Question

This problem has two parts, Part A and Part B. Part A: Are all of the roots of the polynomial P(x)=x3+3x2−11x−5 rational numbers? Why or why not? Part B: What are the rational roots of P(x)=x3+3x2−11x−5? Select one answer for Part A and select all answers that apply for Part B. A: The answer cannot be determined, because the Rational Root Theorem does not tell you the roots of P(x). B: 5 B: −1 B: −5 B: 1 B: 1 +√2 A: Yes, if you apply the Rational Root Theorem and the definition of a root, you find 3 distinct rational roots. Since P(x) is a third-degree polynomial, these must be all of the roots of P(x). B: 1−√2 A: No, if you apply the Rational Root Theorem and the definition of a root, you find only 1 distinct rational root. Since P(x) is a third-degree polynomial, there must also be a conjugate pair of irrational roots.

Answer 1

Given:

This problem has two parts, Part A and Part B.  

To find:

Part A: Are all of the roots of the polynomial P(x)=x3+3x2−11x−5 rational numbers? Why or why not?  

Part B: What are the rational roots of P(x)=x3+3x2−11x−5?

Select one answer for Part A and select all answers that apply for Part B.  

Solution:

From given, we have,

The polynomial P(x) = x3+3x2−11x−5

Let x³ + 3x² - 11x - 5 = 0 to find it's roots.

x³ + 3x² - 11x - 5 = 0

(x + 5) (x² - 2x - 1) = 0

x + 5 = 0   and (x² - 2x - 1) = 0

consider, x + 5 = 0

x = -5

consider (x² - 2x - 1) = 0

x = -(-2) ± √[(-2)² - 4(1)(-1)] / 2(1)

x = 1 ± √2

Part A: The roots of the P(x)=x3+3x2−11x−5  are 5, 1 ± √2.

Only one root of P(x) is rational, that is, x = 5

The other roots are 1 ± √2 since √2 is irrational, so these roots are irrational.

Part B: 5 is the rational root of P(x).