8 6 Choosing a Factoring Method Lesson Objectives

8 -6 Choosing a Factoring Method Lesson Objectives: I will be able to … • Choose an appropriate method for factoring a polynomial • Combine methods for factoring a polynomial Language Objective: I will be able to … • Read, write, and listen about vocabulary, key concepts, and examples Holt Algebra 1

8 -6 Choosing a Factoring Method Recall that a polynomial is in its fully factored form when it is written as a product that cannot be factored further. Fully Factored Example: 4 x 2(3 x – 2) Not Fully Factored Example: 3 x(2 x + 4) (2 can also be factored from 2 x and 4. ) Holt Algebra 1

8 -6 Choosing a Factoring Method Page 23 Example 1: Determining Whether a Polynomial is Completely Factored Tell whether each polynomial is completely factored. If not factor it. A. 3 x 2(6 x – 4) 6 x – 4 can be further factored. 6 x 2(3 x – 2) Factor out 2, the GCF of 6 x and – 4. 6 x 2(3 x – 2) is completely factored. B. (x 2 + 1)(x – 5) Neither x 2 +1 nor x – 5 can be factored further. (x 2 + 1)(x – 5) is completely factored. Holt Algebra 1

8 -6 Choosing a Factoring Method Caution x 2 + 4 is a sum of squares, and cannot be factored. Holt Algebra 1

8 -6 Choosing a Factoring Method Page 24 Your Turn 1 Tell whether the polynomial is completely factored. If not, factor it. A. 5 x 2(x – 1) Neither 5 x 2 nor x – 1 can be factored further. 5 x 2(x – 1) is completely factored. B. (4 x + 4)(x + 1) 4 x + 4 can be further factored. 4(x + 1) Factor out 4, the GCF of 4 x and 4. 4(x + 1)2 is completely factored. Holt Algebra 1

8 -6 Choosing a Factoring Method To factor a polynomial completely, you may need to use more than one factoring method. Use the steps below to factor a polynomial completely. Page 22 Holt Algebra 1

8 -6 Choosing a Factoring Method Example 2: Factoring by GCF and Recognizing Patterns Factor the polynomial completely. Check your answer. Page 24 10 x 2 + 48 x + 32 2(5 x 2 + 24 x + 16) Factor out the GCF. 2(5 x + 4)(x + 4) Factor remaining trinomial. Check 2(5 x + 4)(x + 4) = 2(5 x 2 + 20 x + 4 x + 16) = 10 x 2 + 40 x + 8 x + 32 = 10 x 2 + 48 x + 32 Holt Algebra 1

8 -6 Choosing a Factoring Method Page 25 Example 3: Factoring by GCF and Recognizing Patterns Factor the polynomial completely. Check your answer. 8 x 6 y 2 – 18 x 2 y 2 2 x 2 y 2(4 x 4 – 9) Factor out the GCF. 4 x 4 – 9 is a perfect-square trinomial of the form a 2 – b 2. 2 x 2 y 2(2 x 2 – 3)(2 x 2 + 3) a = 2 x, b = 3 Check 2 x 2 y 2(2 x 2 – 3)(2 x 2 + 3) = 2 x 2 y 2(4 x 4 – 9) = 8 x 6 y 2 – 18 x 2 y 2 Holt Algebra 1

8 -6 Choosing a Factoring Method Page 25 Your Turn 3 Factor the polynomial completely. Check your answer. 4 x 3 + 16 x 2 + 16 x 4 x(x 2 + 4 x + 4) 4 x(x + 2)2 Factor out the GCF. x 2 + 4 x + 4 is a perfect-square trinomial of the form a 2 + 2 ab + b 2. a = x, b = 2 Check 4 x(x + 2)2 = 4 x(x 2 + 2 x + 4) = 4 x(x 2 + 4 x + 4) = 4 x 3 + 16 x 2 + 16 x Holt Algebra 1

8 -6 Choosing a Factoring Method If none of the factoring methods work, the polynomial is said to be unfactorable. Helpful Hint For a polynomial of the form ax 2 + bx + c, if there are no numbers whose sum is b and whose product is ac, then the polynomial is unfactorable. In other words, if the diamond problem is unsolvable, then the polynomial is unfactorable. Holt Algebra 1

8 -6 Choosing a Factoring Method Example 4: Factoring by Multiple Methods Page 26 Factor each polynomial completely. 9 x 2 + 3 x – 2 ( x+ ) Factors of 9 Factors of 2 1 and – 2 1 and 9 1 and – 2 3 and 3 – 1 and 2 3 and 3 (3 x – 1)(3 x + 2) Holt Algebra 1 The GCF is 1 and there is no pattern. a = 9 and c = – 2; Outer + Inner = 3 Outer + Inner 1(– 2) + 1(9) = 7 3(– 2) + 1(3) = – 3 3(2) + 3(– 1) = 3

8 -6 Choosing a Factoring Method Page 26 Example 5: Factoring by Multiple Methods Factor the polynomial completely. 12 b 3 + 48 b 2 + 48 b The GCF is 12 b; (b 2 + 4 b + 4) is a perfect-square 12 b(b 2 + 4 b + 4) trinomial in the form of (b + ) a 2 + 2 ab + b 2. Factors of 4 Sum 1 and 4 5 2 and 2 4 a = 2 and c = 2 12 b(b + 2)2 Holt Algebra 1

8 -6 Choosing a Factoring Method Page 27 Example 6: Factoring by Multiple Methods Factor the polynomial completely. 4 y 2 + 12 y – 72 4(y 2 + 3 y – 18) (y + ) Factor out the GCF. There is no pattern. b = 3 and c = – 18; look for factors of – 18 whose sum is 3. Factors of – 18 Sum – 1 and 18 17 – 2 and 9 7 – 3 and 6 3 The factors needed are – 3 and 6 4(y – 3)(y + 6) Holt Algebra 1

8 -6 Choosing a Factoring Method Page 27 Example 7: Factoring by Multiple Methods. Factor the polynomial completely. (x 4 – x 2) x 2(x 2 – 1) Factor out the GCF. x 2(x + 1)(x – 1) x 2 – 1 is a difference of two squares. Holt Algebra 1

8 -6 Choosing a Factoring Method Page 23 Holt Algebra 1

8 -6 Choosing a Factoring Method Your Turn 7 Page 28 Factor the polynomial completely. 2 p 5 + 10 p 4 – 12 p 3(p 2 + 5 p – 6) (p + ) Factor out the GCF. There is no pattern. b = 5 and c = – 6; look for factors of – 6 whose sum is 5. Factors of – 6 Sum – 1 and 6 5 The factors needed are – 1 and 6 2 p 3(p + 6)(p – 1) Holt Algebra 1

8 -6 Choosing 8 a Factoring Chapter Quick. Method Review Assignment #16 1. Write the prime factorization of 128. 2. Factor out the GCF: 6 x 5 y 2 – 15 xy 3 3. Factor by grouping: 6 a 3 – 4 a 2 – 15 a + 10 4. Factor each trinomial. a) x 2 – 11 x + 28 b) 10 x 2 – 31 x + 15 5. Determine whether each polynomial is a perfect square trinomial, difference of two squares, or neither. If it is a special product, factor it. If not, explain why. a) 4 – 16 x 4 b) 9 x 2 + 6 x + 1 2 – 3 d + 6 6. Factor completely: -18 d Holt Algebra 1

8 -6 Choosing a Factoring Method Homework Assignment #16 • Holt 8 -6 #19 – 30, 40 – 45 • 8 -6 Practice C Worksheet Chapter 8 Group Quiz next class! Holt Algebra 1