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)
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
Consequently, i is a complex number for which P(i)=0;
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
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)
where the degree of Q2(x)
This process can be continued until P(x) is completely factored as
where a is the leading coefficient and r1,r2,…,rn
In general, the numbers r1,r2,…,rn
We have the following theorem:
Find a polynomial P(x) of degree 4 with a leading coefficient of 2 and zeros −1, 3, i,
In completely factored form.
By expanding the product found in part a.
Because P(x) has degree 4, we write
Now replace the leading coefficient a with 2 and the zeros r1,r2,r3,
P(x)=2(x+1)(x−3)(x−i)(x+i)From part a=2(x+1)(x−3)(x2+1)Expand (x−i)(x+i).=2(x+1)(x3−3x2+x−3)Expand (x−3)(x2+1).=2(x4−2x3−2x2−2x−3)Expand (x+1)(x3−3x2+x−3).=2x4−4x3−4x2−4x−6Distributive property
Find a polynomial P(x) of degree 4 with a leading coefficient of 3 and zeros −2, 1, 1+i, 1−i.
In completely factored form.
By expanding the product found in part a.
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,
To prove the Conjugate Pairs Theorem, let
where an,an−1,…,a1,a0
We see that if P(z)=0,
This theorem has several uses. If, for example, 2−5i
For example, the polynomial P(x)=5x7+2x4+9x−12
A polynomial P(x) of degree 9 with real coefficients has the following zeros: 2, of multiplicity 3;4+5i,
Because complex zeros occur in conjugate pairs, the conjugate 4−5i
A polynomial P(x) of degree 8 with real coefficients has the following zeros: −3
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
The quadratic polynomial x2−2ax+(a2+b2)
Given that 2−i
Because P(x) has real coefficients, the conjugate ¯2−i=2+i
is also a factor of P(x). We divide P(x) by x2−4x+5
The zeros of P(x) are 1 (of multiplicity 2), 2−i,
Given that 2i is a zero of P(x)=x4−3x3+6x2−12x+8,
In Example 3, because x2−4x+5
is a factorization of P(x) into linear factors over the complex numbers.
Find all zeros of the polynomial function P(x)=x4−x3+7x2−9x−18.
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
Testing these possible zeros by synthetic division, we find that 2 is a zero.
Because P(2)=0, (x−2)
We can solve the depressed equation x3+x2+9x+9=0
The four zeros of P(x) are −1, 2, 3i,
Find all zeros of the polynomial
The Fundamental Theorem of Algebra states that a polynomial function of degree n≥1
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
If 2−3i
True or False. A polynomial function of degree n≥1
True or False. A polynomial function P(x) of degree n≥1
where a,r1,r2,…rn
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.
In Exercises 9–14, find all solutions of the equation in the complex number system.
x2+25=0
(x−2)2+9=0
x2+4x+4=−9
x3−8=0
(x−2)(x−3i)(x+3i)=0
(x−1)(x−2i)(2x−6i)(2x+6i)=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
Degree 3; zeros: 2, 2−i
Degree 4; zeros: 0, 1, −5+i
Degree 4; zeros: −i,1−i
Degree 6; zeros: 0,5,i,3i
Degree 6; zeros: 2i,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
Degree 4; leading coefficient −3;
Degree 5; leading coefficient 7; zeros: 5 (multiplicity 2), 1, 3−i
Degree 6; leading coefficient 4; zeros: 3, 0 (multiplicity 3), 2−3i
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)=x4−2x3+x2+2x−2, zero:1−i
P(x)=x5−5x4+2x3+22x2−20x,zero: 3−i
P(x)=2x5−11x4+19x3−17x2+17x−6,zero:i
In Exercises 29–38, find all zeros of each polynomial function.
P(x)=x3−9x2+25x−17
P(x)=x3−5x2+7x+13
P(x)=3x3−2x2+22x+40
P(x)=3x3−x2+12x−4
P(x)=2x4−10x3+23x2−24x+9
P(x)=9x4+30x3+14x2−16x+8
P(x)=x4−4x3−5x2+38x−30
P(x)=x4+x3+7x2+9x−18
P(x)=2x5−11x4+19x3−17x2+17x−6
P(x)=x5−2x4−x3+8x2−10x+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
f has complex zeros i and 2i
f has complex zero −2i
The solutions −1
The solutions of the equation xn=1,
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)=x2+3ix−2, x=−2i
P(x)=x3−(3+i)x2−(4−3i)x+4i, x=i
P(x)=x3−(4+2i)x2+(7+8i)x−14i, x=2i
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).
Degree:3;zeros:2,1+2i;
Degree: 3;zeros:1,2−3i;
Degree:4;zeros:1, −1,3+i;
Degree:4;zeros:1−2i,3−2i;
Show that if r1,r2,…,rn
then the sum of the roots satisfies
and the product of the roots satisfies
[Hint: Factor the polynomial; then multiply it out using r1,r2,…,rn
In his book Ars Magna, Cardano explained how to solve cubic equations. He considered the following example:
Explain why this equation has exactly one real solution.
Cardano explained the method as follows: “I take two cubes v3
Solve the system
to find u and v.
Consider the equation x3+px=q,
Consider an arbitrary cubic equation
Show that the substitution x=y−a3
Use Cardano’s method to solve the equation x3+6x2+10x+8=0.
In Exercises 56–59, find the zeros of each function.
f(x)=x2+2x
g(x)=2x2+x−3
g(x)=x3+3x2−10x
f(x)=x4−x3−12x2
In Exercises 60–63, find the quotient Q(x) and remainder R(x).
x3−3x+1x
x2−x−4x−2
x3+5x2+3
8x4+6x3−52x3+x
In Exercises 64–67, evaluate each expression for the given value of the variable.
−32x+1; x=2
7−x2x2+3x; x=−1
2x+35−2x2; x=−3
x2+4x−19−x3; x=2