College Algebra - Problem Drill 13: Zeros of Polynomial Functions No. 1 of 10 1. Determine which statement is true given f() = 3 + 4. A. f() is irreducible. B. f() has no real roots. C. There is a root of f() with multiplicity 3. D. f() has only one real root. E. f() has all real roots. Apply the rational root test to determine if any rational roots eist. Apply the rational root test to determine if any rational roots eist. f() is not the cube of a polynomial of degree 1. D. Correct! You applied the rational root test to find the rational root, then factored. Apply the rational root test to determine if any rational roots eist, and if so, how many. Apply the rational root test: the polynomial has integral coefficients, therefore the rational roots can be:, -, 1, -1, 4, and -4. By direct evaluation we see that 1, -1, -, 4, -4 are not roots, while is a root. Therefore is the only rational root. We can the factor f() as follows: f() = 3 + 4 = ( )( + ) The second factor is irreducible. In particular, the two other roots of f() are comple conjugates. Therefore, the correct answer is choice D.
No. of 10. The polynomial f() = 3 + + : A. has a real root. B. is divisible by + 1. C. is irreducible. D. has a rational root. E. has all imaginary roots. A. Correct! Graph f() to prove it has a real root. This is false because f(-1) = -1. This is false because f() has a real root. Use the rational root test to show that f() has no rational roots. A graph of f() shows that a real root eists. By the rational root test, the only rational roots can be 1, -1,, or -. None of these values return a true statement when substituted into f(), so f() has no rational roots. Net, you can graph f() to see a real root does eist since the graph intersects the -ais. 4 y 3 1 5 4 3 1 1 3 4 5 1 3 4 Therefore, the correct answer is choice A.
No. 3 of 10 3. Which of the following is a characteristic of the roots of f() = 4 + 1? A. all real roots B. all imaginary roots C. two real and two imaginary roots D. all roots have multiplicity E. cannot be determined Set the equation equal to 0 and solve to prove this is false. B. Correct! You set the equation equal to 0 and solved to find that the four solutions occur in conjugate pairs. Set the equation equal to 0 and solve to prove this is false. Set the equation equal to 0 and solve to prove this is false. Set the equation equal to 0 and solve to prove this is false. Set the equation equal to 0 and solve. Use the quadratic formula. f() = 4 + 1 1 ± ( 1) 4(1)(1) = (1) 1± 3 = 1+ 3 1 3 = ± and ± These four roots are imaginary. Therefore, the correct answer is choice B.
No. 4 of 10 4. Which polynomial is irreducible? A. 3 5 + 8 B. 3 5 3 C. + 6 + 9 D. 9 + 4 E. 6 1 Review the characteristics of an irreducible polynomial and try again. Review the characteristics of an irreducible polynomial and try again. Review the characteristics of an irreducible polynomial and try again. D. Correct! Review the characteristics of an irreducible polynomial and to see that 9 + 4 is irreducible. Review the characteristics of an irreducible polynomial and try again. Set each polynomial equal to 0 to find the polynomial with imaginary roots. 9 + 4 = 0 (3 + i)(3 i) = 0 = ± i/3 This polynomial factors into polynomials with imaginary coefficients for the a 0 term. The correct answer is choice D.
No. 5 of 10 5. The equation y = 5 3: A. has 1 positive root. B. has positive roots. C. has no real roots. D. has all rational roots. E. has all imaginary roots. A. Correct! Graph this equation and apply Descartes rule to find this polynomial has 1 positive root. Apply Descartes rule to prove this is false. Graph the equation to prove this is false. Apply the rational root test to prove this is false. Graph this equation to find this polynomial has a real root. Applying the rational root test shows that rational roots do not eist, because 1, -1 are not roots. Graph the equation to see if any real roots eist. y 4 3 1 4 3 1 1 3 4 5 1 3 4 The graph intersects the positive -ais. By Descartes rule, this equation has at most one positive root. Therefore, the correct answer is choice A.
No. 6 of 10 6. Choose the polynomial where each root has a multiplicity of 1. A. 5 B. C. 6 + 9 D. 3 5 3 E. 3 + 3 + 3 + 1 Set each polynomial equal to 0 and factor to determine the multiplicity of each root. B. Correct! You set this polynomial equal to 0 and factored to determine the multiplicity of each root is 1. Set each polynomial equal to 0 and factor to determine the multiplicity of each root. Set each polynomial equal to 0 and factor to determine the multiplicity of each root. Set each polynomial equal to 0 and factor to determine the multiplicity of each root. In choice A, 5 has a multiplicity of 5 for root = 0. In choice B, = ( 1) has roots = 0 and = 1 with multiplicities of 1. In choice C, 6 + 9 = ( 3)( 3) has root = 3 with multiplicity. In choice D, 3 5 3 = ( 3)( + 1)( + 1) has roots = 3 and = -1 with multiplicities 1 and, respectively. In choice E, 3 + 3 + 3 + 1 = ( + 1)( + 1)( + 1) has root = -1 with multiplicity 3. Therefore, the correct answer is choice B.
No. 7 of 10 7. Choose the true statement about the polynomial. 4 3 64 + + 5 5 A. The polynomial has at least one real root. B. The polynomial has no rational roots. C. The polynomial is irreducible. D. The polynomial has four positive roots. E. The polynomial has five rational roots. A. Correct! Apply the rational root theorem to find this polynomial has at least one real root. Apply the rational root theorem to find this polynomial has a rational root. Graph the equation to determine if this polynomial has a real root. Apply Descartes rule to determine the most positive roots this polynomial can have. A polynomial of degree 4 can have at most 4 roots. Rewrite the polynomial with integer coefficients. 4 3 64 4 3 + + = 5 + + 64 5 5 Now we can apply the rational root theorem. We find that is a root, therefore 4 3 3 5 + + 64 = ( )( 5 8 16 3) By Descartes rule, is the only positive root. Graphing the equation, we find there is another root in the interval (-, -1). 14 y 13 1 11 10 9 8 7 6 5 4 3 1 15 14 13 1 11 10 9 8 7 6 5 4 3 1 1 3 4 5 6 7 8 9 10 11 1 13 14 15 16 1 3 4 5 6 7 8 9 10 11 1 13 14 The correct answer is choice A.
No. 8 of 10 8. Choose the true statement given the following functions. f() = 1 g() = 3 h() = 5 A. The root(s) of f() has multiplicity > 1. B. The root(s) of g() has multiplicity > 1. C. The root(s) of h() has multiplicity > 1. D. g() has 3 unique roots. E. f() has all positive roots. Function f() has two unique roots with multiplicity 1. B. Correct! Function g() has one unique root with multiplicity 3. Function g() has five unique roots with multiplicity 1. The green curve is decreasing and has positive slopes. Set the equation equal to 0, factor and solve for to find the roots are not all positive. Factor each equation to find the roots. Function g() has one unique root with multiplicity 3. Therefore, the correct answer is choice B.
No. 9 of 10 9. Determine which statement is true given f() = 9 + 5. A. f() is irreducible. B. f() has no imaginary roots. C. There is a root of f() with multiplicity 3. D. f() has only one real root. E. f() has all real roots. A. Correct! This function cannot be factored as a product of polynomials where each coefficient is a real number. Apply the rational root test to determine if any rational roots eist. f() is not the cube of a polynomial of degree 1. Apply the rational root test to determine if any rational roots eist. Apply the rational root test to determine if any rational roots eist, and if so, how many. This function cannot be factored as a product of polynomials where each coefficient is a real number. Therefore, the correct answer is choice A.
No. 10 of 10 10. Find the interval containing the real root for g() = 5 + 1 by graphing. A. (1, ) B. (0, 1) C. (-1, 0) D. (-, -1) E. (-3, -) Graph the function to find the interval containing the real root for g(). Graph the function to find the interval containing the real root for g(). C. Correct! The root of a function is defined as the -intercept on the -ais. Graph the function to find the interval containing the real root for g(). The -interval is indeed between -1 and 0 while y = 0. Graph the function to find the interval containing the real root for g(). Graph the function to find the interval containing the real root for g(). Graph the function to find the interval containing the real root for g(). 5 y 4 3 1 5 4 3 1 1 3 4 5 1 3 4 The real root for g() is between -1 and 0. Therefore, the correct answer is choice C.