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Unit 9: Factoring Lesson 4: Factoring Trinomials: A Special Situation Using two different factoring methods. In the previous lesson, we factored trinomials that had a lead coefficient of 1. Of course, many trinomials have a lead coefficient that is greater than 1. We will explore one method of factoring these types of trinomials. Example 1 Factor: 3x2 – 3x – 36 Notice how the lead coefficient is not 1. However, each term is divisible by 3. Step 1: Note: If a trinomial looks difficult to factor, first check to see if a Greatest Common Factor (GCF) can be factored out. Example 2 Factor: 4x3 + 4x2 – 24x Copyright © 2009-2010 Karin Hutchinson – Algebra-class.com Unit 9: Factoring Lesson 4: Factoring Trinomials (A Special Situation) Practice Part 1: Factor Completely 1. 2x2 + 10x – 48 7. 2y5 – 16y4 + 32y3 2. 5y2 – 40y + 60 8. 3x2 - 243 3. 6x3 + 54x2 + 120x 9. 2b3 – 2b2 – 24b 4. 3s3 – 108s 10. 4x3 – 20x2 + 24x 5. 5x4 – 50x3 + 125x2 11. 6w4 – 18w3 – 24w2 6. 2a4 + 12a3 + 18a2 12. 3x2 – 9x - 54 Part 2: Thinking Questions 1. Give at least one example of a trinomial in the form of: x2 +bx +c that can be factored. 2. Give at least one example of a trinomial without a lead coefficient of one that can be factored. 3. Give an example of a prime trinomial. Factor each trinomial. (3 points each) 1. 3x3 +18x2 – 81x Copyright © 2009-2010 Karin Hutchinson – Algebra-class.com 2. 2x3 – 8x2 – 8x Unit 9: Factoring Lesson 4: Factoring Trinomials (A Special Situation) Answers Part 1: Factor Completely 1. 2x2 + 10x – 48 Step 1: Since the lead coefficient is greater than 1, see if there is a GCF for all terms. 2 is the GCF for all terms. 2(x2 + 5x – 24) Step 2: Factor the expression inside of parenthesis. We need a positive and negative number that have a product of -24 and a sum of 5 (8 & 3, 8 must be positive since we need a +5 for a sum). Final Answer: 2(x+ 8) (x-3) 7. 2y5 – 16y4 + 32y3 Step 1: Since the lead coefficient is greater than 1, see if there is a GCF for all terms. 3 2y is the GCF for all terms. 2y3(y2 – 8y + 16) Step 2: Factor the expression inside of parenthesis. We need 2 negative numbers that have a product of 16 and a sum of -8. (-4 & -4). 2y3(y-4) (y-4) Final Answer: 2y3(y-4)2 2. 5y2 – 40y + 60 Step 1: Since the lead coefficient is greater than 1, see if there is a GCF for all terms. 5 is the GCF for all terms. 5(y2 - 8x + 12) Step 2: Factor the expression inside of parenthesis. We need 2 negative numbers that have a product of 12 and a sum of -8 (-6 & -2). 8. 3x2 – 243 Step 1: Since the lead coefficient is greater than 1, see if there is a GCF for all terms. 3 is the GCF for all terms. 3(x2 - 81) Step 2: Factor the expression inside of parenthesis. This is a difference of two squares. So, what number squared is 81? (9) Final Answer: 3(x+ 9) (x-9) Final Answer: 5(y-6) (y-2) Copyright © 2009-2010 Karin Hutchinson – Algebra-class.com Unit 9: Factoring 3. 6x3 + 54x2 + 120x 9. 2b3 – 2b2 – 24b Step 1: Since the lead coefficient is greater than 1, see if there is a GCF for all terms. 6x is the GCF for all terms. 6x(x2 + 9x + 20) Step 2: Factor the expression inside of parenthesis. We need 2 positive numbers that have a product of 20 and a sum of 9 (4 & 5) Step 1: Since the lead coefficient is greater than 1, see if there is a GCF for all terms. 2b is the GCF for all terms. 2b(b2 - b – 12) Step 2: Factor the expression inside of parenthesis. We need a positive and negative number that have a product of -12 and a sum of -1 (4 & 3, 4 must be negative since we need a -1 for a sum). Final Answer: 6x(x+ 4) (x+ 5) Final Answer: 2b(b-4) (b+3) 4. 3s3 – 108s 10. 4x3 – 20x2 + 24x Step 1: Since the lead coefficient is greater than 1, see if there is a GCF for all terms. 3s is the GCF for all terms. 3s(s2 - 36) Step 2: Factor the expression inside of parenthesis. This is a difference of two squares. What number when squared equals 36? (6) Step 1: Since the lead coefficient is greater than 1, see if there is a GCF for all terms. 4x is the GCF for all terms. 4x(x2 – 5x+ 6) Step 2: Factor the expression inside of parenthesis. We need two negative numbers that have a product of 6 and a sum of -5. (-3 & -2) Final Answer: 3s(s+ 6) (s-6) Final Answer: 4x(x-3) (x-2) 5. 5x4 – 50x3 + 125x2 11. 6w4 – 18w3 – 24w2 Step 1: Since the lead coefficient is greater than 1, see if there is a GCF for all terms. 2 5x is the GCF for all terms. 5x2(x2 -10x + 25) Step 2: Factor the expression inside of parenthesis. We need two negative numbers whose product is 25 and sum is -10 (-5 & -5) This is a perfect square. Final Answer: 5x2(x - 5) (x-5) or Step 1: Since the lead coefficient is greater than 1, see if there is a GCF for all terms. 2 6w is the GCF for all terms. 6w2(w2 -3w -4) Step 2: Factor the expression inside of parenthesis. We need a positive and negative number whose product is -4 and whose sum is -3. (-4 & 1) 4 must be negative since the sum is negative. 5x2(x-5)2 Final Answer: 6w2(w -4) (w +1) Copyright © 2009-2010 Karin Hutchinson – Algebra-class.com Unit 9: Factoring 6. 2a4 + 12a3 + 18a2 Step 1: Since the lead coefficient is greater than 1, see if there is a GCF for all terms. 2 2a is the GCF for all terms. 2a2(a2 + 6a + 9) Step 2: Factor the expression inside of parenthesis. This is a perfect square. What number can you square and get 9, and add to get 6? (3) 12. 3x2 – 9x – 54 Step 1: Since the lead coefficient is greater than 1, see if there is a GCF for all terms. 3 is the GCF for all terms. 3(x2 – 3x - 18) Step 2: Factor the expression inside of parenthesis. What positive and negative numbers have a product of -18 and a sum of -3? (-6 & 3) 6 must be negative since the sum is negative. Final Answer: 2a2(a+3)2 Final Answer: 3(x -6) (x +3) Part 2: Thinking Questions 1. Give at least one example of a trinomial in the form of: x2 +bx +c that can be factored. These answers may vary. The easiest way to find a trinomial that can be factored is to work backwards and start with two binomials. You can choose any binomials. For example, I’ll choose (x+4) (x+2) now multiply using foil. (x)(x) +(2)x+(4)x+4(2) x2 +6x + 8 is a trinomial that can be factored. We know this because we started with the two factors and multiplied. 2. Give at least one example of a trinomial without a lead coefficient of one that can be factored. These answers may vary. The easiest way to find a trinomial that can be factored is to work backwards and start with two binomials. You can choose any binomials. Since you want a lead coefficient that is greater than 1, make sure that one of your binomials has a coefficient of x that is greater than 1. For example, I’ll choose (2x+4) (x+2) now multiply using foil. (2x)(x) +(2x)2+(4)x+4(2) 2x2 +8x + 8 is a trinomial that can be factored. We know this because we started with the two factors and multiplied. Copyright © 2009-2010 Karin Hutchinson – Algebra-class.com Unit 9: Factoring 3. Give an example of a prime trinomial. A prime trinomial cannot be factored. For example: x2 + 2x + 2 - We would need two positive numbers whose product is 2 and whose sum is 2. The only way to make a product of 2 is (2 ●1) Since 2+1 = 3, we know that this cannot be factored. The best way to make up a prime trinomial is to use a prime number as your constant. Then make sure the coefficient of x is not the sum of the factors of your prime number. 1. 3x3 +18x2 – 81x Step 1: Since the lead coefficient is greater than 1, see if there is a GCF for all terms. 3x is the GCF for all terms. 3x(x2 + 6x - 27) Step 2: Factor the expression inside of parenthesis. We need a positive and negative number whose product is -27 and whose sum is 6. (9 & -3) (9 must be positive since the sum is positive) Final Answer: 3x(x+9) (x - 3) 2. 2x3 – 8x2 – 8x Step 1: Since the lead coefficient is greater than 1, see if there is a GCF for all terms. 2x is the GCF for all terms. 2x(x2- 4x - 4) Step 2: Factor the expression inside of parenthesis. We need a positive and negative number whose product is -4 and whose sum is -4. The inside of the parenthesis cannot be factored any further because there’s no positive and negative number whose product is -4 and whose sum is -4. Final Answer: 2x(x2 – 4x – 4) Copyright © 2009-2010 Karin Hutchinson – Algebra-class.com