Lesson 1: Solutions to Polynomial Equations

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NYS COMMON CORE MATHEMATICS CURRICULUM Lesson 1 M3 PRECALCULUS AND ADVANCED TOPICS

Student Outcomes . Students determine all solutions of polynomial equations over the set of complex numbers and understand the implications of the fundamental theorem of algebra, especially for quadratic polynomials.

Lesson Notes Students studied polynomial equations and the nature of the solutions of these equations extensively in Algebra II Module 1, extending factoring to the complex realm. The fundamental theorem of algebra indicates that any polynomial function of degree 푛 will have 푛 zeros (including repeated zeros). Establishing the fundamental theorem of algebra was one of the greatest achievements of nineteenth-century mathematics. It is worth spending time further exploring it now that students have a much broader understanding of complex numbers. This lesson reviews what they learned in previous grades and provides additional support for their understanding of what it means to solve polynomial equations over the set of complex numbers. Students work with equations with complex number solutions and apply identities such as 푎2 + 푏2 = (푎 + 푏푖)(푎 − 푏푖) for real numbers 푎 and 푏 to solve equations and explore the implications of the fundamental theorem of algebra (N-CN.C.8 and N-CN.C.9). Throughout the lesson, students vary their reasoning by applying algebraic properties (MP.3) and examining the structure of expressions to support their solutions and make generalizations (MP.7 and MP.8). Relevant definitions introduced in Algebra II are provided in the student materials for this lesson. A note on terminology: Equations have solutions, and functions have zeros. The distinction is subtle but important. For example, the equation (푥 − 1)(푥 − 3) = 0 has solutions 1 and 3, while the polynomial function 푝(푥) = (푥 − 1)(푥 − 3) has zeros at 1 and 3. Zeros of a function are the 푥-intercepts of the graph of the function; they are also known as roots.

Classwork Opening Exercise (3 minutes) Scaffolding: . Remind students of the Use this Opening Exercise to activate prior knowledge about polynomial equations. definition of a polynomial Students may need to be reminded of the definition of a polynomial from previous grades. equation by using a Frayer Have students work this exercise independently and then quickly share their answers with model. For an example, a partner. Lead a short discussion using the questions below. see Module 1 Lesson 5. . Be sure to include Opening Exercise examples 2 How many solutions are there to the equation 풙ퟐ = ퟏ? Explain how you know. (2푥 + 3 = 0, 푥 − 4 = 0, (푥 − 3)(푥 + 2) = 2푥 − 5, There are two solutions to the equation: ퟏ and −ퟏ. I know these are the solutions because they and 푥5 = 1) make the equation true when each value is substituted for 풙. and non-examples 1 (sin(2푥) − 1 = 0, = 5, 푥 and 23푥 − 1 = 5).

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NYS COMMON CORE MATHEMATICS CURRICULUM Lesson 1 M3 PRECALCULUS AND ADVANCED TOPICS

. How do you know that there aren’t any more real number solutions?  If you sketch the graph of 푓(푥) = 푥2 and the graph of the line 푦 = 1, they intersect in exactly two points. The 푥-coordinates of the intersection points are the solutions to the equation. . How can you show algebraically that this equation has just two solutions?  Rewrite the equation 푥2 = 1, and solve it by factoring. Then, apply the zero product property. 푥2 − 1 = 0 or (푥 − 1)(푥 + 1) = 0

푥 − 1 = 0 or 푥 + 1 = 0

푥 = 1 or 푥 = −1

 So, the solutions are 1 and −1. . You just found and justified why this equation has only two real number solutions. How do we know that there aren’t any complex number solutions to 푥2 = 1?  We would have to show that a second-degree polynomial equation has exactly two solutions over the set of complex numbers. (Another acceptable answer would be the fundamental theorem of algebra, which states that a second-degree polynomial equation has at most two solutions. Since we have found two solutions that are real, we know we have found all possible solutions.) . Why do the graphical approach and the algebraic approach not clearly provide an answer to the previous question?  The graphical approach assumes you are working with real numbers. The algebraic approach does not clearly eliminate the possibility of complex number solutions. It only shows two real number solutions.

Example 1 (5 minutes): Prove that a Quadratic Equation Has Only Two Solutions over the Set of Complex Numbers

This example illustrates an approach to showing that 1 and −1 are the only real or complex solutions to the quadratic equation 푥2 = 1. Students may not have seen this approach before. However, they should be very familiar with operations with complex numbers after their work in Algebra II Module 1 and Module 1 of this course. We work with solutions to 푥푛 = 1 in later lessons, and this approach is most helpful.

Example 1: Prove that a Quadratic Equation Has Only Two Solutions over the Set of Complex Numbers

Prove that ퟏ and −ퟏ are the only solutions to the equation 풙ퟐ = ퟏ. Scaffolding: 풙 = 풂 + 풃풊 풙ퟐ = ퟏ Let be a complex number so that . Ask advanced learners to a. Substitute 풂 + 풃풊 for 풙 in the equation 풙ퟐ = ퟏ. develop the proof in Example 1 MP.3 (풂 + 풃풊)ퟐ = ퟏ on their own without the leading questions.

b. Rewrite both sides in standard form for a complex number.

풂ퟐ + ퟐ풂풃풊 + 풃ퟐ풊ퟐ = ퟏ (풂ퟐ − 풃ퟐ) + ퟐ풂풃풊 = ퟏ + ퟎ풊

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NYS COMMON CORE MATHEMATICS CURRICULUM Lesson 1 M3 PRECALCULUS AND ADVANCED TOPICS

c. Equate the real parts on each side of the equation and equate the imaginary parts on each side of the equation.

풂ퟐ − 풃ퟐ = ퟏ and ퟐ풂풃 = ퟎ

d. Solve for 풂 and 풃, and find the solutions for 풙 = 풂 + 풃풊.

MP.3 For ퟐ풂풃 = ퟎ, either 풂 = ퟎ or 풃 = ퟎ. If 풂 = ퟎ, then −풃ퟐ = ퟏ, which gives us 풃 = 풊 or 풃 = −풊.

Substituting into 풙 = 풂 + 풃풊 gives us ퟎ + 풊 ∙ 풊 = −ퟏ or ퟎ − 풊 ∙ 풊 = ퟏ.

If 풃 = ퟎ, then 풂ퟐ = ퟏ, which gives us 풂 = ퟏ or 풂 = −ퟏ.

Substituting into 풙 = 풂 + 풃풊 gives us ퟏ + ퟎ ∙ 풊 = ퟏ or −ퟏ + ퟎ ∙ 풊 = −ퟏ.

Thus, the only complex number solutions to this equation are the complex numbers ퟏ + ퟎ풊 or −ퟏ + ퟎ풊.

The quadratic formula also proves that the only solutions to the equation are 1 and −1 even if we solve the equation over the set of complex numbers. If this did not come up earlier in this lesson as students shared their thinking on the Opening Exercise, discuss it now. . What is the quadratic formula?  The formula that provides the solutions to a quadratic equation when it is written in the form 푎푥2 + 푏푥 + 푐 = 0 where 푎 ≠ 0.

−푏+√푏2−4푎푐 −푏−√푏2−4푎푐  If 푎푥2 + 푏푥 + 푐 = 0 and 푎 ≠ 0, then 푥 = or 푥 = , and the solutions to the 2푎 2푎

−푏+√푏2−4푎푐 −푏−√푏2−4푎푐 equation are the complex numbers and . 2푎 2푎 . How does the quadratic formula guarantee that a quadratic equation has at most two solutions over the set of complex numbers?  The quadratic formula is a general solution to the equation 푎푥2 + 푏푥 + 푐 = 0, where 푎 ≠ 0. The −푏+√푏2−4푎푐 −푏−√푏2−4푎푐 formula shows two solutions: 푥 = and 푥 = . There is only one solution 2푎 2푎

if 푏2 − 4푎푐 = 0. There are two distinct real number solutions when 푏2 − 4푎푐 > 0, and there are two distinct complex number solutions when 푏2 − 4푎푐 < 0.

Exercises 1–6 (5 minutes) Allow students time to work these exercises individually, and then discuss as a class. Note that this is very similar to the Opening Exercise used in Algebra II Module 1 Lesson 40, with an added degree of difficulty since the coefficients of the polynomial are also complex. Exercises 1–6 are a review of patterns in the factors of the polynomials below for real numbers 푎 and 푏. 푎2 − 푏2 = (푎 + 푏)(푎 − 푏) 푎2 + 푏2 = (푎 + 푏푖)(푎 − 푏푖) Exercises 1 and 2 ask students to multiply binomials that have a product that is a sum or difference of squares. Exercises 3–6 ask students to factor polynomials that are sums and differences of squares.

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NYS COMMON CORE MATHEMATICS CURRICULUM Lesson 1 M3 PRECALCULUS AND ADVANCED TOPICS

Exercises

Find the product.

1. (풛 − ퟐ)(풛 + ퟐ) 풛ퟐ − ퟒ

2. (풛 + ퟑ풊)(풛 − ퟑ풊)

풛ퟐ + ퟗ

Write each of the following quadratic expressions as the product of two linear factors.

3. 풛ퟐ − ퟒ (풛 + ퟐ)(풛 − ퟐ)

4. 풛ퟐ + ퟒ (풛 + ퟐ풊)(풛 − ퟐ풊)

5. 풛ퟐ − ퟒ풊 ퟏ+풊 (풛 + ퟐ√풊)(풛 − ퟐ√풊) or (풛 + (ퟏ + 풊)√ퟐ)(풛 − (ퟏ + 풊)√ퟐ) [using the fact that √풊 = ] √ퟐ

6. 풛ퟐ + ퟒ풊 ퟏ+풊 (풛 + ퟐ풊√풊)(풛 − ퟐ풊√풊) or (풛 + (풊 − ퟏ)√ퟐ)(풛 − (풊 − ퟏ)√ퟐ) [using the fact that √풊 = ] √ퟐ

Students may be curious about the square roots of a complex number. They may recall from Precalculus Module 1, that we studied these when considering the polar form of a complex number. This question is also addressed in Lesson 2 from an algebraic perspective. . How did we know that each quadratic expression could be factored into two linear terms?  The fundamental theorem of algebra guarantees that a polynomial of degree 2 can be factored into 2 linear factors. We proved this was true for quadratic expressions by using the solutions produced with the quadratic formula to write the expression as two linear factors. . Does the fundamental theorem of algebra apply even if the coefficients are non-real numbers?  It still held true for Exercises 3 and 4, so it seems to, at least if the constant is a non-real number.

Exercises 7–10 (10 minutes) Students should work the following exercises in small groups. Have different groups come to the board and present their solutions to the class.

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NYS COMMON CORE MATHEMATICS CURRICULUM Lesson 1 M3 PRECALCULUS AND ADVANCED TOPICS

7. Can a quadratic polynomial equation with real coefficients have one real solution and one complex solution? If so, give an example of such an equation. If not, explain why Scaffolding: not. Provide several factored quadratic The quadratic formula shows that if the discriminant 풃ퟐ − ퟒ풂풄 is negative, both polynomial equations, and ask solutions are complex numbers that are complex conjugates. If it is positive, both students to identify the solutions solutions are real. If it is zero, there is one (repeated) real solution. We cannot have a and write them in standard form. real solution coupled with a complex solution because complex solutions occur in conjugate pairs. (푥 − 2)(푥 + 2) = 0 (2푥 − 3)(2푥 + 3) = 0 Recall from Algebra II that every quadratic expression can be written as a product of two (푥 − 2푖)(푥 + 2푖) = 0 linear factors, that is, (2푥 − 푖)(2푥 + 푖) = 0 ퟐ ( )( ) 풂풙 + 풃풙 + 풄 = 풂 풙 − 풓ퟏ 풙 − 풓ퟐ , (푥 − 1 + 푖)(푥 − 1 − 푖) = 0 ퟐ where 풓ퟏ and 풓ퟐ are solutions of the polynomial equation 풂풙 + 풃풙 + 풄 = ퟎ. (푥 − (1 + 2푖))(푥 − (1 − 2푖)) = 0

(푥 − 1 + √3)(푥 − 1 − √3) = 0 8. Solve each equation by factoring, and state the solutions. a. 풙ퟐ + ퟐퟓ = ퟎ (풙 + ퟓ풊)(풙 − ퟓ풊) = ퟎ The solutions are ퟓ풊 and −ퟓ풊.

b. 풙ퟐ + ퟏퟎ풙 + ퟐퟓ = ퟎ (풙 + ퟓ)(풙 + ퟓ) = ퟎ The solution is −ퟓ.

9. Give an example of a quadratic equation with ퟐ + ퟑ풊 as one of its solutions.

We know that if ퟐ + ퟑ풊 is a solution of the equation, then its conjugate ퟐ − ퟑ풊 must also be a solution.

(풙 − (ퟐ + ퟑ풊))(풙 − (ퟐ − ퟑ풊)) = ((풙 − ퟐ) − ퟑ풊)((풙 − ퟐ) + ퟑ풊)

Using the structure of this expression, we have (풂 − 풃풊)(풂 + 풃풊) where 풂 = 풙 − ퟐ and 풃 = ퟑ. Since (풂 + 풃풊)(풂 − 풃풊) = 풂ퟐ + 풃ퟐ for all real numbers 풂 and 풃,

(풙 − ퟐ − ퟑ풊)(풙 − ퟐ + ퟑ풊) = (풙 − ퟐ)ퟐ + ퟑퟑ = 풙ퟐ − ퟒ풙 + ퟒ + ퟗ = 풙ퟐ − ퟒ풙 + ퟏퟑ.

10. A quadratic polynomial equation with real coefficients has a complex solution of the form 풂 + 풃풊 with 풃 ≠ ퟎ. What must its other solution be and why?

The other solution is 풂 − 풃풊. If the polynomial equation must have real coefficients, then (풙 − (풂 + 풃풊))(풙 − (풂 − 풃풊)) when multiplied must yield an expression with real number coefficients.

(풙 − (풂 + 풃풊))(풙 − (풂 − 풃풊)) = (풙 − 풂 − 풃풊)(풙 − 풂 + 풃풊) = (풙 − 풂)ퟐ − 풃ퟐ풊ퟐ = (풙 − 풂)ퟐ + 풃ퟐ = 풙ퟐ − ퟐ풂풙 + 풂ퟐ + 풃ퟐ

Since 풂 and 풃 are real numbers, this expression always has real number coefficients.

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Debrief these exercises by having different groups share their approaches. Give students enough time to struggle with these exercises in their small groups. Use the results of these exercises to further plan for reteaching if students cannot recall what they learned in Algebra I and Algebra II.

Discussion (5 minutes) Use this discussion to help students recall the fundamental theorem of algebra first introduced in Algebra II Module 1 Lesson 40. . What are the solutions to the polynomial equation (푥 − 1)(푥 + 2푖)(푥 − 2푖) = 0? What is the degree of this equation?  The solutions are 1, 2푖, and −2푖. This is a third-degree equation. . What are the solutions to the polynomial equation (푥 − 1)(푥 + 1)(푥 − 2푖)(푥 + 2푖) = 0? What is the degree of this equation?  The solutions are 1, −1, 2푖, and −2푖. This is a fourth-degree equation. . Predict how many solutions the equation 푥5 − 3푥3 + 2푥 = 0 has. Justify your response.  It should have at most 5 solutions. It is a fifth-degree polynomial. . To find the solutions, we need to write 푥5 − 3푥3 + 2푥 as a product of 5 linear MP.7 Scaffolding: factors. Explain how to factor this polynomial. & Have advanced learners factor MP.8 4 2  Factor out 푥, giving us 푥(푥 − 3푥 + 2). 푥4 − 3푥2 + 2 without the hint. . How can you factor (푥4 − 3푥2 + 2)?  We know 푥4 = (푥2)2, so let 푢 = 푥2. This gives us a polynomial 푢2 − 3푢 + 2 in terms of 푢, which factors into (푢 − 1)(푢 − 2). Now substitute 푥2 for 푢, and we have (푥2 − 1)(푥2 − 2). . What is the factored form of the equation?  푥(푥2 − 1)(푥2 − 2) = 푥(푥 + 1)(푥 − 1)(푥 + √2)(푥 − √2) = 0 . What are the solutions to the equation 푥5 − 3푥3 + 2푥 = 0?  The solutions are 0, 1, −1, √2, and −√2. . How many solutions does a degree 푛 polynomial equation have? Explain your reasoning.  If every polynomial equation can be written as the product of 푛 linear factors, then there are at most 푛 solutions. This is an appropriate point to reintroduce the fundamental theorem of algebra. For more details or to provide additional background information, refer back to Algebra II Module 1.

Fundamental Theorem of Algebra 1. Every polynomial function of degree 푛 ≥ 1 with real or complex coefficients has at least one real or complex zero. 2. Every polynomial of degree 푛 ≥ 1 with real or complex coefficients can be factored into 푛 linear terms with real or complex coefficients.

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Continue the discussion, and have students write any examples and their summaries in the space below each question. Have students discuss both of these questions in their small groups before leading a whole-group discussion. . Could a polynomial function of degree 푛 have more than 푛 zeros? Explain your reasoning.  If a polynomial function has 푛 zeros, then it has 푛 linear factors, which means the degree is 푛. . Could a polynomial function of degree 푛 have less than 푛 zeros? Explain your reasoning.  Yes. If a polynomial has repeated linear factors, then it has less than 푛 distinct zeros. For example, 푝(푥) = (푥 − 1)푛 has only one zero: the number 1.

Exercises 11–15 (10 minutes) Give students time to work on the exercises either individually or with a partner, and then share answers as a class. On Exercises 11 and 12, students need to recall polynomial division from Algebra II. Students divided polynomials using both the reverse tabular method and long division. Consider reviewing one or both of these methods with students. In Exercise 11, students may recall the sum and difference of cube formulas derived through polynomial long division in Algebra II. Students should have access to technology to aid with problems such as Exercise 12. However, in Exercise 12, consider asking students to verify that 푥 = 2 is a zero of 푝 rather than having them use technology to locate the zero.

11. Write the left side of each equation as a product of linear factors, and state the solutions. a. 풙ퟑ − ퟏ = ퟎ Scaffolding:

ퟎ = 풙ퟑ − ퟏ = (풙 − ퟏ)(풙ퟐ + 풙 + ퟏ) Display the sum and difference of cube formulas on the board −ퟏ±√ퟏ−ퟒ −ퟏ±풊√ퟑ 풙 = = Using the quadratic formula, ퟐ ퟐ , so the quadratic expression once students have completed 풙ퟐ + 풙 + ퟏ factors into Exercise 11 as a reminder for future work. −ퟏ + 풊√ퟑ −ퟏ − 풊√ퟑ 풙ퟐ + 풙 + ퟏ = (풙 − ( )) (풙 − ( )) . 푎3 − 푏3 = ퟐ ퟐ (푎 − 푏)(푎2 + 푎푏 + 푏2) Then, the factored form of the equation is 푎3 + 푏3 = −ퟏ + 풊√ퟑ −ퟏ − 풊√ퟑ (푎 + 푏)(푎2 − 푎푏 + 푏2) (풙 − ퟏ) (풙 − ( )) (풙 − ( )) = ퟎ. ퟐ ퟐ

−ퟏ+풊√ퟑ −ퟏ−풊√ퟑ The solutions are ퟏ, , and . ퟐ ퟐ

b. 풙ퟑ + ퟖ = ퟎ

풙ퟑ + ퟖ = (풙 + ퟐ)(풙ퟐ − ퟐ풙 + ퟒ)

ퟐ±√ퟒ−ퟏퟔ ퟐ±√−ퟏퟐ ퟐ±ퟐ√−ퟑ 풙 = = = = ퟏ ± 풊 ퟑ Using the quadratic formula, ퟐ ퟐ ퟐ √ , so the quadratic expression 풙ퟐ − ퟐ풙 + ퟒ factors into

풙ퟐ ± ퟐ + ퟒ = (풙 − (ퟏ + 풊√ퟑ)) (풙 − (ퟏ − 풊√ퟑ)) .

Then, the factored form of the equation is

(풙 + ퟐ) (풙 − (ퟏ + 풊√ퟑ)) (풙 − (ퟏ − 풊√ퟑ)) = ퟎ.

The solutions are −ퟐ, ퟏ + 풊 √ퟑ, and ퟏ − 풊√ퟑ.

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c. 풙ퟒ + ퟕ풙ퟐ + ퟏퟎ = ퟎ

(풙ퟐ + ퟓ)(풙ퟐ + ퟐ) = (풙 + 풊√ퟓ)(풙 − 풊√ퟓ)(풙 + 풊√ퟐ)(풙 − 풊√ퟐ) = ퟎ

The solutions are ±풊√ퟓ and ±풊√ퟐ.

12. Consider the polynomial 풑(풙) = 풙ퟑ + ퟒ풙ퟐ + ퟔ풙 − ퟑퟔ. a. Graph 풚 = 풙ퟑ + ퟒ풙ퟐ + ퟔ풙 − ퟑퟔ, and find the real zero of polynomial 풑.

MP.5 The graph of 풑 has an 풙-intercept at 풙 = ퟐ. Therefore, 풙 = ퟐ is a zero of 풑.

b. Write 풑 as a product of linear factors.

풑(풙) = (풙 − ퟐ)(풙 + ퟑ − ퟑ풊)(풙 + ퟑ + ퟑ풊)

c. What are the solutions to the equation 풑(풙) = ퟎ?

The solutions are ퟐ, −ퟑ + ퟑ풊, and −ퟑ − ퟑ풊.

13. Malaya was told that the volume of a box that is a cube is ퟒ, ퟎퟗퟔ cubic inches. She knows the formula for the volume of a cube with side length 풙 is 푽(풙) = 풙ퟑ, so she models the volume of the box with the equation 풙ퟑ − ퟒퟎퟗퟔ = ퟎ. a. Solve this equation for 풙.

풙ퟑ − ퟒퟎퟗퟔ = (풙 − ퟏퟔ)(풙ퟐ + ퟏퟔ풙 + ퟐퟓퟔ) = ퟎ 풙 = ퟏퟔ, 풙 = −ퟖ + ퟖ√ퟑ풊, 풙 = −ퟖ − ퟖ√ퟑ풊

b. Malaya shows her work to Tiffany and tells her that she has found three different values for the side length of the box. Tiffany looks over Malaya’s work and sees that it is correct but explains to her that there is only one valid answer. Help Tiffany explain which answer is valid and why.

Since we are looking for the dimensions of a box, only real solutions are acceptable, so the answer is ퟏퟔ inches.

14. Consider the polynomial 풑(풙) = 풙ퟔ − ퟐ풙ퟓ + ퟕ풙ퟒ − ퟏퟎ풙ퟑ + ퟏퟒ풙ퟐ − ퟖ풙 + ퟖ. a. Graph 풚 = 풙ퟔ − ퟐ풙ퟓ + ퟕ풙ퟒ − ퟏퟎ풙ퟑ + ퟏퟒ풙ퟐ − ퟖ풙 + ퟖ, and state the number of real zeros of 풑.

There are no real zeros.

b. Verify that 풊 is a zero of 풑.

풑(풊) = 풊ퟔ − ퟐ풊ퟓ + ퟕ풊ퟒ − ퟏퟎ풊ퟑ + ퟏퟒ풊ퟐ − ퟖ풊 + ퟖ = −ퟏ − ퟐ풊 + ퟕ + ퟏퟎ풊 − ퟏퟒ − ퟖ풊 + ퟖ = ퟎ

c. Given that 풊 is a zero of 풑, state another zero of 풑.

Another zero is −풊.

d. Given that ퟐ풊 and ퟏ + 풊 are also zeros of 풑, explain why polynomial 풑 cannot possibly have any real zeros.

MP.3 Since ퟐ풊 and ퟏ + 풊 are zeros, −ퟐ풊 and ퟏ − 풊 must also be zeros of 풑. The fundamental theorem of algebra tells us that since 풑 is a degree-ퟔ polynomial it can be written as a product of ퟔ linear factors. We now know that 풑 has ퟔ complex zeros and therefore cannot have any real zeros.

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NYS COMMON CORE MATHEMATICS CURRICULUM Lesson 1 M3 PRECALCULUS AND ADVANCED TOPICS

e. What is the solution set to the equation 풑(풙) = ퟎ?

The solution set is {풊, −풊, ퟐ풊, −ퟐ풊, ퟏ + 풊, ퟏ − 풊}.

15. Think of an example of a sixth-degree polynomial equation that, when written in standard form, has integer coefficients, four real number solutions, and two imaginary number solutions. How can you be sure your equation will have integer coefficients?

One correct response is (풙 − ퟏ)(풙 + ퟏ)(풙 − ퟐ)(풙 + ퟐ)(풙 − ퟐ풊)(풙 + ퟐ풊) = ퟎ. By selecting an imaginary number 풃풊, where 풃 is an integer, and its conjugate as solutions, we know by the identity (풂 + 풃풊)(풂 − 풃풊) = 풂ퟐ + 풃ퟐ that these factors produce a quadratic expression with integer coefficients. If we choose the real solutions to also be integers, then, when written in standard form, the polynomial equation has integer coefficients.

Closing (4 minutes) Review the information in the Lesson Summary box by asking students to choose one vocabulary term, theorem, or identity and paraphrase it with a partner. Select students to share their paraphrasing with the class.

Lesson Summary

Relevant Vocabulary

POLYNOMIAL FUNCTION: Given a polynomial expression in one variable, a polynomial function in one variable is a function 풇: ℝ → ℝ such that for each real number 풙 in the domain, 풇(풙) is the value found by substituting the number 풙 into all instances of the variable symbol in the polynomial expression and evaluating.

It can be shown that if a function 풇: ℝ → ℝ is a polynomial function, then there is some nonnegative integer 풏 and

collection of real numbers 풂ퟎ, 풂ퟏ, 풂ퟐ,… , 풂풏 with 풂풏 ≠ ퟎ such that the function satisfies the equation

풏 풏−ퟏ 풇(풙) = 풂풏풙 + 풂풏−ퟏ풙 + ⋯ + 풂ퟏ풙 + 풂ퟎ, for every real number 풙 in the domain, which is called the standard form of the polynomial function. The function 풇(풙) = ퟑ풙ퟑ + ퟒ풙ퟐ + ퟒ풙 + ퟕ, where 풙 can be any real number, is an example of a function written in standard form.

DEGREE OF A POLYNOMIAL FUNCTION: The degree of a polynomial function is the degree of the polynomial expression used to define the polynomial function. The degree is the highest degree of its terms.

The degree of 풇(풙) = ퟖ풙ퟑ + ퟒ풙ퟐ + ퟕ풙 + ퟔ is ퟑ, but the degree of 품(풙) = (풙 + ퟏ)ퟐ − (풙 − ퟏ)ퟐ is ퟏ because when 품 is put into standard form, it is 품(풙) = ퟒ풙.

ZEROS OR ROOTS OF A FUNCTION: A zero (or root) of a function 풇: ℝ → ℝ is a number 풙 of the domain such that 풇(풙) = ퟎ. A zero of a function is an element in the solution set of the equation 풇(풙) = ퟎ.

Given any two polynomial functions 풑 and 풒, the solution set of the equation 풑(풙)풒(풙) = ퟎ can be quickly found by solving the two equations 풑(풙) = ퟎ and 풒(풙) = ퟎ and combining the solutions into one set.

A number 풂 is zero of a polynomial function 풑 with multiplicity 풎 if the factored form of 풑 contains (풙 − 풂)풎.

Every polynomial function of degree 풏, for 풏 ≥ ퟏ, has 풏 zeros over the complex numbers, counted with multiplicity. Therefore, such polynomials can always be factored into 풏 linear factors.

Exit Ticket (3 minutes)

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NYS COMMON CORE MATHEMATICS CURRICULUM Lesson 1 M3 PRECALCULUS AND ADVANCED TOPICS

Name Date

Lesson 1: Solutions to Polynomial Equations

Exit Ticket

1. Find the solutions of the equation 푥4 − 푥2 − 12. Show your work.

2. The number 1 is a zero of the polynomial 푝(푥) = 푥3 − 3푥2 + 7푥 − 5. a. Write 푝(푥) as a product of linear factors.

b. What are the solutions to the equation 푥3 − 3푥2 + 7푥 − 5 = 0?

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NYS COMMON CORE MATHEMATICS CURRICULUM Lesson 1 M3 PRECALCULUS AND ADVANCED TOPICS

Exit Ticket Sample Solutions

1. Find the solutions of the equation 풙ퟒ − 풙ퟐ − ퟏퟐ. Show your work.

풙ퟒ − 풙ퟐ − ퟏퟐ = (풙ퟐ + ퟑ)(풙ퟐ − ퟒ) = (풙 − 풊√ퟑ)(풙 + 풊√ퟑ)(풙 − ퟐ)(풙 + ퟐ) = ퟎ

The solutions of the equation are 풊√ퟑ, −풊√ퟑ, ퟐ, −ퟐ.

2. The number ퟏ is a zero of the polynomial 풑(풙) = 풙ퟑ − ퟑ풙ퟐ + ퟕ풙 − ퟓ. a. Write 풑(풙) as a product of linear factors.

(풙 − ퟏ)(풙ퟐ − ퟐ풙 + ퟓ)

(풙 − ퟏ)(풙 − (ퟏ + ퟐ풊))(풙 − (ퟏ − ퟐ풊))

b. What are the solutions to the equation 풙ퟑ − ퟑ풙ퟐ + ퟕ풙 − ퟓ = ퟎ?

The solutions are ퟏ, ퟏ + ퟐ풊, and ퟏ − ퟐ풊.

Problem Set Sample Solutions

1. Find all solutions to the following quadratic equations, and write each equation in factored form. a. 풙ퟐ + ퟐퟓ = ퟎ 풙 = ±ퟓ풊, (풙 + ퟓ풊)(풙 − ퟓ풊) = ퟎ

b. −풙ퟐ − ퟏퟔ = −ퟕ 풙 = ±ퟑ풊, (풙 + ퟑ풊)(풙 − ퟑ풊) = ퟎ

c. (풙 + ퟐ)ퟐ + ퟏ = ퟎ

−ퟒ ± √ퟏퟔ − ퟐퟎ −ퟒ ± ퟐ풊 풙ퟐ + ퟒ풙 + ퟓ = ퟎ, 풙 = = = −ퟐ ± 풊 ퟐ ퟐ (풙 + ퟐ + 풊)(풙 + ퟐ − 풊) = ퟎ

d. (풙 + ퟐ)ퟐ = 풙

−ퟑ ± √ퟗ − ퟏퟔ −ퟑ ± √ퟑ풊 ퟑ √ퟑ풊 풙ퟐ + ퟑ풙 + ퟒ = ퟎ, 풙 = = = − ± ퟐ ퟐ ퟐ ퟐ ퟑ √ퟑ풊 ퟑ √ퟑ풊 (풙 + + ) (풙 + − ) = ퟎ ퟐ ퟐ ퟐ ퟐ

e. (풙ퟐ + ퟏ)ퟐ + ퟐ(풙ퟐ + ퟏ) − ퟖ = ퟎ

(풙ퟐ + ퟏ + ퟒ)(풙ퟐ + ퟏ − ퟐ) = ퟎ, 풙ퟐ + ퟓ = ퟎ, 풙 = ±√ퟓ풊, 풙ퟐ − ퟏ = ퟎ, 풙 = ±ퟏ

(풙 + √ퟓ풊)(풙 − √ퟓ풊)(풙 + ퟏ)(풙 − ퟏ) = ퟎ

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NYS COMMON CORE MATHEMATICS CURRICULUM Lesson 1 M3 PRECALCULUS AND ADVANCED TOPICS

f. (ퟐ풙 − ퟏ)ퟐ = (풙 + ퟏ)ퟐ − ퟑ

ퟒ풙ퟐ − ퟒ풙 + ퟏ − 풙ퟐ − ퟐ풙 − ퟏ + ퟑ = ퟎ, ퟑ풙ퟐ − ퟔ풙 + ퟑ = ퟎ, 풙ퟐ − ퟐ풙 + ퟏ = ퟎ,

(풙 − ퟏ)ퟐ = ퟎ, 풙 = ±ퟏ (풙 + ퟏ)(풙 − ퟏ) = ퟎ

g. 풙ퟑ + 풙ퟐ − ퟐ풙 = ퟎ

풙(풙ퟐ + 풙 − ퟐ) = ퟎ, 풙(풙 + ퟐ)(풙 − ퟏ) = ퟎ, 풙 = ퟎ, 풙 = −ퟐ, 풙 = ퟏ 풙(풙 + ퟐ)(풙 − ퟏ) = ퟎ

h. 풙ퟑ − ퟐ풙ퟐ + ퟒ풙 − ퟖ = ퟎ

풙ퟐ(풙 − ퟐ) + ퟒ(풙 − ퟐ) = ퟎ, (풙 − ퟐ)(풙ퟐ + ퟒ) = ퟎ, 풙 = ퟐ, 풙 = ±ퟐ풊 (풙 − ퟐ)(풙 + ퟐ풊)(풙 − ퟐ풊) = ퟎ

2. The following cubic equations all have at least one real solution. Find the remaining solutions. a. 풙ퟑ − ퟐ풙ퟐ − ퟓ풙 + ퟔ = ퟎ

One real solution is −ퟐ; then 풙ퟑ − ퟐ풙ퟐ − ퟓ풙 + ퟔ = (풙 + ퟐ)(풙ퟐ − ퟒ풙 + ퟑ) = (풙 + ퟐ)(풙 − ퟏ)(풙 − ퟑ).

The solutions are −ퟐ, ퟏ, ퟑ.

b. 풙ퟑ − ퟒ풙ퟐ + ퟔ풙 − ퟒ = ퟎ

One real solution is ퟐ; then 풙ퟑ − ퟒ풙ퟐ + ퟔ풙 − ퟒ = (풙 − ퟐ)(풙ퟐ − ퟐ풙 + ퟐ). Using the quadratic formula on 풙ퟐ − ퟐ풙 + ퟐ = ퟎ gives 풙 = ퟏ ± 풊.

The solutions are ퟐ, ퟏ + 풊, ퟏ − 풊.

c. 풙ퟑ + 풙ퟐ + ퟗ풙 + ퟗ = ퟎ

One real solution is −ퟏ; then 풙ퟑ + 풙ퟐ + ퟗ풙 + ퟗ = (풙 + ퟏ)(풙ퟐ + ퟗ) = (풙 + ퟏ)(풙 − ퟑ풊)(풙 + ퟑ풊).

The solutions are −ퟏ, ퟑ풊, −ퟑ풊.

d. 풙ퟑ + ퟒ풙 = ퟎ

One real solution is ퟎ; then (풙ퟑ + ퟒ풙) = 풙(풙ퟐ + ퟒ) = 풙(풙 + ퟐ풊)(풙 − ퟐ풊).

The solutions are ퟎ, ퟐ풊, −ퟐ풊.

e. 풙ퟑ + 풙ퟐ + ퟐ풙 + ퟐ = ퟎ

One real solution is 풙 = −ퟏ; then 풙ퟑ + 풙ퟐ + ퟐ풙 + ퟐ = (풙 + ퟏ)(풙ퟐ + ퟐ) = (풙 + ퟏ)(풙 − 풊√ퟐ)(풙 + 풊√ퟐ).

The solutions are −ퟏ, 풊√ퟐ, −풊√ퟐ.

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NYS COMMON CORE MATHEMATICS CURRICULUM Lesson 1 M3 PRECALCULUS AND ADVANCED TOPICS

3. Find the solutions of the following equations. a. ퟒ풙ퟒ − 풙ퟐ − ퟏퟖ = ퟎ

Set 풖 = 풙ퟐ. Then

ퟒ풙ퟒ − 풙ퟐ − ퟏퟖ = ퟒ풖ퟐ − 풖 − ퟏퟖ = (ퟒ풖 − ퟗ)(풖 + ퟐ) = (ퟒ풙ퟐ − ퟗ)(풙ퟐ + ퟐ) = (ퟐ풙 − ퟑ)(ퟐ풙 + ퟑ)(풙 + 풊√ퟐ)(풙 − 풊√ퟐ)

ퟑ ퟑ 풊 ퟐ −풊 ퟐ − The solutions are √ , √ , ퟐ, ퟐ.

b. 풙ퟑ − ퟖ = ퟎ (풙ퟑ − ퟖ) = (풙 − ퟐ)(풙ퟐ + ퟐ풙 + ퟒ) = (풙 − ퟐ) (풙 − (−ퟏ − 풊√ퟑ)) (풙 − (−ퟏ + 풊√ퟑ))

The solutions are ퟐ, −ퟏ + 풊√ퟑ, −ퟏ − 풊√ퟑ.

c. ퟖ풙ퟑ − ퟐퟕ = ퟎ (ퟖ풙ퟑ − ퟐퟕ) = (ퟐ풙 − ퟑ)(ퟒ풙ퟐ + ퟔ풙 + ퟗ)

−ퟑ − ퟑ풊√ퟑ −ퟑ + ퟑ풊√ퟑ = (ퟐ풙 − ퟑ) (ퟐ풙 − ( )) (ퟐ풙 − ( )) ퟐ ퟐ

ퟑ ퟑ ퟑ풊√ퟑ ퟑ ퟑ풊√ퟑ − + − − The solutions are ퟐ, ퟒ ퟒ , ퟒ ퟒ .

d. 풙ퟒ − ퟏ = ퟎ (풙ퟒ − ퟏ) = (풙ퟐ − ퟏ)(풙ퟐ + ퟏ) = (풙 − ퟏ)(풙 + ퟏ)(풙 + 풊)(풙 − 풊)

The solutions are ퟏ, −ퟏ, 풊, −풊.

e. ퟖퟏ풙ퟒ − ퟔퟒ = ퟎ (ퟖퟏ풙ퟒ − ퟔퟒ) = (ퟗ풙ퟐ − ퟖ)(ퟗ풙ퟐ + ퟖ) = (ퟑ풙 − ퟐ√ퟐ)(ퟑ풙 + ퟐ√ퟐ)(ퟑ풙 + ퟐ풊√ퟐ)(ퟑ풙 − ퟐ풊√ퟐ )

ퟐ√ퟐ ퟐ√ퟐ ퟐ√ퟐ ퟐ√ퟐ 풊 − 풊 − The solutions are ퟑ , ퟑ , ퟑ , ퟑ .

f. ퟐퟎ풙ퟒ + ퟏퟐퟏ풙ퟐ − ퟐퟓ = ퟎ (ퟐퟎ풙ퟒ + ퟏퟐퟏ풙ퟐ − ퟐퟓ) = (ퟒ풙ퟐ + ퟐퟓ)(ퟓ풙ퟐ − ퟏ) = (ퟐ풙 + ퟓ풊)(ퟐ풙 − ퟓ풊)(√ퟓ풙 − ퟏ)(√ퟓ풙 + ퟏ)

ퟓ ퟓ √ퟓ √ퟓ 풊 − 풊 − The solutions are ퟐ , ퟐ , ퟓ , ퟓ .

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NYS COMMON CORE MATHEMATICS CURRICULUM Lesson 1 M3 PRECALCULUS AND ADVANCED TOPICS

g. ퟔퟒ풙ퟑ + ퟐퟕ = ퟎ (ퟔퟒ풙ퟑ + ퟐퟕ) = (ퟒ풙 + ퟑ)(ퟏퟔ풙ퟐ − ퟏퟐ풙 + ퟗ) = (ퟒ풙 + ퟑ) (ퟖ풙 − ퟑ(ퟏ − 풊√ퟑ)) (ퟖ풙 − ퟑ(ퟏ + 풊√ퟑ))

ퟑ ퟑ풊√ퟑ ퟑ ퟑ ퟑ풊√ퟑ + − − The solutions are ퟖ ퟖ , ퟒ, ퟖ ퟖ .

h. 풙ퟑ + ퟏퟐퟓ = ퟎ (풙ퟑ + ퟏퟐퟓ) = (풙 + ퟓ)(풙ퟐ − ퟓ풙 + ퟐퟓ) ퟓ(ퟏ − 풊√ퟑ) ퟓ(ퟏ + 풊√ퟑ) = (풙 + ퟓ) (풙 − ) (풙 − ) ퟐ ퟐ

ퟓ ퟓ풊√ퟑ ퟓ ퟓ풊√ퟑ The solutions are −ퟓ, + , − . ퟐ ퟐ ퟐ ퟐ

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