Polynomials and Polynomial Functions *Chapter 5 Polynomials and Polynomial Functions

Chapter Sections 5.1 – Addition and Subtraction of Polynomials 5.2 – Multiplication of Polynomials 5.3 – Division of Polynomials and Synthetic Division 5.4 – Factoring a Monomial from a Polynomial and Factoring by Grouping 5.5 – Factoring Trinomials 5.6 – Special Factoring Formulas 5.7-A General Review of Factoring 5.8- Polynomial Equations Chapter 1 Outline

Recall: Special Products: (a + b)2 = (a + b)(a + b) = a2 + 2ab + b2 (a – b)2 = (a – b)(a – b) = a2 – 2ab + b2 (a + b)(a – b) = a2 – b2

5.5 Factoring Trinomials Objectives: 1. Factor trinomials of the form x2 + bx + c. Factor out a monomial GCF, then factor the trinomial of the form x2 + bx + c. 3. Factor trinomials of the form ax2 + bx + c, where a  1, by trial. 4. Factor trinomials of the form ax2 + bx + c, where a  1, by grouping.

Form: x2 + bx + c Factoring Trinomials The first step in factoring: factor out the GCF. Form: x2 + bx + c Factoring (Reverse process of multiplication) We know: (x + 3)(x + 8) = x2 + 11x + 24 Therefore x2 + 11x + 24 = (x = 3)(x + 8). Find 2 Numbers: c = product & b = sum 2. Result: (x + 1st # ) (x + 2nd #) Numbers go here. x2 + bx + c = (x +?)(x +?)

Example 1: Factor. a)x2 – 6x + 8 x2 + 6x + 8 = ? x2 – 2x – 8 = ? Solution: Find a pair of numbers: Product = c = 8 Sum = b = –6 Factor. a)x2 – 6x + 8 x2 – 2x – 8 = ? x2 + 2x – 8 = ? Product Sum (–1)(–8) = 8 –1 + (–8) = –9 (–2)(–4) = 8 –2 + (–4) = –6 x2 + 6x – 8 = ? This is the correct combination. x2 – 6x + 8 = (x – 2)(x – 4) c) Factor x2 – 10x – 25. prime polynomial (it cannot be factored.) b) Factor x2 + 8x + 15. x2 + 8x + 15 = (x + ?) (x + ?) = (x + 3) (x + 5)

Example 2: Factor = (x – 5y)(x + 3y) a)3x4 – 6x3 – 72x2 c) 2z5 + 16z4 + 30z3 = 2z3(z2 + 8z + 15) = 2z3(z + 3)(z + 5) b) 2a2 – 12a – 32 = 3x2(x2 – 2x – 24) = 2(a2 – 6a – 16) = 3x2(x – 6)(x + 4) = 2(a –8)(a + 2) e) x2 – 2xy – 15y2 d) a2 – ab – 20b2 = (a + ?b)(a + ?b) = (x – 5y)(x + 3y) = (a – 5b)(a + 4b)

Form: ax2 + bx + c, a ≠ 1 Trial and Error Method Write all pairs of factors of a. Write all pairs of factors of the constant, c. Try various combinations of these factors until the correct middle term, bx, is found. Example 4: Factor 3t2 – 13t + 10. There is no GCF a = 3 and the only factors of 3 are 1 and 3. Therefore: 3t2 – 13t + 10 = (3t )(t ) Possible Factors Sum of the Products of the Inner and Outer Terms (3t – 1)(t – 10) -31t (3t – 10)(t – 1) -13t (3t – 2)(t – 5) -17t (3t – 5)(t – 2) -11t 3t2 – 13t + 10 = (3t – 10)(t – 1)

Trial and Error Method Grouping: a = 2, b = 9, c = 4 There is no GCF. Example 5: Factor 2x2 + 9x + 4. Find two numbers such that: product = a · c & sum = b Factors of ac = 8 (1)(8) (2)(4) Sum of Factors 9 6 There is no GCF.  Since the first term is 2x2, one factor must contain 2x and the other an x: (2x + ?)(x + ?) 2. Rewrite bx using these numbers. 2x2 + 1x + 8x + 4 = Factors of 4 Possible Factors of Trinomial Sum of the Products of the Inner and Outer Terms 1(4) (2x + 1)(x + 4) 9x 2(2) (2x + 2)(x + 2) 6x 4(1) (2x + 4)(x + 1) 3. Factor by grouping. x(2x + 1) + 4(2x + 1) (2x + 1) (x + 4) 2x2 + 9x + 4 = (2x + 1)(x + 4)

Example 6: Factor Factor. Solution a = 2, b = –15, c = 7, and ac = (2)(7) = 14. Factors of ac=14 Sum of Factors of ac (–2)(–7) = 14 –2 + (–7) = –9 (–1)(–14) = 14 –1 + (– 14) = –15 Correct –15x –x – 14x

Example 7: Factor a.) 5x2 + 19x – 4 5x2 + 20x – x – 4 b.) 12x4 + 14x3 – 6x2 ac = – 20 ac = – 18 2x + 9x = 2x2(6x2 + 7x – 3) =

Factor Trinomials Using Substitution Example 8: Factor 3z4 – 17z2 – 28. Let x = z2. Then the trinomial can be written

Example 3: a)x2 + 5x + 6 b) x2 –7x + 12 c) x2 – 5x – 24 Example 2: (Time)Factor. x2 + 2x – 24 Factor. x2 – 6x – 27 Factor. x2 – x – 12 = (x – 9)(x + 3) Example 3: a)x2 + 5x + 6 b) x2 –7x + 12 c) x2 – 5x – 24 = (x + 3)(x – 4)