Explanation Given information: The polynomial is P ( x ) = x 5 − 3 x 4 − x 3 + 11 x 2 − 12 x + 4 . Concept used: Descartes' Rule of Signs: Let P be a polynomial with real coefficients then, The number of positive real zeros of P either is the number of chances of sign in the coefficients of P ( x ) or is less than that by an even number. The number of negative real zeros of P either is the number of changes of sign in the coefficients of P ( − x ) or is less than that by an even number. Theorems used: Rational Zeros Theorem: If a polynomial Q ( x ) = a n x n + a n − 1 x n + ... + a 1 x 1 + a 0 x 0 has integer coefficients (here a n ≠ 0 and a 0 ≠ 0 ), then every rational zero of Qis of the form, p q Here, pis a factor of the constant coefficient a 0 and q is a factor of the leading coefficient a n . Factor theorem: If x = a is a zero of the polynomial P ( x ) , then ( x − a ) is a factor of the polynomial P ( x ) . Calculation: The degree of the polynomial P ( x ) = x 5 − 3 x 4 − x 3 + 11 x 2 − 12 x + 4 is 5 ,so it has five zeroes. To obtain possible number of positive zeroes proceed as follows, P ( x ) = x 5 − 3 x 4 − x 3 + 11 x 2 − 12 x + 4 Since, the polynomial P ( x ) has four variations in sign, so by Descartes' rule of sign it has four positive zeroes. Now, to obtain possible number of negative zeros substitute − x for x in P ( x ) = x 5 − 3 x 4 − x 3 + 11 x 2 − 12 x + 4 . P ( x ) = x 5 − 3 x 4 − x 3 + 11 x 2 − 12 x + 4 P ( − x ) = − x 5 − 3 x 4 + x 3 + 11 x 2 + 12 x + 4 Since, P ( − x ) has one variation in sign, so it has one negative zeros. Now, obtain the roots of the polynomial P ( x ) = x 5 − 3 x 4 − x 3 + 11 x 2 − 12 x + 4 by rational zeros theorem. In the polynomial P ( x ) = x 5 − 3 x 4 − x 3 + 11 x 2 − 12 x + 4 , the constant coefficient term is 4 and the leading coefficient is 1. So, possible rational zero of P(x) are obtained in the form , factor of 4 factor of 1 The factors of 4 are ± 1 , ± 2 , ± 4 and factors of 1 are ± 1 . Thus, the possible rational zeros of P ( x ) are ± 1 , ± 2 , ± 4 . Now, to check the actual zeros out of the possible zeros, use synthetic division. Test whether 1 is a zero of P ( x ) or not. 1 1 − 3 − 1 11 − 12 4 1 − 2 − 3 8 − 4 _ 1 − 2 − 3 8 − 4 0 _ The remainder is 0 ,and the quotient is x 4 − 2 x 3 − 3 x 2 + 8 x − 4 . Hence 1 is a zero or ( x − 1 ) is a factor of P ( x ) = x 5 − 3 x 4 − x 3 + 11 x 2 − 12 x + 4

7th Edition

ISBN: 9781305115545

Author: James Stewart, Lothar Redlin, Saleem Watson

Publisher: Cengage Learning

See similar textbooks

Concept explainers

Quadratic Equation

When it comes to the concept of polynomial equations, quadratic equations can be said to be a special case. What does solving a quadratic equation mean? We will understand the quadratics and their types once we are familiar with the polynomial equations …

Demand and Supply Function

The concept of demand and supply is important for various factors. One of them is studying and evaluating the condition of an economy within a given period of time. The analysis or evaluation of the demand side factors are important for the suppliers to …

Videos

Question

Chapter 3, Problem 63E

To determine

To find:

All rational, irrational, and complex zeroes of the polynomial P(x)=x5−3x4−x3+11x2−12x+4 and state their multiplicities.

Expert Solution & Answer

Want to see the full answer?

Check out a sample textbook solution

See solution

Chapter 3, Problem 62E

Chapter 3, Problem 64E

Chapter 3 Solutions

College Algebra

Ch. 3.1 - To put the quadratic function f(x)=ax2+bx+c in...Ch. 3.1 - The quadratic function f(x)=a(xh)2+k is in...Ch. 3.1 - The graph of f(x)=3(x2)26 is a parabola that...Ch. 3.1 - Prob. 4E Ch. 3.1 - Graphs of Quadratic Functions The graph of a...Ch. 3.1 - Prob. 6E Ch. 3.1 - Graphs of Quadratic Functions The graph of a...Ch. 3.1 - Prob. 8E Ch. 3.1 - Graphing Quadratic Functions A quadratic function...Ch. 3.1 - Graphing Quadratic Functions A quadratic function...

Knowledge Booster

$Algebra$

Learn more about

Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, algebra and related others by exploring similar questions and additional content below.

All rational, irrational, and complex zeroes of the polynomial P ( x ) = x 5 − 3 x 4 − x 3 + 11 x 2 − 12 x + 4 and state their multiplicities.

Solution Summary: The author explains the rational, irrational, and complex zeroes of the polynomial P(x) and their multiplicities.

Concept explainers

Videos

To find:

All rational, irrational, and complex zeroes of the polynomial P(x)=x5−3x4−x3+11x2−12x+4 and state their multiplicities.

Want to see the full answer?

Chapter 3 Solutions

All rational, irrational, and complex zeroes of the polynomial P ( x ) = x 5 − 3 x 4 − x 3 + 11 x 2 − 12 x + 4 and state their multiplicities.

Solution Summary: The author explains the rational, irrational, and complex zeroes of the polynomial P(x) and their multiplicities.

Concept explainers

Videos

To find: All rational, irrational, and complex zeroes of the polynomial P(x)=x5−3x4−x3+11x2−12x+4 and state their multiplicities.

Want to see the full answer?

Chapter 3 Solutions

To find:

All rational, irrational, and complex zeroes of the polynomial P(x)=x5−3x4−x3+11x2−12x+4 and state their multiplicities.