Upcoming SlideShare
×

Like this presentation? Why not share!

# 5 5factoring trinomial iii

## on Jun 23, 2011

• 924 views

### Views

Total Views
924
Views on SlideShare
924
Embed Views
0

Likes
0
10
0

No embeds

### Report content

• Comment goes here.
Are you sure you want to

## 5 5factoring trinomial iiiPresentation Transcript

• Factoring Trinomials III
• Factoring Trinomials III In this section we give a formula that enables us to tell if a trinomial is factorable or not.
• Factoring Trinomials III In this section we give a formula that enables us to tell if a trinomial is factorable or not. Also we give the ac-method for factoring.
• Factoring Trinomials III In this section we give a formula that enables us to tell if a trinomial is factorable or not. Also we give the ac-method for factoring. Theorem: If b2 – 4ac = 0, 1, 4, 9, 16, 25, 36, .. i.e. is a squared number, then the trinomial is factorable.
• Factoring Trinomials III In this section we give a formula that enables us to tell if a trinomial is factorable or not. Also we give the ac-method for factoring. Theorem: If b2 – 4ac = 0, 1, 4, 9, 16, 25, 36, .. i.e. is a squared number, then the trinomial is factorable. Otherwise, it is not factorable.
• Factoring Trinomials III In this section we give a formula that enables us to tell if a trinomial is factorable or not. Also we give the ac-method for factoring. Theorem: If b2 – 4ac = 0, 1, 4, 9, 16, 25, 36, .. i.e. is a squared number, then the trinomial is factorable. Otherwise, it is not factorable. Example I. Use the b2 – 4ac to see if the trinomial is factorable. If it is, factor it. a. 3x2 – 7x – 2
• Factoring Trinomials III In this section we give a formula that enables us to tell if a trinomial is factorable or not. Also we give the ac-method for factoring. Theorem: If b2 – 4ac = 0, 1, 4, 9, 16, 25, 36, .. i.e. is a squared number, then the trinomial is factorable. Otherwise, it is not factorable. Example I. Use the b2 – 4ac to see if the trinomial is factorable. If it is, factor it. a. 3x2 – 7x – 2 b2 – 4ac = (–7)2 – 4(3)(–2)
• Factoring Trinomials III In this section we give a formula that enables us to tell if a trinomial is factorable or not. Also we give the ac-method for factoring. Theorem: If b2 – 4ac = 0, 1, 4, 9, 16, 25, 36, .. i.e. is a squared number, then the trinomial is factorable. Otherwise, it is not factorable. Example I. Use the b2 – 4ac to see if the trinomial is factorable. If it is, factor it. a. 3x2 – 7x – 2 b2 – 4ac = (–7)2 – 4(3)(–2) = 49 + 24
• Factoring Trinomials III In this section we give a formula that enables us to tell if a trinomial is factorable or not. Also we give the ac-method for factoring. Theorem: If b2 – 4ac = 0, 1, 4, 9, 16, 25, 36, .. i.e. is a squared number, then the trinomial is factorable. Otherwise, it is not factorable. Example I. Use the b2 – 4ac to see if the trinomial is factorable. If it is, factor it. a. 3x2 – 7x – 2 b2 – 4ac = (–7)2 – 4(3)(–2) = 49 + 24 = 73 is not a square,
• Factoring Trinomials III In this section we give a formula that enables us to tell if a trinomial is factorable or not. Also we give the ac-method for factoring. Theorem: If b2 – 4ac = 0, 1, 4, 9, 16, 25, 36, .. i.e. is a squared number, then the trinomial is factorable. Otherwise, it is not factorable. Example I. Use the b2 – 4ac to see if the trinomial is factorable. If it is, factor it. a. 3x2 – 7x – 2 b2 – 4ac = (–7)2 – 4(3)(–2) = 49 + 24 = 73 is not a square, hence it is prime.
• Factoring Trinomials III In this section we give a formula that enables us to tell if a trinomial is factorable or not. Also we give the ac-method for factoring. Theorem: If b2 – 4ac = 0, 1, 4, 9, 16, 25, 36, .. i.e. is a squared number, then the trinomial is factorable. Otherwise, it is not factorable. Example I. Use the b2 – 4ac to see if the trinomial is factorable. If it is, factor it. a. 3x2 – 7x – 2 b2 – 4ac = (–7)2 – 4(3)(–2) = 49 + 24 = 73 is not a square, hence it is prime. b. 3x2 – 7x + 2
• Factoring Trinomials III In this section we give a formula that enables us to tell if a trinomial is factorable or not. Also we give the ac-method for factoring. Theorem: If b2 – 4ac = 0, 1, 4, 9, 16, 25, 36, .. i.e. is a squared number, then the trinomial is factorable. Otherwise, it is not factorable. Example I. Use the b2 – 4ac to see if the trinomial is factorable. If it is, factor it. a. 3x2 – 7x – 2 b2 – 4ac = (–7)2 – 4(3)(–2) = 49 + 24 = 73 is not a square, hence it is prime. b. 3x2 – 7x + 2 b2 – 4ac = (–7)2 – 4(3)(2)
• Factoring Trinomials III In this section we give a formula that enables us to tell if a trinomial is factorable or not. Also we give the ac-method for factoring. Theorem: If b2 – 4ac = 0, 1, 4, 9, 16, 25, 36, .. i.e. is a squared number, then the trinomial is factorable. Otherwise, it is not factorable. Example I. Use the b2 – 4ac to see if the trinomial is factorable. If it is, factor it. a. 3x2 – 7x – 2 b2 – 4ac = (–7)2 – 4(3)(–2) = 49 + 24 = 73 is not a square, hence it is prime. b. 3x2 – 7x + 2 b2 – 4ac = (–7)2 – 4(3)(2) = 49 – 24
• Factoring Trinomials III In this section we give a formula that enables us to tell if a trinomial is factorable or not. Also we give the ac-method for factoring. Theorem: If b2 – 4ac = 0, 1, 4, 9, 16, 25, 36, .. i.e. is a squared number, then the trinomial is factorable. Otherwise, it is not factorable. Example I. Use the b2 – 4ac to see if the trinomial is factorable. If it is, factor it. a. 3x2 – 7x – 2 b2 – 4ac = (–7)2 – 4(3)(–2) = 49 + 24 = 73 is not a square, hence it is prime. b. 3x2 – 7x + 2 b2 – 4ac = (–7)2 – 4(3)(2) = 49 – 24 = 25 which is a squared number,
• Factoring Trinomials III In this section we give a formula that enables us to tell if a trinomial is factorable or not. Also we give the ac-method for factoring. Theorem: If b2 – 4ac = 0, 1, 4, 9, 16, 25, 36, .. i.e. is a squared number, then the trinomial is factorable. Otherwise, it is not factorable. Example I. Use the b2 – 4ac to see if the trinomial is factorable. If it is, factor it. a. 3x2 – 7x – 2 b2 – 4ac = (–7)2 – 4(3)(–2) = 49 + 24 = 73 is not a square, hence it is prime. b. 3x2 – 7x + 2 b2 – 4ac = (–7)2 – 4(3)(2) = 49 – 24 = 25 which is a squared number, hence it is factorable.
• Factoring Trinomials III In this section we give a formula that enables us to tell if a trinomial is factorable or not. Also we give the ac-method for factoring. Theorem: If b2 – 4ac = 0, 1, 4, 9, 16, 25, 36, .. i.e. is a squared number, then the trinomial is factorable. Otherwise, it is not factorable. Example I. Use the b2 – 4ac to see if the trinomial is factorable. If it is, factor it. a. 3x2 – 7x – 2 b2 – 4ac = (–7)2 – 4(3)(–2) = 49 + 24 = 73 is not a square, hence it is prime. b. 3x2 – 7x + 2 b2 – 4ac = (–7)2 – 4(3)(2) = 49 – 24 = 25 which is a squared number, hence it is factorable. In fact 3x2 – 7x + 2 = (3x – 1)(x – 2)
• Factoring Trinomials III The ac-Method
• Factoring Trinomials III The ac-Method The product of two binomials has four terms.
• Factoring Trinomials III The ac-Method The product of two binomials has four terms. We may use grouping-method to work backward to obtain the two binomials.
• Factoring Trinomials III The ac-Method The product of two binomials has four terms. We may use grouping-method to work backward to obtain the two binomials. Example A. Factor x2 – x – 6 by grouping.
• Factoring Trinomials III The ac-Method The product of two binomials has four terms. We may use grouping-method to work backward to obtain the two binomials. Example A. Factor x2 – x – 6 by grouping. x2 – x – 6 = x2 – 3x + 2x – 6
• Factoring Trinomials III The ac-Method The product of two binomials has four terms. We may use grouping-method to work backward to obtain the two binomials. Example A. Factor x2 – x – 6 by grouping. x2 – x – 6 = x2 – 3x + 2x – 6 Put them into two groups = (x2 – 3x) + (2x – 6)
• Factoring Trinomials III The ac-Method The product of two binomials has four terms. We may use grouping-method to work backward to obtain the two binomials. Example A. Factor x2 – x – 6 by grouping. x2 – x – 6 = x2 – 3x + 2x – 6 Put them into two groups = (x2 – 3x) + (2x – 6) Take out the common factors = x(x – 3) + 2(x – 3)
• Factoring Trinomials III The ac-Method The product of two binomials has four terms. We may use grouping-method to work backward to obtain the two binomials. Example A. Factor x2 – x – 6 by grouping. x2 – x – 6 = x2 – 3x + 2x – 6 Put them into two groups = (x2 – 3x) + (2x – 6) Take out the common factors = x(x – 3) + 2(x – 3) Take out the common (x – 3) = (x – 3)(x + 2)
• Factoring Trinomials III The ac-Method The product of two binomials has four terms. We may use grouping-method to work backward to obtain the two binomials. Example A. Factor x2 – x – 6 by grouping. x2 – x – 6 = x2 – 3x + 2x – 6 Put them into two groups = (x2 – 3x) + (2x – 6) Take out the common factors = x(x – 3) + 2(x – 3) Take out the common (x – 3) = (x – 3)(x + 2) But how do we know to write x2 – x – 6 as x2 – 3x + 2x – 6 in the first place?
• Factoring Trinomials III The ac-Method The product of two binomials has four terms. We may use grouping-method to work backward to obtain the two binomials. Example A. Factor x2 – x – 6 by grouping. x2 – x – 6 = x2 – 3x + 2x – 6 Put them into two groups = (x2 – 3x) + (2x – 6) Take out the common factors = x(x – 3) + 2(x – 3) Take out the common (x – 3) = (x – 3)(x + 2) But how do we know to write x2 – x – 6 as x2 – 3x + 2x – 6 in the first place? This leads to the ac-method.
• Factoring Trinomials III The ac-Method The product of two binomials has four terms. We may use grouping-method to work backward to obtain the two binomials. Example A. Factor x2 – x – 6 by grouping. x2 – x – 6 = x2 – 3x + 2x – 6 Put them into two groups = (x2 – 3x) + (2x – 6) Take out the common factors = x(x – 3) + 2(x – 3) Take out the common (x – 3) = (x – 3)(x + 2) But how do we know to write x2 – x – 6 as x2 – 3x + 2x – 6 in the first place? This leads to the ac-method. The ac-method tells us how to rewrite the trinomial so we may use the grouping method.
• Factoring Trinomials III The ac-Method The product of two binomials has four terms. We may use grouping-method to work backward to obtain the two binomials. Example A. Factor x2 – x – 6 by grouping. x2 – x – 6 = x2 – 3x + 2x – 6 Put them into two groups = (x2 – 3x) + (2x – 6) Take out the common factors = x(x – 3) + 2(x – 3) Take out the common (x – 3) = (x – 3)(x + 2) But how do we know to write x2 – x – 6 as x2 – 3x + 2x – 6 in the first place? This leads to the ac-method. The ac-method tells us how to rewrite the trinomial so we may use the grouping method. If it isn't possible to rewrite the trinomial according to the ac-method, then the trinomial is prime.
• Factoring Trinomials III ac-Method:
• Factoring Trinomials III ac-Method: We assume that there is no common factor for the trinomial ax2 + bx + c.
• Factoring Trinomials III ac-Method: We assume that there is no common factor for the trinomial ax2 + bx + c. 1. Calculate ac,
• Factoring Trinomials III ac-Method: We assume that there is no common factor for the trinomial ax2 + bx + c. 1. Calculate ac, and find two numbers u and v such that uv is ac,
• Factoring Trinomials III ac-Method: We assume that there is no common factor for the trinomial ax2 + bx + c. 1. Calculate ac, and find two numbers u and v such that uv is ac, and u + v = b.
• Factoring Trinomials III ac-Method: We assume that there is no common factor for the trinomial ax2 + bx + c. 1. Calculate ac, and find two numbers u and v such that uv is ac, and u + v = b. 2. Write ax2 + bx + c as ax2 + ux + vx +cme.
• Factoring Trinomials III ac-Method: We assume that there is no common factor for the trinomial ax2 + bx + c. 1. Calculate ac, and find two numbers u and v such that uv is ac, and u + v = b. 2. Write ax2 + bx + c as ax2 + ux + vx +c 3. Use the grouping method to factor (ax2 + ux) + (vx + c)
• Factoring Trinomials III ac-Method: We assume that there is no common factor for the trinomial ax2 + bx + c. 1. Calculate ac, and find two numbers u and v such that uv is ac, and u + v = b. 2. Write ax2 + bx + c as ax2 + ux + vx +c 3. Use the grouping method to factor (ax2 + ux) + (vx + c) If step 1 can’t be done, then the expression is prime.
• Factoring Trinomials III ac-Method: We assume that there is no common factor for the trinomial ax2 + bx + c. 1. Calculate ac, and find two numbers u and v such that uv is ac, and u + v = b. 2. Write ax2 + bx + c as ax2 + ux + vx +c 3. Use the grouping method to factor (ax2 + ux) + (vx + c) If step 1 can’t be done, then the expression is prime. Example B. Factor 3x2 – 4x – 20 using the ac-method.
• Factoring Trinomials III ac-Method: We assume that there is no common factor for the trinomial ax2 + bx + c. 1. Calculate ac, and find two numbers u and v such that uv is ac, and u + v = b. 2. Write ax2 + bx + c as ax2 + ux + vx +c 3. Use the grouping method to factor (ax2 + ux) + (vx + c) If step 1 can’t be done, then the expression is prime. Example B. Factor 3x2 – 4x – 20 using the ac-method. Because a = 3, c = –20, we’ve ac = 3(–20) = –60.
• Factoring Trinomials III ac-Method: We assume that there is no common factor for the trinomial ax2 + bx + c. 1. Calculate ac, and find two numbers u and v such that uv is ac, and u + v = b. 2. Write ax2 + bx + c as ax2 + ux + vx +c 3. Use the grouping method to factor (ax2 + ux) + (vx + c) If step 1 can’t be done, then the expression is prime. Example B. Factor 3x2 – 4x – 20 using the ac-method. Because a = 3, c = –20, we’ve ac = 3(–20) = –60. We need two numbers u and v such that uv = –60
• Factoring Trinomials III ac-Method: We assume that there is no common factor for the trinomial ax2 + bx + c. 1. Calculate ac, and find two numbers u and v such that uv is ac, and u + v = b. 2. Write ax2 + bx + c as ax2 + ux + vx +c 3. Use the grouping method to factor (ax2 + ux) + (vx + c) If step 1 can’t be done, then the expression is prime. Example B. Factor 3x2 – 4x – 20 using the ac-method. Because a = 3, c = –20, we’ve ac = 3(–20) = –60. We need two numbers u and v such that uv = –60 and u + v = –4.
• Factoring Trinomials III ac-Method: We assume that there is no common factor for the trinomial ax2 + bx + c. 1. Calculate ac, and find two numbers u and v such that uv is ac, and u + v = b. 2. Write ax2 + bx + c as ax2 + ux + vx +c 3. Use the grouping method to factor (ax2 + ux) + (vx + c) If step 1 can’t be done, then the expression is prime. Example B. Factor 3x2 – 4x – 20 using the ac-method. Because a = 3, c = –20, we’ve ac = 3(–20) = –60. We need two numbers u and v such that uv = –60 and u + v = –4. A good way to do a thorough search is to make a table of all the ways to factor 60.
• Factoring Trinomials III ac-Method: We assume that there is no common factor for the trinomial ax2 + bx + c. 1. Calculate ac, and find two numbers u and v such that uv is ac, and u + v = b. 2. Write ax2 + bx + c as ax2 + ux + vx +c 3. Use the grouping method to factor (ax2 + ux) + (vx + c) If step 1 can’t be done, then the expression is prime. Example B. Factor 3x2 – 4x – 20 using the ac-method. Because a = 3, c = –20, we’ve ac = 3(–20) = –60. We need two numbers u and v such that uv = –60 and u + v = –4. A good way to do a thorough search is to make a table of all the ways to factor 60. u v 1 60 2 30 3 20 4 15 5 12 6 10
• Factoring Trinomials III ac-Method: We assume that there is no common factor for the trinomial ax2 + bx + c. 1. Calculate ac, and find two numbers u and v such that uv is ac, and u + v = b. 2. Write ax2 + bx + c as ax2 + ux + vx +c 3. Use the grouping method to factor (ax2 + ux) + (vx + c) If step 1 can’t be done, then the expression is prime. Example B. Factor 3x2 – 4x – 20 using the ac-method. Because a = 3, c = –20, we’ve ac = 3(–20) = –60. We need two numbers u and v such that uv = –60 and u + v = –4. A good way to do a thorough search is to make a table of all the ways to factor 60. From the list we got 6 and (–10). u v 1 60 2 30 3 20 4 15 5 12 6 10 6*(–10) = – 60 6 + (–10) = –4
• Factoring Trinomials III Write 3x2 – 4x – 20 = 3x2 + 6x –10x – 20 u v 1 60 2 30 3 20 4 15 5 12 6 10 6*(–10) = – 60 6 + (–10) = –4
• Factoring Trinomials III Write 3x2 – 4x – 20 = 3x2 + 6x –10x – 20 put in two groups = (3x2 + 6x ) + (–10x – 20) u v 1 60 2 30 3 20 4 15 5 12 6 10 6*(–10) = – 60 6 + (–10) = –4
• Factoring Trinomials III Write 3x2 – 4x – 20 = 3x2 + 6x –10x – 20 put in two groups = (3x2 + 6x ) + (–10x – 20) pull out common factor = 3x(x + 2) – 10 (x + 2) u v 1 60 2 30 3 20 4 15 5 12 6 10 6*(–10) = – 60 6 + (–10) = –4
• Factoring Trinomials III Write 3x2 – 4x – 20 = 3x2 + 6x –10x – 20 put in two groups = (3x2 + 6x ) + (–10x – 20) pull out common factor = 3x(x + 2) – 10 (x + 2) pull out common factor = (3x – 10)(x + 2) u v 1 60 2 30 3 20 4 15 5 12 6 10 6*(–10) = – 60 6 + (–10) = –4
• Factoring Trinomials III Write 3x2 – 4x – 20 = 3x2 + 6x –10x – 20 put in two groups = (3x2 + 6x ) + (–10x – 20) pull out common factor = 3x(x + 2) – 10 (x + 2) pull out common factor = (3x – 10)(x + 2) Example C. Factor 3x2 – 6x – 20 using the ac-method. u v 1 60 2 30 3 20 4 15 5 12 6 10 6*(–10) = – 60 6 + (–10) = –4
• Factoring Trinomials III Write 3x2 – 4x – 20 = 3x2 + 6x –10x – 20 put in two groups = (3x2 + 6x ) + (–10x – 20) pull out common factor = 3x(x + 2) – 10 (x + 2) pull out common factor = (3x – 10)(x + 2) Example C. Factor 3x2 – 6x – 20 using the ac-method. a = 3, c = –20, hence ac = 3(-20) = –60. u v 1 60 2 30 3 20 4 15 5 12 6 10 6*(–10) = – 60 6 + (–10) = –4
• Factoring Trinomials III Write 3x2 – 4x – 20 = 3x2 + 6x –10x – 20 put in two groups = (3x2 + 6x ) + (–10x – 20) pull out common factor = 3x(x + 2) – 10 (x + 2) pull out common factor = (3x – 10)(x + 2) Example C. Factor 3x2 – 6x – 20 using the ac-method. a = 3, c = –20, hence ac = 3(-20) = –60. Need two numbers u and v such that uv = –60 and u + v = –6. u v 1 60 2 30 3 20 4 15 5 12 6 10 6*(–10) = – 60 6 + (–10) = –4
• Factoring Trinomials III Write 3x2 – 4x – 20 = 3x2 + 6x –10x – 20 put in two groups = (3x2 + 6x ) + (–10x – 20) pull out common factor = 3x(x + 2) – 10 (x + 2) pull out common factor = (3x – 10)(x + 2) Example C. Factor 3x2 – 6x – 20 using the ac-method. a = 3, c = –20, hence ac = 3(-20) = –60. Need two numbers u and v such that uv = –60 and u + v = –6. After searching all possibilities we found that it's impossible. Hence 3x2 – 6x – 20 is prime. u v 1 60 2 30 3 20 4 15 5 12 6 10 6*(–10) = – 60 6 + (–10) = –4
• Factoring Trinomials III Exercise A. Use the ac–method, factor the trinomial or demonstrate that it’s not factorable. 1. 3x2 – x – 2 2. 3x2 + x – 2 3. 3x2 – 2x – 1 4. 3x2 + 2x – 1 5. 2x2 – 3x + 1 6. 2x2 + 3x – 1 7. 2x2 + 3x – 2 8. 2x2 – 3x – 2 9. 5x2 – 3x – 2 11. 3x2 + 5x + 2 12. 3x2 – 5x + 2 15. 6x2 + 5x – 6 10. 5x2 + 9x – 2 13. 3x2 – 5x + 2 14. 6x2 – 5x – 6 16. 6x2 – x – 2 17. 6x2 – 13x + 2 18. 6x2 – 13x + 2 19. 6x2 + 7x + 2 20. 6x2 – 7x + 2 21. 6x2 – 13x + 6 22. 6x2 + 13x + 6 23. 6x2 – 5x – 4 24. 6x2 – 13x + 8 25. 6x2 – 13x – 8 25. 4x2 – 9 26. 4x2 – 49 27. 25x2 – 4 28. 4x2 + 9 29. 25x2 + 9 B. Factor. Factor out the GCF, the “–”, and arrange the terms in order first. 30. – 6x2 – 5xy + 6y2 31. – 3x2 + 2x3– 2x 32. –6x3 – x2 + 2x 33. –15x3 – 25x2 – 10x 34. 12x3y2 –14x2y2 + 4xy2