Download presentation

Presentation is loading. Please wait.

Published byBlanche Hawkins Modified over 4 years ago

1
Introduction You have learned several methods for solving polynomial equations by determining the factors, but not all equations are factorable. In this lesson, you will learn how to find the roots of a polynomial equation with integer coefficients that is not factorable. 1 2.3.4: The Rational Root Theorem

2
Key Concepts You can find the possible rational roots of a polynomial equation with integer coefficients using the Rational Root Theorem. This theorem states that if a polynomial has rational roots, then those roots will be the ratio of the factors of the constant term to the factors of the leading coefficient written in lowest terms. Be sure to include all factors of both the constant term and the leading coefficient, including both the positive and negative factors. It is not necessary to repeat duplicate ratios. 2 2.3.4: The Rational Root Theorem

3
Key Concepts, continued 3 2.3.4: The Rational Root Theorem Rational Root Theorem If the polynomial p(x) has integer coefficients, then every rational root of the polynomial equation p(x) = 0 can be written in the form, where p is a factor of the constant term of p(x) and q is a factor of the leading coefficient of p(x).

4
Key Concepts, continued It is possible for polynomial equations to have no rational roots; polynomial equations could also have irrational roots as stated in the Irrational Root Theorem. 4 2.3.4: The Rational Root Theorem Irrational Root Theorem If a polynomial p(x) has rational coefficients and is a root of the polynomial equation p(x) = 0, where a and b are rational and is irrational, then is also a root of p(x) = 0.

5
Key Concepts, continued Recall that it is also possible for an equation to have imaginary roots in the form a + bi and a – bi, referred to as complex conjugates. If the imaginary number a + bi is a root of a polynomial equation with real coefficients, then the conjugate a – bi is also a root. 5 2.3.4: The Rational Root Theorem

6
Common Errors/Misconceptions misinterpreting the Rational Root Theorem as stating that all zeros of a polynomial function must be rational not identifying both the positive and negative factors of the constant term and leading coefficients not finding all factors of a polynomial function 6 2.3.4: The Rational Root Theorem

7
Guided Practice Example 2 Find all of the roots to the polynomial equation 4x 3 – x 2 + 36x – 9 = 0. 7 2.3.4: The Rational Root Theorem

8
Guided Practice: Example 2, continued 1.List all the possible rational roots. The Rational Root Theorem states that the only possible rational roots of the equation must be ratios of the factors of the constant term and the factors of the leading coefficient. The constant term is –9 and the leading coefficient is 4. The factors of the constant term, –9, are ±1, ±3, and ±9. The factors of the leading coefficient, 4, are ±1, ±2, and ±4. 8 2.3.4: The Rational Root Theorem

9
Guided Practice: Example 2, continued Use these factors to set up the ratios. 9 2.3.4: The Rational Root Theorem Ratio of the factors of the constant term and the factors of the leading coefficient Write a separate ratio for each of the factors.

10
Guided Practice: Example 2, continued Simplify each ratio. 10 2.3.4: The Rational Root Theorem

11
Guided Practice: Example 2, continued 2.Determine the number of possible roots of the polynomial equation. 4x 3 – x 2 + 36x – 9 = 0 is a third-degree polynomial. Therefore, this equation will have either 3 real roots or 1 real root and 2 complex roots. 11 2.3.4: The Rational Root Theorem

12
Guided Practice: Example 2, continued 3.Test each possible rational root to find one root. Begin by selecting either a positive or negative possible rational root. Substitute the chosen possibility into the original polynomial. If the result is 0, the number substituted is a rational root. If the result is not 0, then the chosen number is not a rational root. 12 2.3.4: The Rational Root Theorem

13
Guided Practice: Example 2, continued First, try the possible root 1. 4x 3 – x 2 + 36x – 9Original polynomial 4(1) 3 – (1) 2 + 36(1) – 9 Substitute 1 for x. 4 – 1 + 36 – 9 Simplify. 30 Substituting 1 for x does not result in the polynomial equaling 0; therefore, 1 is not a rational root of the polynomial equation. 13 2.3.4: The Rational Root Theorem

14
Guided Practice: Example 2, continued Next, try the possible root. 4x 3 – x 2 + 36x – 9Original polynomial Substitute for x. Simplify. 14 2.3.4: The Rational Root Theorem

15
Guided Practice: Example 2, continued Substituting for x does not result in the polynomial equaling 0; therefore, is not a rational root of the polynomial equation. Next, try the possible root. 15 2.3.4: The Rational Root Theorem

16
Guided Practice: Example 2, continued 4x 3 – x 2 + 36x – 9Original polynomial Substitute for x. Simplify. 0 Substituting for x does result in the polynomial equaling 0; therefore, is a rational root of the polynomial equation. 16 2.3.4: The Rational Root Theorem

17
Guided Practice: Example 2, continued 4.Use synthetic division to find the roots of the depressed polynomial equation. The divisor is. The depressed polynomial is 4x 2 + 36. Find the roots of 4x 2 + 36 = 0. 17 2.3.4: The Rational Root Theorem

18
Guided Practice: Example 2, continued 4x 2 + 36 = 0Depressed polynomial 4(x 2 + 9) = 0Factor out the greatest common factor, 4. x 2 + 9 = 0Divide both sides by 4. x 2 = –9Subtract 9 from both sides. Take the square root of both sides. Simplify. x = ±3i The roots of the depressed polynomial equation 4x 2 + 36 = 0 are ±3i. 18 2.3.4: The Rational Root Theorem

19
Guided Practice: Example 2, continued 5.List the roots of the original polynomial equation. The roots of the equation 4x 3 – x 2 + 36x – 9 = 0 are, 3i, and –3i. 19 2.3.4: The Rational Root Theorem ✔

20
Guided Practice: Example 2, continued 20 2.3.4: The Rational Root Theorem

21
Guided Practice Example 3 Find a third-degree polynomial with rational coefficients that has the roots 6 and 3 – i. 21 2.3.4: The Rational Root Theorem

22
Guided Practice: Example 3, continued 1.Use the Imaginary Root Theorem to find the remaining root. 3 – i is one root. The complex conjugate of 3 – i is 3 + i. 22 2.3.4: The Rational Root Theorem

23
Guided Practice: Example 3, continued 2.Write the polynomial expression. The factors of the polynomial are (x – 6), x – (3 – i), and x – (3 + i). (x – 6)[x – (3 – i)][x – (3 + i)] Write the polynomial in factored form. (x – 6)[x 2 – x(3 + i ) – x(3 – i ) Distribute. + (3 – i )(3 + i)] (x – 6)(x 2 – 3x – ix – 3x + Simplify. ix + 9 + 3i – 3i – i 2 ) 23 2.3.4: The Rational Root Theorem

24
Guided Practice: Example 3, continued (x – 6)[x 2 – 6x + 9 – (–1)]Replace i 2 with –1. (x – 6)(x 2 – 6x + 10) Simplify. x 3 – 12x 2 + 46x – 60 Distribute. x 3 – 12x 2 + 46x – 60 is a third-degree polynomial with rational coefficients and the roots 6 and 3 – i. 24 2.3.4: The Rational Root Theorem ✔

25
Guided Practice: Example 3, continued 25 2.3.4: The Rational Root Theorem

Similar presentations

© 2019 SlidePlayer.com Inc.

All rights reserved.

To make this website work, we log user data and share it with processors. To use this website, you must agree to our Privacy Policy, including cookie policy.

Ads by Google