing a Polynomial Whose Zeros Are Given

Find a polynomial P(x) of degree 4 with a leading coefficient of 2 and zeros −1, 3, i, $- 1, 3, i,$ and −i. $- i .$ Write P(x)

In completely factored form.
By expanding the product found in part a.

Solution

Because P(x) has degree 4, we write

$P (x) = a (x - r 1) (x - r 2) (x - r 3) (x - r 4) Completely factored form$ $P (x) = a (x - r_{1}) (x - r_{2}) (x - r_{3}) (x - r_{4}) Completely factored form$

Now replace the leading coefficient a with 2 and the zeros r1,r2,r3, $r_{1}, r_{2}, r_{3},$ and r4 $r_{4}$ with −1,3,i, $- 1, 3, i,$ and −i $- i$ (in any order). Then

$P (x) = = 2 [x - (- 1)] (x - 3) (x - i) [x - (- i)] 2 (x + 1) (x - 3) (x - i) (x + i) Simplify .$ $\begin{array}{rcll} P (x) & = & 2 [x - (- 1)] (x - 3) (x - i) [x - (- i)] \\ = & 2 (x + 1) (x - 3) (x - i) (x + i) & Simplify . \end{array}$
P(x)=====2(x+1)(x−3)(x−i)(x+i)2(x+1)(x−3)(x2+1)2(x+1)(x3−3x2+x−3)2(x4−2x3−2x2−2x−3)2x4−4x3−4x2−4x−6From part aExpand (x−i)(x+i).Expand (x−3)(x2+1).Expand (x+1)(x3−3x2+x−3).Distributive property $\begin{array}{rcll} P (x) & = & 2 (x + 1) (x - 3) (x - i) (x + i) & From part a \\ = & 2 (x + 1) (x - 3) (x^{2} + 1) & Expand (x - i) (x + i) . \\ = & 2 (x + 1) (x^{3} - 3 x^{2} + x - 3) & Expand (x - 3) (x^{2} + 1) . \\ = & 2 (x^{4} - 2 x^{3} - 2 x^{2} - 2 x - 3) & Expand (x + 1) (x^{3} - 3 x^{2} + x - 3) . \\ = & 2 x^{4} - 4 x^{3} - 4 x^{2} - 4 x - 6 & Distributive property \end{array}$

Practice Problem 1

Find a polynomial P(x) of degree 4 with a leading coefficient of 3 and zeros −2, 1, 1+i, 1−i. $- 2, 1, 1 + i, 1 - i .$ Write P(x)
1. In completely factored form.
2. By expanding the product found in part a.

Conjugate Pairs Theorem

2 Use the Conjugate Pairs Theorem to find zeros of polynomials.

In Example 1, we constructed a polynomial that had both i and its conjugate, −i, $- i,$ among its zeros. Our next result says that for all polynomials with real coefficients, nonreal zeros occur in conjugate pairs.

To prove the Conjugate Pairs Theorem, let

P (x) = a n x n + a n - 1 x n - 1 + \dots + a 1 x + a 0,

$P (x) = a_{n} x^{n} + a_{n - 1} x^{n - 1} + \dots + a_{1} x + a_{0},$

where an,an−1,…,a1,a0 $a_{n}, a_{n - 1}, \dots, a_{1}, a_{0}$ are real numbers. Now suppose that P(z)=0. $P (z) = 0 .$ Then we must show that P(z¯)=0 $P (\bar{z}) = 0$ also. We will use the fact that the conjugate of a real number a is identical to the real number a (because a¯=a−0i=a $\bar{a} = a - 0 i = a$ ) and then use the conjugate sum and product properties z1+z2¯¯¯¯¯¯¯¯¯¯=z1¯¯¯+z2¯¯¯ $\bar{z_{1} + z_{2}} = \bar{z_{1}} + \bar{z_{2}}$ and z1z2¯¯¯¯¯¯=z1¯¯¯z2¯¯¯. $\bar{z_{1} z_{2}} = \bar{z_{1}} \bar{z_{2}} .$

P (z) P (z ¯) = = = = = = a n z n + a n - 1 z n - 1 + \dots + a 1 z + a 0 a n (z ¯) n + a n - 1 (z ¯) n - 1 + \dots + a 1 z ¯ + a 0 a ¯ n z n ¯ ¯ ¯ ¯ + a ¯ n - 1 z n - 1 ¯ ¯ ¯ ¯ ¯ ¯ ¯ + \dots + a ¯ 1 z ¯ + a ¯ 0 a n z n ¯ ¯ ¯ ¯ ¯ ¯ ¯ + a n - 1 z n - 1 ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ + \dots + a 1 z ¯ ¯ ¯ ¯ ¯ + a ¯ 0 - - - - - - - - - - - - - - - - - - - - - - - - - - a n z n + a n - 1 z n - 1 + \dots + a 1 z + a 0 P (z) ¯ ¯ ¯ ¯ ¯ ¯ ¯ = 0 ¯ = 0 Replace x with z . Replace z with z ¯ . (z ¯) k = z k ¯ ¯ ¯, a k ¯ ¯ ¯ = a k z 1 z 2 ¯ ¯ ¯ ¯ ¯ ¯ = z 1 z 2 ¯ ¯ ¯ ¯ ¯ ¯ z 1 ¯ ¯ ¯ + z 2 ¯ ¯ ¯ = z 1 + z 2 ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ P (z) = 0

$\begin{array}{rcll} P (z) & = & a_{n} z^{n} + a_{n - 1} z^{n - 1} + \dots + a_{1} z + a_{0} & Replace x with z . \\ P (\bar{z}) & = & a_{n} {(\bar{z})}^{n} + a_{n - 1} {(\bar{z})}^{n - 1} + \dots + a_{1} \bar{z} + a_{0} & Replace z with \bar{z} . \\ = & {\bar{a}}_{n} \bar{z^{n}} + {\bar{a}}_{n - 1} \bar{z^{n - 1}} + \dots + {\bar{a}}_{1} \bar{z} + {\bar{a}}_{0} & {(\bar{z})}^{k} = \bar{z^{k}}, \bar{a_{k}} = a_{k} \\ = & \underline{\bar{a_{n} z^{n}} + \bar{a_{n - 1} z^{n - 1}} + \dots + \bar{a_{1} z} + {\bar{a}}_{0}} & \bar{z_{1} z_{2}} = \bar{z_{1} z_{2}} \\ = & a_{n} z^{n} + a_{n - 1} z^{n - 1} + \dots + a_{1} z + a_{0} & \bar{z_{1}} + \bar{z_{2}} = \bar{z_{1} + z_{2}} \\ = & \bar{P (z)} = \bar{0} = 0 & P (z) = 0 \end{array}$

We see that if P(z)=0, $P (z) = 0,$ then P(z¯)=0, $P (\bar{z}) = 0,$ and the theorem is proved.

This theorem has several uses. If, for example, 2−5i $2 - 5 i$ is a zero of a polynomial with real coefficients, then we know that 2+5i $2 + 5 i$ is also a zero. Because nonreal zeros occur in conjugate pairs, we know that there will always be an even number of nonreal zeros. Therefore, any polynomial of odd degree with real coefficients has at least one zero that is a real number.

For example, the polynomial P(x)=5x7+2x4+9x−12 $P (x) = 5 x^{7} + 2 x^{4} + 9 x - 12$ has at least one zero that is a real number because P(x) has degree 7 (odd degree) and has real numbers as coefficients.

Example 2 Using the Conjugate Pairs Theorem

A polynomial P(x) of degree 9 with real coefficients has the following zeros: 2, of multiplicity 3;4+5i, $3; 4 + 5 i,$ of multiplicity 2; and 3−7i. $3 - 7 i .$ Write all nine zeros of P(x).

Solution

Because complex zeros occur in conjugate pairs, the conjugate 4−5i $4 - 5 i$ of 4+5i $4 + 5 i$ is a zero of multiplicity 2, and the conjugate 3+7i $3 + 7 i$ of 3−7i $3 - 7 i$ is a zero of P(x). The nine zeros of P(x), including the repetitions, are

2, 2, 2, 4 + 5 i, 4 + 5 i, 4 - 5 i, 4 - 5 i, 3 + 7 i, and 3 - 7 i .

$2, 2, 2, 4 + 5 i, 4 + 5 i, 4 - 5 i, 4 - 5 i, 3 + 7 i, and 3 - 7 i .$

Practice Problem 2

A polynomial P(x) of degree 8 with real coefficients has the following zeros: −3 $- 3$ and 2−3i, $2 - 3 i,$ each of multiplicity 2, and i. Write all eight zeros of P(x).

Side Note

In problems such as Example 2, once you determine all of the zeros, you can use the Factorization Theorem to write a polynomial with those zeros.

Every polynomial of degree n has exactly n zeros and can be factored into a product of n linear factors. If the polynomial has real coefficients, then, by the Conjugate Pairs Theorem, its nonreal zeros occur as conjugate pairs. Consequently, if r=a+bi $r = a + b i$ is a zero, then so is r¯=a−bi, $\bar{r} = a - b i,$ and both x−r $x - r$ and x−r¯ $x - \bar{r}$ are linear factors of the polynomial. Multiplying these factors, we have

(x - r) (x - r ¯) = = x 2 - (r + r ¯) x + r r ¯ x 2 - 2 a x + (a 2 + b 2) Use FOIL . r + r ¯ = 2 a; r r ¯ = a 2 + b 2

$\begin{array}{rcll} (x - r) (x - \bar{r}) & = & x^{2} - (r + \bar{r}) x + r \bar{r} & Use FOIL . \\ = & x^{2} - 2 a x + (a^{2} + b^{2}) & r + \bar{r} = 2 a; r \bar{r} = a^{2} + b^{2} \end{array}$

The quadratic polynomial x2−2ax+(a2+b2) $x^{2} - 2 a x + (a^{2} + b^{2})$ has real coefficients and is irreducible (cannot be factored any further) over the real numbers. This result shows that each pair of nonreal conjugate zeros can be combined into one irreducible quadratic factor.

Example 3 Finding the Zeros of a Polynomial from a Given Complex Zero

Given that 2−i $2 - i$ is a zero of P(x)=x4−6x3+14x2−14x+5, $P (x) = x^{4} - 6 x^{3} + 14 x^{2} - 14 x + 5,$ find the remaining zeros.

Solution

Because P(x) has real coefficients, the conjugate 2−i¯¯¯¯¯¯¯=2+i $\bar{2 - i} = 2 + i$ is also a zero. By the Factorization Theorem, the linear factors [x−(2−i)] $[x - (2 - i)]$ and [x−(2+i)] $[x - (2 + i)]$ appear in the factorization of P(x). Consequently, their product

[x - (2 - i)] [x - (2 + i)] = = = = = (x - 2 + i) (x - 2 - i) [(x - 2) + i] [(x - 2) - i] Regroup . (x - 2) 2 - i 2 (A + B) (A - B) = A 2 - B 2 x 2 - 4 x + 4 + 1 Expand (x - 2) 2, i 2 = - 1. x 2 - 4 x + 5 Simplify .

$\begin{array}{rcl} [x - (2 - i)] [x - (2 + i)] & = & (x - 2 + i) (x - 2 - i) \\ = & [(x - 2) + i] [(x - 2) - i] Regroup . \\ = & {(x - 2)}^{2} - i^{2} (A + B) (A - B) = A^{2} - B^{2} \\ = & x^{2} - 4 x + 4 + 1 Expand {(x - 2)}^{2}, i^{2} = - 1. \\ = & x^{2} - 4 x + 5 Simplify . \end{array}$

is also a factor of P(x). We divide P(x) by x2−4x+5 $x^{2} - 4 x + 5$ to find the other factor.

x 2 - 2 x + 1 Divisor \to x 2 - 4 x + 5 x 4 - 6 x 3 + 14 x 2 - 14 x + 5 x 4 - 4 x 3 + 5 x 2 - - - - - - - - - - - - - - 2 x 3 + 9 x 2 - 14 x - 2 x 3 + 8 x 2 - 10 x - - - - - - - - - - - - - - x 2 - 4 x + 5 x 2 - 4 x + 5 - - - - - - - - - - 0 \leftarrow - - - - - - - - - - Quotient \leftarrow - - - - - - - - - - Dividend \leftarrow - - - - - - - - - - Remainder

$\begin{aligned} x^{2} - 2 x + 1 & \leftarrow Quotient \\ Divisor \to x^{2} - 4 x + 5 x^{4} - 6 x^{3} + 14 x^{2} - 14 x + 5 & \leftarrow Dividend \\ \underline{x^{4} - 4 x^{3} + 5 x^{2}} \\ - 2 x^{3} + 9 x^{2} - 14 x \\ \underline{- 2 x^{3} + 8 x^{2} - 10 x} \\ x^{2} - 4 x + 5 \\ \underline{x^{2} - 4 x + 5} \\ 0 & \leftarrow Remainder \end{aligned}$

P (x) = = = (x 2 - 2 x + 1) (x 2 - 4 x + 5) (x - 1) (x - 1) (x 2 - 4 x + 5) (x - 1) (x - 1) [x - (2 - i)] [x - (2 + i)] P (x) = Quotient \cdot Divisor Factor x 2 - 2 x + 1. Factor x 2 - 4 x + 5.

$\begin{array}{rcll} P (x) & = & (x^{2} - 2 x + 1) (x^{2} - 4 x + 5) & P (x) = Quotient \cdot Divisor \\ = & (x - 1) (x - 1) (x^{2} - 4 x + 5) & Factor x^{2} - 2 x + 1. \\ = & (x - 1) (x - 1) [x - (2 - i)] [x - (2 + i)] & Factor x^{2} - 4 x + 5. \end{array}$

The zeros of P(x) are 1 (of multiplicity 2), 2−i, $2 - i,$ and 2+i. $2 + i .$

Practice Problem 3

Given that 2i is a zero of P(x)=x4−3x3+6x2−12x+8, $P (x) = x^{4} - 3 x^{3} + 6 x^{2} - 12 x + 8,$ find the remaining zeros.

In Example 3, because x2−4x+5 $x^{2} - 4 x + 5$ is irreducible over the real number system, the form P(x)=(x−1)(x−1)(x2−4x+5) $P (x) = (x - 1) (x - 1) (x^{2} - 4 x + 5)$ is a factorization of P(x) into linear and irreducible quadratic factors. However, the form

P (x) = (x - 1) (x - 1) [x - (2 - i)] [x - (2 + i)]

$P (x) = (x - 1) (x - 1) [x - (2 - i)] [x - (2 + i)]$

is a factorization of P(x) into linear factors over the complex numbers.

Example 4 Finding the Zeros of a Polynomial

Find all zeros of the polynomial function P(x)=x4−x3+7x2−9x−18. $P (x) = x^{4} - x^{3} + 7 x^{2} - 9 x - 18 .$

Solution

Because the degree of P(x) is 4, P(x) has four zeros. The Rational Zeros Theorem tells us that the possible rational zeros are

\pm 1, \pm 2, \pm 3, \pm 6, \pm 9, \pm 18 .

$\pm 1, \pm 2, \pm 3, \pm 6, \pm 9, \pm 18 .$

Testing these possible zeros by synthetic division, we find that 2 is a zero.

Because P(2)=0, (x−2) $P (2) = 0, (x - 2)$ is a factor of P(x) and the quotient is x3+x2+9x+9. $x^{3} + x^{2} + 9 x + 9 .$

We can solve the depressed equation x3+x2+9x+9=0 $x^{3} + x^{2} + 9 x + 9 = 0$ by factoring by grouping.

x 2 (x + 1) + 9 (x + 1) = 0 (x 2 + 9) (x + 1) = 0 x 2 + 9 x 2 x = = = 0 or x + 1 = 0 - 9 or x = - 1 \pm - 9 - - - \sqrt = \pm 3 i Group terms, distributive property Distributive property Zero-product property Solve each equation . Solve x 2 = - 9 for x .

$\begin{array}{l} \begin{array}{r} x^{2} (x + 1) + 9 (x + 1) = 0 \\ (x^{2} + 9) (x + 1) = 0 \end{array} & \begin{array}{l} Group terms, distributive property \\ Distributive property \end{array} \\ \begin{array}{rcl} x^{2} + 9 & = & 0 or x + 1 = 0 \\ x^{2} & = & - 9 or x = - 1 \\ x & = & \pm \sqrt{- 9} = \pm 3 i \end{array} & \begin{array}{l} Zero-product property \\ Solve each equation . \\ Solve x^{2} = - 9 for x . \end{array} \end{array}$

The four zeros of P(x) are −1, 2, 3i, $- 1, 2, 3 i,$ and −3i. $- 3 i .$ The complete factorization of P(x) is

P (x) = x 4 - x 3 + 7 x 2 - 9 x - 18 = (x + 1) (x - 2) (x - 3 i) (x + 3 i) .

$P (x) = x^{4} - x^{3} + 7 x^{2} - 9 x - 18 = (x + 1) (x - 2) (x - 3 i) (x + 3 i) .$

Practice Problem 4

Find all zeros of the polynomial

$P (x) = x 4 - 8 x 3 + 22 x 2 - 28 x + 16 .$ $P (x) = x^{4} - 8 x^{3} + 22 x^{2} - 28 x + 16 .$

Section 3.5 Exercises

Concepts and Vocabulary

The Fundamental Theorem of Algebra states that a polynomial function of degree n≥1 $n \geq 1$ has at least one zero.
The Number of Zeros Theorem states that a polynomial function of degree n has exactly zeros, provided a zero of multiplicity k is counted times.
If P is a polynomial function with real coefficients and if z=a+bi $z = a + b i$ is a zero of P, then is also a zero of P.
If 2−3i $2 - 3 i$ is a zero of a polynomial function with real coefficients, then so is .
True or False. A polynomial function of degree n≥1 $n \geq 1$ with real coefficients has at least one real zero and, at most, n real zeros.
True or False. A polynomial function P(x) of degree n≥1 $n \geq 1$ with real coefficients can be factored into n factors (not necessarily distinct) as

$P (x) = a (x - r 1) (x - r 2) \dots (x - r n),$ $P (x) = a (x - r_{1}) (x - r_{2}) \dots (x - r_{n}),$

where a,r1,r2,…rn $a, r_{1}, r_{2}, \dots r_{n}$ are real numbers.
True or False. Every polynomial of odd degree with real coefficients has at least one real zero.
True or False. A polynomial of degree 4 must have four distinct zeros.

Building Skills

In Exercises 9–14, find all solutions of the equation in the complex number system.

x2+25=0 $x^{2} + 25 = 0$
(x−2)2+9=0 ${(x - 2)}^{2} + 9 = 0$
x2+4x+4=−9 $x^{2} + 4 x + 4 = - 9$
x3−8=0 $x^{3} - 8 = 0$
(x−2)(x−3i)(x+3i)=0 $(x - 2) (x - 3 i) (x + 3 i) = 0$
(x−1)(x−2i)(2x−6i)(2x+6i)=0 $(x - 1) (x - 2 i) (2 x - 6 i) (2 x + 6 i) = 0$

In Exercises 15–20, find the remaining zeros of a polynomial P(x) with real coefficients and with the specified degree and zeros.

Degree 3; zeros: 2, 3+i $3 + i$
Degree 3; zeros: 2, 2−i $2 - i$
Degree 4; zeros: 0, 1, −5+i $- 5 + i$
Degree 4; zeros: −i,1−i $- i, 1 - i$
Degree 6; zeros: 0,5,i,3i $0, 5, i, 3 i$
Degree 6; zeros: 2i,4+i,i−1 $2 i, 4 + i, i - 1$

In Exercises 21–24, find the polynomial P(x) with real coefficients having the specified degree, leading coefficient, and zeros.

Degree 4; leading coefficient 2; zeros: 5−i, 3i $5 - i, 3 i$
Degree 4; leading coefficient −3; $- 3;$ zeros: 2+3i, 1−4i $2 + 3 i, 1 - 4 i$
Degree 5; leading coefficient 7; zeros: 5 (multiplicity 2), 1, 3−i $3 - i$
Degree 6; leading coefficient 4; zeros: 3, 0 (multiplicity 3), 2−3i $2 - 3 i$

In Exercises 25–28, use the given zero to write P(x) as a product of linear and irreducible quadratic factors.

P(x)=x4+x3+9x2+9x,zero: 3i $P (x) = x^{4} + x^{3} + 9 x^{2} + 9 x, zero : 3 i$
P(x)=x4−2x3+x2+2x−2, zero:1−i $P (x) = x^{4} - 2 x^{3} + x^{2} + 2 x - 2, zero : 1 - i$
P(x)=x5−5x4+2x3+22x2−20x,zero: 3−i $P (x) = x^{5} - 5 x^{4} + 2 x^{3} + 22 x^{2} - 20 x, zero : 3 - i$
P(x)=2x5−11x4+19x3−17x2+17x−6,zero:i $P (x) = 2 x^{5} - 11 x^{4} + 19 x^{3} - 17 x^{2} + 17 x - 6, zero : i$

In Exercises 29–38, find all zeros of each polynomial function.

P(x)=x3−9x2+25x−17 $P (x) = x^{3} - 9 x^{2} + 25 x - 17$
P(x)=x3−5x2+7x+13 $P (x) = x^{3} - 5 x^{2} + 7 x + 13$
P(x)=3x3−2x2+22x+40 $P (x) = 3 x^{3} - 2 x^{2} + 22 x + 40$
P(x)=3x3−x2+12x−4 $P (x) = 3 x^{3} - x^{2} + 12 x - 4$
P(x)=2x4−10x3+23x2−24x+9 $P (x) = 2 x^{4} - 10 x^{3} + 23 x^{2} - 24 x + 9$
P(x)=9x4+30x3+14x2−16x+8 $P (x) = 9 x^{4} + 30 x^{3} + 14 x^{2} - 16 x + 8$
P(x)=x4−4x3−5x2+38x−30 $P (x) = x^{4} - 4 x^{3} - 5 x^{2} + 38 x - 30$
P(x)=x4+x3+7x2+9x−18 $P (x) = x^{4} + x^{3} + 7 x^{2} + 9 x - 18$
P(x)=2x5−11x4+19x3−17x2+17x−6 $P (x) = 2 x^{5} - 11 x^{4} + 19 x^{3} - 17 x^{2} + 17 x - 6$
P(x)=x5−2x4−x3+8x2−10x+4 $P (x) = x^{5} - 2 x^{4} - x^{3} + 8 x^{2} - 10 x + 4$

In Exercises 39–42, find an equation of a polynomial function of least degree having the given complex zeros, intercepts, and graph.

f has complex zero 3i
f has complex zero −i $- i$
f has complex zeros i and 2i
f has complex zero −2i $- 2 i$

Beyond the Basics

The solutions −1 $- 1$ and 1 of the equation x2=1 $x^{2} = 1$ are called the square roots of 1. The solutions of the equation x3=1 $x^{3} = 1$ are called the cube roots of 1. Find the cube roots of 1. How many are there?
The solutions of the equation xn=1, $x^{n} = 1,$ where n is a positive integer, are called the nth roots of 1 or “nth roots of unity.” Explain the relationship between the solutions of the equation xn=1 $x^{n} = 1$ and the zeros of the complex polynomial P(x)=xn−1. $P (x) = x^{n} - 1 .$ How many nth roots of 1 are there?

In Exercises 45–48, use the given zero and synthetic division to determine all of the zeros of P(x). Then factor the depressed equation to write P(x) as a product of linear factors.

P(x)=x2+(i−2)x−2i, x=−i $P (x) = x^{2} + (i - 2) x - 2 i, x = - i$
P(x)=x2+3ix−2, x=−2i $P (x) = x^{2} + 3 i x - 2, x = - 2 i$
P(x)=x3−(3+i)x2−(4−3i)x+4i, x=i $P (x) = x^{3} - (3 + i) x^{2} - (4 - 3 i) x + 4 i, x = i$
P(x)=x3−(4+2i)x2+(7+8i)x−14i, x=2i $P (x) = x^{3} - (4 + 2 i) x^{2} + (7 + 8 i) x - 14 i, x = 2 i$

In Exercises 49–52, find an equation with real coefficients of a polynomial function f that has the given characteristics. Then write the end behavior of the graph of y=f(x). $y = f (x) .$

Degree:3;zeros:2,1+2i; $Degree : 3; zeros : 2, 1 + 2 i;$ y-intercept: 40
Degree: 3;zeros:1,2−3i; $Degree : 3; zeros : 1, 2 - 3 i;$ y-intercept: −26 $- 26$
Degree:4;zeros:1, −1,3+i; $Degree : 4; zeros : 1, - 1, 3 + i;$ y-intercept: 20
Degree:4;zeros:1−2i,3−2i; $Degree : 4; zeros : 1 - 2 i, 3 - 2 i;$ y-intercept: 130

Critical Thinking / Discussion / Writing

Show that if r1,r2,…,rn $r_{1}, r_{2}, \dots, r_{n}$ are the roots of the equation

$a n x n + a n - 1 x n - 1 + \dots + a 1 x + a 0 = 0 (a n \neq 0),$ $a_{n} x^{n} + a_{n - 1} x^{n - 1} + \dots + a_{1} x + a_{0} = 0 (a_{n} \neq 0),$

then the sum of the roots satisfies

$r 1 + r 2 + \dots + r n = - a n - 1 a n$ $r_{1} + r_{2} + \dots + r_{n} = - \frac{a_{n - 1}}{a_{n}}$

and the product of the roots satisfies

$r 1 \cdot r 2 \cdot \dots \cdot r n = (- 1) n a 0 a n .$ $r_{1} \cdot r_{2} \cdot \dots \cdot r_{n} = {(- 1)}^{n} \frac{a_{0}}{a_{n}} .$

[Hint: Factor the polynomial; then multiply it out using r1,r2,…,rn $r_{1}, r_{2}, \dots, r_{n}$ and compare coefficients with those of the original polynomial.]
In his book Ars Magna, Cardano explained how to solve cubic equations. He considered the following example:

$x 3 + 6 x = 20$ $x^{3} + 6 x = 20$
1. Explain why this equation has exactly one real solution.
2. Cardano explained the method as follows: “I take two cubes v3 $v^{3}$ and u3 $u^{3}$ whose difference is 20 and whose product is 2, that is, a third of the coefficient of x. Then, I say that x=v−u $x = v - u$ is a solution of the equation.” Show that if v3−u3=20 $v^{3} - u^{3} = 20$ and vu=2, $v u = 2,$ then x=v−u $x = v - u$ is indeed the solution of the equation x3+6x=20. $x^{3} + 6 x = 20 .$
3. Solve the system
  
  $v 3 - u 3 u v = = 202$ $\begin{array}{rcl} v^{3} - u^{3} & = & 20 \\ u v & = & 2 \end{array}$
  
  to find u and v.
4. Consider the equation x3+px=q, $x^{3} + p x = q,$ where p is a positive number. Using your work in parts (a), (b), and (c) as a guide, show that the unique solution of this equation is
  
  $x = 3 q 2 + (q 2) 2 + (p 3) 3 - - - - - - - - - - - \sqrt - - - - - - - - - - - - - - - - - - \sqrt - 3 - q 2 + (q 2) 2 + (p 3) 3 - - - - - - - - - - - \sqrt - - - - - - - - - - - - - - - - - - - - \sqrt .$ $x = \sqrt{3 \frac{q}{2} + \sqrt{{(\frac{q}{2})}^{2} + {(\frac{p}{3})}^{3}}} - \sqrt{3 - \frac{q}{2} + \sqrt{{(\frac{q}{2})}^{2} + {(\frac{p}{3})}^{3}}} .$
5. Consider an arbitrary cubic equation
  
  $x 3 + a x 2 + b x + c = 0 .$ $x^{3} + a x^{2} + b x + c = 0 .$
  
  Show that the substitution x=y−a3 $x = y - \frac{a}{3}$ allows you to write the cubic equation as
  
  $y 3 + p y = q .$ $y^{3} + p y = q .$
Use Cardano’s method to solve the equation x3+6x2+10x+8=0. $x^{3} + 6 x^{2} + 10 x + 8 = 0.$

Getting Ready for the Next Section

In Exercises 56–59, find the zeros of each function.

f(x)=x2+2x $f (x) = x^{2} + 2 x$
g(x)=2x2+x−3 $g (x) = 2 x^{2} + x - 3$
g(x)=x3+3x2−10x $g (x) = x^{3} + 3 x^{2} - 10 x$
f(x)=x4−x3−12x2 $f (x) = x^{4} - x^{3} - 12 x^{2}$

In Exercises 60–63, find the quotient Q(x) and remainder R(x).

x3−3x+1x $\frac{x^{3} - 3 x + 1}{x}$
x2−x−4x−2 $\frac{x^{2} - x - 4}{x - 2}$
x3+5x2+3 $\frac{x^{3} + 5}{x^{2} + 3}$
8x4+6x3−52x3+x $\frac{8 x^{4} + 6 x^{3} - 5}{2 x^{3} + x}$

In Exercises 64–67, evaluate each expression for the given value of the variable.

−32x+1; x=2 $\frac{- 3}{2 x + 1}; x = 2$
7−x2x2+3x; x=−1 $\frac{7 - x}{2 x^{2} + 3 x}; x = - 1$
2x+35−2x2; x=−3 $\frac{2 x + 3}{5 - 2 x^{2}}; x = - 3$
x2+4x−19−x3; x=2 $\frac{x^{2} + 4 x - 1}{9 - x^{3}}; x = 2$

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Table of Contents for Section 3.5 The Complex Zeros of a Polynomial Function

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Table of Contents for
Section 3.5 The Complex Zeros of a Polynomial Function