5 5factoring trinomial iii

Factoring Trinomials III

Factoring Trinomials IIIIn this section we give a formula that enables us to tell if a trinomial is factorable or not.

Factoring Trinomials IIIIn 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.

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





a. 3x2 – 7x – 2 b2 – 4ac = (–7)2 – 4(3)(–2)





a. 3x2 – 7x – 2 b2 – 4ac = (–7)2 – 4(3)(–2) = 49 + 24





a. 3x2 – 7x – 2 b2 – 4ac = (–7)2 – 4(3)(–2) = 49 + 24 = 73 is not a square,





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






b. 3x2 – 7x + 2

b2 – 4ac = (–7)2 – 4(3)(2)






b. 3x2 – 7x + 2

b2 – 4ac = (–7)2 – 4(3)(2) = 49 – 24






b. 3x2 – 7x + 2

b2 – 4ac = (–7)2 – 4(3)(2) = 49 – 24 = 25 which is a squared number,






b. 3x2 – 7x + 2

b2 – 4ac = (–7)2 – 4(3)(2) = 49 – 24 = 25 which is a squared number, hence it is factorable.






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)

The ac-MethodFactoring Trinomials III

The ac-MethodThe product of two binomials has four terms.


The ac-MethodThe 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




x2 – x – 6 = x2 – 3x + 2x – 6 Put them into two groups= (x2 – 3x) + (2x – 6)




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)




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?





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.





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.





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 IIIac-Method:

Factoring Trinomials IIIac-Method: We assume that there is no common factor for the trinomial ax2 + bx + c.


1. Calculate ac,


1. Calculate ac, and find two numbers u and v such that uv is ac,


1. Calculate ac, and find two numbers u and v such that uv is ac, and u + v = b.


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.


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 +c3. 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’veac = 3(–20) = –60.





Because a = 3, c = –20, we’veac = 3(–20) = –60. We need two numbers u and v such that uv = –60





Because a = 3, c = –20, we’veac = 3(–20) = –60. We need two numbers u and v such that uv = –60 and u + v = –4.





Because a = 3, c = –20, we’veac = 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.





Because a = 3, c = –20, we’veac = 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





Because a = 3, c = –20, we’veac = 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

6 106*(–10) = – 60

6 + (–10) = –4

Write 3x2 – 4x – 20 = 3x2 + 6x –10x – 20


u v

1 60

2 30

3 20

4 15

5 12

6 106*(–10) = – 60

6 + (–10) = –4

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

6 106*(–10) = – 60

6 + (–10) = –4

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

6 106*(–10) = – 60

6 + (–10) = –4

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

6 106*(–10) = – 60

6 + (–10) = –4



Example C. Factor 3x2 – 6x – 20 using the ac-method.

u v

1 60

2 30

3 20

4 15

5 12

6 106*(–10) = – 60

6 + (–10) = –4




a = 3, c = –20, hence ac = 3(-20) = –60.

u v

1 60

2 30

3 20

4 15

5 12

6 106*(–10) = – 60

6 + (–10) = –4




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

6 106*(–10) = – 60

6 + (–10) = –4




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

6 106*(–10) = – 60

6 + (–10) = –4

1. 3x2 – x – 2 2. 3x2 + x – 2 3. 3x2 – 2x – 14. 3x2 + 2x – 1 5. 2x2 – 3x + 1 6. 2x2 + 3x – 1

8. 2x2 – 3x – 27. 2x2 + 3x – 2

15. 6x2 + 5x – 610. 5x2 + 9x – 2

B. Factor. Factor out the GCF, the “–”, and arrange the terms in order first.

9. 5x2 – 3x – 212. 3x2 – 5x + 211. 3x2 + 5x + 2

14. 6x2 – 5x – 613. 3x2 – 5x + 216. 6x2 – x – 2 17. 6x2 – 13x + 2 18. 6x2 – 13x + 219. 6x2 + 7x + 2 20. 6x2 – 7x + 2

21. 6x2 – 13x + 6

22. 6x2 + 13x + 6 23. 6x2 – 5x – 4 24. 6x2 – 13x + 825. 6x2 – 13x – 8


25. 4x2 – 9 26. 4x2 – 4927. 25x2 – 4 28. 4x2 + 9 29. 25x2 + 9

30. – 6x2 – 5xy + 6y2 31. – 3x2 + 2x3– 2x 32. –6x3 – x2 + 2x

33. –15x3 – 25x2 – 10x 34. 12x3y2 –14x2y2 + 4xy2

Exercise A. Use the ac–method, factor the trinomial or demonstrate that it’s not factorable.

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5 5factoring trinomial iii