226 CHAPTER 4 Polynomial and Rational Functions Historical Feature Formulas for the solution of third- and fourth-degree polynomial equations exist, and while not very practical, they do have an interesting history. In the 1500s in Italy, mathematical contests were a popular pastime, and people who possessed methods for solving problems kept them secret. (Solutions that were published were already common knowledge.) Niccolo of Brescia (1499—1557) had the secret for solving cubic (third-degree) equations, which gave him a decided advantage in the contests. Girolamo Cardano (1501—1576) learned that Tartaglia had the secret, and, being interested in cubics, he requested it from Tartaglia. The reluctant Tartaglia hesitated for some time, but finally, swearing Cardano to secrecy with midnight oaths by candlelight, told him the secret. Cardano then published the solution in his book Ars Magna (1545), giving Tartaglia the credit but rather compromising the secrecy. Tartaglia exploded into bitter recriminations, and each wrote pamphlets that reflected on the other’s mathematics, moral character, and ancestry. The quartic (fourth-degree) equation was solved by Cardano’s student Lodovico Ferrari, and this solution also was included, with credit and this time with permission, in the Ars Magna . Attempts were made to solve the fifth-degree equation in similar ways, all of which failed. In the early 1800s, P. Ruffini, Niels Abel, and Evariste Galois all found ways to show that it is not possible to solve fifth-degree equations by formula, but the proofs required the introduction of new methods. Galois’s methods eventually developed into a large part of modern algebra. Problems 1—8 develop the Tartaglia–Cardano solution of the cubic equation and show why it is not altogether practical. 1. Show that the general cubic equation y by cy d 0 3 2 + + + = can be transformed into an equation of the form x px q 0 3 + + = by using the substitution y x b 3 = − . 2. In the equation x px q 0, 3 + + = replace x by H K+ . Let HK p 3 , =− and show that + =− H K q 3 3 . 3. Based on Problem 2, we have the two equations =− + =− HK p and H K q 3 3 3 Solve for K in HK p 3 =− and substitute into + =− H K q 3 3 . Then show that H q q p 2 4 27 2 3 3 = − + + [ Hint : Look for an equation that is quadratic in form.] 4. Use the solution for H from Problem 3 and the equation + =− H K q 3 3 to show that K q q p 2 4 27 2 3 3 = − − + 5. Use the results from Problems 2 to 4 to show that the solution of x px q 0 3 + + = is x q q p q q p 2 4 27 2 4 27 2 3 3 2 3 3 = − + + + − − + 6. Use the result of Problem 5 to solve the equation x x6 9 0 3 − − = . 7. Use a calculator and the result of Problem 5 to solve the equation x x3 14 0 3 + − = . 8. Use the methods of this section to solve the equation x x3 14 0 3 + − = . Historical Problems ‘Are You Prepared?’ Answers are given at the end of these exercises. If you get a wrong answer, read the pages listed in red. 4.3 Assess Your Understanding 1. Find f f x x x 1 if 2 2 ( ) ( ) − = − . (pp. 65–67) 2. Factor the expression x x 6 2 2 + − . (pp. A27–A28) 3. Find the quotient and remainder when x x x 3 5 7 4 4 3 − + − is divided by x 3 − . (pp. A26–A27 or A31–A34) 4. Solve x x 3 2 = − . (pp. A48–A54) 1. Now Work 1. Modeling 1.ExplainingConcepts Calculus Preview 1.InteractiveFigure 5. Multiple Choice If f x q x g x r x , ( ) ( ) ( ) ( ) = + the function r x( ) is called the . (a) remainder (b) dividend (c) quotient (d) divisor 6. When a polynomial function f is divided by x c, − the remainder is . 7. Multiple Choice Given f x x x x 3 2 7 2, 4 3 ( ) = − + − how many sign changes are there in the coefficients of f x ? ( ) − (a) 0 (b) 1 (c) 2 (d) 3 8. True or False Every polynomial function of degree 3 with real coefficients has exactly three real zeros. 9. If f is a polynomial function and x 4 − is a factor of f , then f 4( ) = . 10. True or False If f is a polynomial function of degree 4 and if f 2 5, ( ) = then f x x p x x 2 5 2 ( ) ( ) − = + − where p x( ) is a polynomial of degree 3. Concepts and Vocabulary
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