Application of Algebra

Collaborative Learning, Experiential Learning, Problem-based Learning, Project-Based Learning and Talk for Learning
Activity : Zeros of a Polynomial function. Use Problem-based Learning, Collaborative Learning and Talk for Learning Approaches: Working with a partner using think-pair-share, learners recall and discuss some facts about polynomial functions.
Example
- Factoring Polynomials: Terms are factors of a polynomial if, when they are multiplied, they equal that polynomial: e.g., 𝑥 ) + 2 𝑥 − 15 = (𝑥 − 3)(𝑥 + 5) and so, (𝑥 − 3) and (𝑥 + 5) are Factors of the polynomial 𝑥 ) + 2 𝑥 − 15.
- Solving a Polynomial Equation By rearranging the terms to have zero on one side: e.g., 𝑥 ) + 2 𝑥 = 15 ⇒ 𝑥 ) + 2 𝑥 − 15 = 0 therefore, ( 𝑥 + 5)( 𝑥 - 3) = 0 and (𝑥 + 5) = 0 and (𝑥 − 3) = 0; so 𝑥 = −5 or 𝑥 = 3
- Solutions/Roots a Polynomial: Setting the Factors of a Polynomial Expression equal to zero gives the Solutions to the Equation when the polynomial expression equals zero. Another name for the Solutions of a Polynomial is the Roots of a Polynomial!
- The Zeros of a Polynomial Function are the solutions to the equation you get when you set the polynomial equal to zero.
Activity 2: Graphical solutions: Factors, Roots and Zeros of Polynomial functions.
Collaborative Learning, Experiential Learning and Talk for Learning Approaches: Learners work in pairs: explore by hand and by use of appropriate ICT tools (e.g. use of GeoGebra) to establish and recognise that the Solutions/Roots of Polynomial Equations are the x-coordinates for the xintercepts of the Polynomial Graph.
Example 1. Here is a graph of our polynomial function: 

Conclusion For our Polynomial Function: 𝑦 = 𝑥 ) + 2 𝑥 − 15 The Factors are: (𝑥 + 5) & (𝑥 − 3) The Roots/Solutions are: 𝑥 = −5 and 3
The Zeros are at: (−5, 0) and (3, 0) 2. The graph shows a polynomial function. Study it carefully and answer the questions that follow. a. Write down the zeros of the polynomial. b. Identify the roots or solution for the polynomial. c. State the factors of the polynomial. d. Write the polynomial function for the graph.
Activity 3: Application of Rational Zeros Theorem
Collaborative Learning, Experiential Learning and Talk for Learning Approaches: Work with a partner to investigate the possible rational to zeros of a polynomial function and use the theorem to establish the zeros, factors and roots of a polynomial function.
Note: If the polynomial 𝑃 ( 𝑥 ) = 𝑎 𝑥 % + 𝑎 − 1 𝑥 %*! + 𝑎 − 2 𝑥 %*) + ... + 𝑎1+ 𝑎0+ has integer coefficients, then every rational + zero of 𝑃(𝑥) is of the form ` ; where 𝑝 is a factor of the constant coefficient 𝑎0; and 𝑞 is a factor of the
leading coefficient 𝑎 (Rational Zeros Theorem)
Example
- List all possible rational zeros given by the Rational Zeros Theorem of 𝑃(𝑥) = 6𝑥 , + 7𝑥 " − 4 (but don't check to see which actually are zeros)
Solution: Factors of 𝑎0 =4= ±1, ±2, 𝑎 ± 4 Factors of 𝑎 =6 = ±1, ±2, ±3, ±6 + ±! ±! ±! ±! ±) ±) ±) ±) ±, ±, ±, ±, Possible ` = ±! , ±) , ±" , ±/ , ±! , ±) , ±" , ±/ , ±! , ±) , ±" , ±/ + By simplifying the fractions and eliminating duplicates, we get the following list of possible values for `
±! ±! ±) ±, ±! ±1, ±2, ±4, ±) , ±" , ±" , ±" , ±/
Find all real zeros of the polynomial P(x) = 2x 4 + x 3 - 6x 2 - 7x - 2.
Solution: ±! Applying the Rational Zeros Theorem, we have ±1, ±2, , as the possible rational zeros. ±) Using synthetic division, We now check if 1 is a root by dividing 𝑃(𝑥) = 2𝑥 , + 𝑥 " - 6𝑥 ) - 7𝑥- 2 by 𝑥 − 1 −1 2 1 −6 −7 −2 −2 1 5 2 2 − 1 −5 −2 0 Since the remainder is zero, 1 is a zero This also tells us that P factors as: 2𝑥 , + 𝑥 " − 6𝑥 ) - 7𝑥- 2 = (𝑥 + 1)(2𝑥 " - 𝑥 ) - 5𝑥- 2)
Applying the Rational Zeros Theorem again, and again P factor as: 2𝑥 , + 𝑥 " - 6𝑥 ) - 7𝑥- 2 = (𝑥 + 1) ) (𝑥- 2)(2𝑥 + 1) ! Thus, the zeros of 𝑃(𝑥) = 2𝑥 , + 𝑥 " - 6𝑥 ) - 7𝑥 - 2 are: (-1,0), (2, 0) and (− ) , 0)
Activity 4: Application of Descartes' rule of signs Theorem.
Collaborative Learning, Experiential Learning and Talk for Learning Approaches: Work with a partner: investigate using Descartes' rule of signs Theorem to the nature of the roots of a polynomial function.
Note: Let P be a polynomial with real coefficients:
- The number of positive real zeros of P(x) is either equal to the number of variations in sign in P(x) or is less than that by an even whole number.
- The number of negative real zeros of P(x) is either equal to the number of variations in sign in P(- x) or is less than that by an even whole number. (Missing terms (those with 0 coefficients) are counted as no change in sign and can be ignored.
Example
- Use Descartes' Rule of Signs to determine how many positive and how many negative real zeros 𝑃( 𝑥) = 6𝑥 " + 17𝑥 ) - 31𝑥 - 12 can have. Then determine the possible total number of real zeros
Solution: P ( x ) = 6 x 3 + 17 x 2 − 31x −12 .
P(x) has one positive real zero. Also; P ( − x ) = − 6 x 3 + 17 x 2 + 31x −12 .
P(-x) has two variations in sign. Combining the findings, P(x) has either one or three real zeros
Positive Negative Real 1 0 1 1 2 3
- Find the maximum number of positive and negative real zeros of the polynomial function 𝑓(𝑥) = 2𝑥 , + 𝑥 " − 6𝑥 ) − 7𝑥 + 1.
Activity 5: Application of the Fundamental Theorem of Algebra.
Collaborative Learning, Experiential Learning and Problem-based Learning Approaches: Learners work in pairs to investigate and apply The Fundamental Theorem of Algebra to establish the zeros, factors and roots of a polynomial function
Note 1:
- Every polynomial equation with a degree higher than zero has at least one root in the set of complex numbers.
- A polynomial equation of the form 𝑃(𝑥) = 0 of degree '𝑛' with complex coefficients has exactly '𝑛' Roots in the set of complex numbers.
Example
- Investigate whether if a polynomial has '𝑛' complex roots, will its graph necessarily have '𝑛' xintercepts?
For instance: 

- In this example, the degree n = 3; if we factor the polynomial, the roots are x = -2, 0, 2. We can also see from the graph that there are three x-intercepts. 𝑦 = 𝑥 " − 4𝑥
- In this example, however, the degree is still n = 3, but there is only one Real x-intercept or root at x = -1. The other 2 roots must have imaginary components. 𝑦 = 𝑥 " − 2𝑥 ) + 𝑥 + 4
Conclusion: Just because a polynomial has 'n' complex roots does not mean they are all Real!

Note 2: Every polynomial 𝑃 ( 𝑥 ) = 𝑎 𝑥 % + 𝑎 − 1 𝑥 %*! + 𝑎 − 2 𝑥 %*) + ... + 𝑎1+ 𝑎0 (n ≥ 1, 𝑎 ≠ 0) with complex coefficients has at least one complex zero, hence for every polynomial 𝑃 ( 𝑥 ), there is a complex number c 1 such that P (c 1 ) = 0. From the Factor Theorem, this tells us that x - c 1 is a factor of P(x). Thus, we can write P ( x ) = (x − c 1 ) Q ( x), where Q(x) has degree n - 1.
Example:
- Factor the polynomial P(x) = 4x 5 - 324x completely and find all its zeros. State the multiplicity of each zero.
Solution: 𝑃 ( 𝑥 ) = 4𝑥 ( 𝑥 − 3)( 𝑥 + 3)(𝑥2 + 9) = 4𝑥 ( 𝑥 − 3) ( 𝑥 + 3) ( 𝑥 − 3𝑖 ) ( 𝑥 + 3𝑖) Therefore, the zeros of P are 0, 3, - 3, 3𝑖 and - 3𝑖. Since each factor occurs only once, all the zeros are of multiplicity 1, and the total number of zeros is five.
- Find all zeros of the polynomial P(x) = x 4 + x 3 + 3x 2 - 5x.
Activity 6: Complex Conjugates Theorem
Collaborative Learning, Experiential Learning and Problem-based learning Approaches: Work with a partner to investigate and apply the Complex Conjugate Theorem to solve problems related to the zeros, factors and roots of a polynomial function.
Note:
- Roots/Zeros that are not Real are Complex with an Imaginary component. Complex roots with Imaginary components always exist in Conjugate Pairs.
- If a + 𝑏 (b ≠ 0) is a zero of a polynomial function, then its conjugate, a − 𝑏, is also a zero of the function.
Example Find all the roots of 𝑓 ( 𝑥) = 𝑥 " − 5𝑥 ) − 7𝑥 + 51. If one root is 4 - i.
Solution: Applying the Complex Conjugate Theorem, and Descartes' rules we have the factor [𝑥 − (4 − 𝑖)], and [𝑥 − (4 + 𝑖)]. The product of factors gives [𝑥 − (4 − 𝑖)]. [𝑥 − (4 + 𝑖)] = 𝑥 ) − 8𝑥 + 17 Since the product of the two non-real factors is 𝑥 ) − 8𝑥 + 17, then the third factor (that gives us the negative real root) is the quotient of 𝑃(𝑥) divided by 𝑥 ) − 8𝑥 + 17, which is 𝑥 = −3 Therefore, the roots of P(x) are -3, -4i and 4i.
Activity 7: Linear and Quadratic Factors Theorem.
Collaborative Learning, Experiential Learning and Problem-based Learning Approaches: Learners work in pairs to investigate and establish that every polynomial with real coefficients can be factored into a product of linear and irreducible quadratic factors with real coefficients.
Example: Given the polynomial 𝑃(𝑥) = 𝑥 , + 2𝑥 ) - 63.
- Factor P into linear and irreducible quadratic factors with real coefficients.
- Factor P completely into linear factors with complex coefficients.
Solution: a) 𝑃 ( 𝑥 ) = (𝑥 − √7 )( 𝑥 + √7)(𝑥 ) + 9) 𝑏) 𝑃 ( 𝑥 ) = (𝑥 − √7 )( 𝑥 + √7)( 𝑥 − 3𝑖 ) ( 𝑥 + 3𝑖)
Teaching and Learning Resources:
- Graph paper
- Ruler
- A scientific calculator
Assessment (2.1.1.AS.10). The document marks these depth-of-knowledge levels for this indicator: Level 2 Skills of conceptual understanding.