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Polynomials and Factoring CHAPTER 9. Introduction This chapter presents a number of skills necessary prerequisites to solving equations. These skills.

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Presentation on theme: "Polynomials and Factoring CHAPTER 9. Introduction This chapter presents a number of skills necessary prerequisites to solving equations. These skills."— Presentation transcript:

1 Polynomials and Factoring CHAPTER 9

2 Introduction This chapter presents a number of skills necessary prerequisites to solving equations. These skills involve combining, simplifying, and factoring polynomials. Factoring will be made easier if students can recognize patterns that frequently occur in products.

3 Adding and Subtracting Polynomials (9.1)

4

5 Polynomial: A monomial or the sum or difference of two or more monomials. Example: 3x 4 + 5x 2 – 7x +1 This polynomial is written in standard form. Standard form of a polynomial: This is the form of a polynomial when the degrees of its monomial terms are written so that they decrease from left to tight. Degree of a polynomial: This will be the same as the monomial with the greatest exponent.

6 Adding and Subtracting Polynomials (9.1) After simplifying a polynomial by combining like terms, we can name the polynomial based on its degree or the number of monomials it contains. PolynomialDegreeName Using Degree Number of Terms Name Using Number of Terms 7x+41Linear2Binomial 3x 2 +2x+12Quadratic3Trinomial 4x 3 3Cubic1Monomial 9x 4 +11x4Fourth degree2Binomial 50Constant1Monomial

7 Adding and Subtracting Polynomials (9.1) Sample Problem Write each polynomial in standard form. Then name each polynomial based on its degree and the number of its terms. a)5 – 2x b)3x 4 – 4 + 2x 2 + 5x 4

8 Adding and Subtracting Polynomials (9.1) Adding and subtracting polynomials simply require that we add or subtract the like terms. Group the like terms then add or subtract. Remember, subtraction means to add the opposite, which means we change the signs of each term then add (Chapter1). Sample Problem Simplify the following polynomials a)(4x 2 + 6x + 7) + (2x 2 – 9x +1) b)(2x 3 + 5x 2 – 3x) – (x 3 – 8x 2 + 11)

9 Multiplying and Factoring (9.2) Multiplying polynomials require that we remember and use the Distributive Property (Chapter 1). Example: 2x(3x + 1) = (2x)(3x) + (2x)(1) = 6x 2 + 2x We can also use the Distributive Property for multiplying powers with the same base when multiplying by a monomial.

10 Multiplying and Factoring (9.2) Factoring requires us to reverse the multiplication process. To factor a monomial from a polynomial, we have to first find the greatest common factor (GCF) of its terms. The greatest common factor is the greatest factor that divides evenly into each term of an expression. To factor a polynomial completely, you must factor until there are no more common factors other than one.

11 Multiplying and Factoring (9.2)

12 Multiplying Binomials (9.3) We can also use the Distributive Property to multiply two binomials together. Sample Problem Simplify (2x + 3) (x + 4)

13 Multiplying Binomials (9.3) One way to organize multiplying two binomials is to use the acronym FOIL. F: First O: Outside I: Inside L: Last Sample Problem Simplify (3x – 5)(2x + 7)

14 Multiplying Binomials (9.3) FOIL is useful for multiplying two binomials, but it does not work when multiplying a trinomial and a binomial. In these cases, we can use the either the horizontal method or the vertical method of distribution to help us organize our work. Sample Problem Simplify the product (4x 2 + x – 6)(2x – 3)

15 Multiplying Special Cases (9.4) To square a binomial you can either use the above rule or use the FOIL method. Let’s see hoe the FOIL method will give us the above expressions. The Square of a Binomial (a + b) 2 = a 2 + 2ab + b 2 (a – b) 2 = a 2 – 2ab + b 2

16 Multiplying Special Cases (9.4) Sample Problem Find each square. a)(x + 7) 2 b)(4k – 3) 2

17 Multiplying Special Cases (9.4) Let’s use FOIL to derive the above rule. The Difference of Squares (a + b) (a – b) = a 2 – b 2

18 Multiplying Special Cases (9.4) Sample Problem Find the product of (t 3 – 6)(t 3 + 6)

19 Factoring Trinomials of the Type x 2 + bx +c (9.5) To factor a trinomial of the form x 2 + bx + c, you must: Find two numbers that have the sum of b, Find two numbers that have the product of c. At first this process will be trial and error, but as you do more problems patterns will arise that you will recognize that will speed up the factoring process. The end result from factoring will be two binomials.

20 Factoring Trinomials of the Type x 2 + bx +c (9.5) Sample Problem: Factor x 2 + 7x + 12

21 Factoring Trinomials of the Type x 2 + bx +c (9.5) To factor trinomials of the form x 2 – bx + c: Following the same method as listed above, however, inspect the negative factors of c to make sure that the sum will be a negative number. Sample Problem Factor d 2 – 17d + 42

22 Factoring Trinomials of the Type x 2 + bx +c (9.5) To factor trinomials of the form x 2 + bx – c or x 2 – bx – c Following the same method as listed above, however, inspect the negative factors that are both positive and negative Sample Problem Factor: a)m 2 + 6m – 27 b)p 2 – 3p – 18

23 Factoring Trinomials of the Type x 2 + bx +c (9.5) Factoring trinomials that have more than one variable is also possible. The first term will be the square of the first variable. The middle term includes both variables. The last term includes the square of the second variable. Sample Problem Factor: a)h 2 – 4hk – 77k 2

24 Factoring Trinomials of the Type ax 2 + bx +c (9.6) To factor trinomials of the form ax 2 + bx +c involves a little more work, but is not much more difficult. To find the binomials that are the factor of the above trinomial: Look for two numbers that will give the product a. Look for two numbers that will give the product c. The sum of the two products will give b.

25 Factoring Trinomials of the Type ax 2 + bx +c (9.6) Sample Problem Factor 6n 2 + 23n + 7

26 Factoring Trinomials of the Type ax 2 + bx +c (9.6) Sample Problem Factor 7x 2 + 26x – 8

27 Factoring Trinomials of the Type ax 2 + bx +c (9.6) Some polynomials require more than one factoring step. In this case, continue factoring until there are no common factors other than 1. If a trinomial has a common monomial factor, factor it out before trying to find the binomial factors. Sample Problem Factor completely: a)20x 2 + 80x + 35

28 Factoring Special Cases (9.7) Perfect square trinomial: Any trinomial that takes the form a 2 + 2ab + b 2 or a 2 – 2ab + b 2. Perfect square trinomials factor to give identical binomial factors. Factoring perfect square trinomials is made a lot easier if you learn to recognize when a perfect square trinomial appears. Perfect-Square Trinomials a 2 + 2ab + b 2 = (a + b)(a + b) = (a + b) 2 a 2 – 2ab + b 2 = (a – b)(a – b) = (a – b) 2

29 Factoring Special Cases (9.7) To recognize a perfect square trinomial: The first and the last terms can both be written as the product of two identical factors. The middle term is twice the product of one factor from the first term and one factor from the last term. Example: 4x 2 + 12x + 9 First term: 2x 2x Last term: 3 3 Middle term: 2(2x 3)

30 Factoring Special Cases (9.7) Sample Problem Factor each expression: a)x 2 – 8x + 16 b)9g 2 + 12g + 4

31 Factoring Special Cases (9.7) Factoring can also be made easier when you recognize that you have a difference of two squares: Difference of Two Squares a 2 – b 2 = (a + b)(a – b) Sample Problem Factor: x 2 – 64

32 Factoring Special Cases (9.7) Sometimes, to recognize that we have a difference of square, we will have to factor out the greatest common factor first. Sample Problem Factor: 10x 2 – 40

33 Factoring by Grouping (9.8) We can use the Distributive Property to factor by grouping if two groups of terms have the same factor. To factor by grouping: Group terms and use the Distributive Property to see if a common term exits; Look for a common binomial factor of two pairs of terms; Factor out the common binomial. This method works for larger polynomials (i.e. four term polynomials).

34 Factoring by Grouping (9.8) Sample Problem Factor 4n 3 + 8n 2 – 5n – 10

35 Factoring by Grouping (9.8) Sometimes before factoring by grouping, you may need to factor the GCF of all terms of a polynomial. Sample Problem Factor 12p 4 + 10p 3 – 36p 2 – 30p.

36 Factoring by Grouping (9.8) We can also use factor by grouping to factor trinomials of the form ax 2 + bx + c. We can use this method of factoring in situations in which we cannot quickly factor a trinomial using the methods discussed in Lesson 9.6. To factor a trinomial by grouping: Find the product ac; Find the two factors of ac that have the sum b; Rewrite the trinomial using this sum; Factor by grouping.

37 Factoring by Grouping (9.8) Sample Problem Factor 24q 2 + 25q – 25.

38 Factoring by Grouping (9.8) Summary of Factoring Polynomials: 1.Factor out the GCF. 2.If the polynomial has two terms or three terms, look for a difference of two squares, a product of two squares, or a pair of binomial factors. 3.If there are four or more terms, group terms and factor to find common binomial factors. 4.As a final check, make sure there are no common factors other than 1.

39 Polynomials and Factoring CHAPTER 9 THE END


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