10 more on factoring trinomials and factoring by formulas

Factoring Trinomials II

Factoring Trinomials IINow let’s try to factor trinomials of the form ax2 + bx + c.

Factoring Trinomials IINow let’s try to factor trinomials of the form ax2 + bx + c. We’ll give two methods.

Factoring Trinomials IINow let’s try to factor trinomials of the form ax2 + bx + c. We’ll give two methods. One is short but not reliable.

Factoring Trinomials IINow let’s try to factor trinomials of the form ax2 + bx + c. We’ll give two methods. One is short but not reliable. The second one takes more steps but gives definite answers.

Factoring Trinomials IINow let’s try to factor trinomials of the form ax2 + bx + c. We’ll give two methods. One is short but not reliable. The second one takes more steps but gives definite answers. Reversed FOIL Method

Factoring Trinomials IINow let’s try to factor trinomials of the form ax2 + bx + c. We’ll give two methods. One is short but not reliable. The second one takes more steps but gives definite answers. For this method, we need to find four numbers that fit certain descriptions.

Reversed FOIL Method

Factoring Trinomials IINow let’s try to factor trinomials of the form ax2 + bx + c. We’ll give two methods. One is short but not reliable. The second one takes more steps but gives definite answers. For this method, we need to find four numbers that fit certain descriptions. The following are examples of the task to be accomplished.


Factoring Trinomials IINow let’s try to factor trinomials of the form ax2 + bx + c. We’ll give two methods. One is short but not reliable. The second one takes more steps but gives definite answers. For this method, we need to find four numbers that fit certain descriptions. The following are examples of the task to be accomplished. Example A. Let {1, 3} and {1, 2} be two pairs of numbers. Is it possible to split the {1, 2 }, put them in the boxes that makes the equality true?



a. 1* (± ) + 3*(± ) = 5.



a. 1* (± ) + 3*(± ) = 5.

Yes, 1* (2) + 3 * (1) = 5



a. 1* (± ) + 3*(± ) = 5.

Yes, 1* (2) + 3 * (1) = 5

b. 1* (± ) + 3* (± ) = –5.



a. 1* (± ) + 3*(± ) = 5.

Yes, 1* (2) + 3 * (1) = 5

b. 1* (± ) + 3* (± ) = –5.Yes, 1* (1) + 3* (–2) = –5



a. 1* (± ) + 3*(± ) = 5.

Yes, 1* (2) + 3 * (1) = 5

b. 1* (± ) + 3* (± ) = –5.Yes, 1* (1) + 3* (–2) = –5 or 1* (–2) + 3* (–1) = –5



a. 1* (± ) + 3*(± ) = 5.

Yes, 1* (2) + 3 * (1) = 5

b. 1* (± ) + 3* (± ) = –5.Yes, 1* (1) + 3* (–2) = –5 or 1* (–2) + 3* (–1) = –5

c. 1* (± ) + 3* (± ) = 8.



a. 1* (± ) + 3*(± ) = 5.

Yes, 1* (2) + 3 * (1) = 5

b. 1* (± ) + 3* (± ) = –5.Yes, 1* (1) + 3* (–2) = –5 or 1* (–2) + 3* (–1) = –5

c. 1* (± ) + 3* (± ) = 8.No, since the most we can obtain is 1* (1) + 3* (2) = 7.


Factoring Trinomials II(Reversed FOIL Method)

Factoring Trinomials II(Reversed FOIL Method) Let’s see how the above examples are related to factoring.

Example B. Factor 3x2 + 5x + 2.



The only way to get 3x2 is (3x ± #)(1x ± #).




The #’s must be 1 and 2




The #’s must be 1 and 2 to get the constant term +2.





We need to place 1 and 2 as the #'s so the product will yield the correct middle term +5x.






That is, (3x ± #)(1x ± #) must yields +5x,


3(± # ) +1(± #) = 5 where the #’s are 1 and 2.






That is, (3x ± #)(1x ± #) must yields +5x, or that


Since 3(1) +1(2) = 5,








Since 3(1) +1(2) = 5,








Since 3(1) +1(2) = 5,








Since 3(1) +1(2) = 5, we see that

3x2 + 5x + 2 = (3x + 2)(1x + 1).








Since 3(1) +1(2) = 5, we see that

3x2 + 5x + 2 = (3x + 2)(1x + 1).5x







Factoring Trinomials IIExample C. Factor 3x2 – 7x + 2.


We start with (3x ± #)(1x ± #).


We start with (3x ± #)(1x ± #). We need to fill in 1 and 2 as #'s

3(± # ) + 1(± # ) = –7.


We start with (3x ± #)(1x ± #). We need to fill in 1 and 2 as #'s so that

3(± # ) + 1(± # ) = –7.It's 3(–2) + 1(–1) = –7.



3(± # ) + 1(± # ) = –7.It's 3(–2) + 1(–1) = –7.So 3x2 – 7x + 2 = (3x –1)(1x – 2)



3(± # ) + 1(± # ) = –7.It's 3(–2) + 1(–1) = –7.So 3x2 – 7x + 2 = (3x –1)(1x – 2)

Example D. Factor 3x2 + 5x – 2.



3(± # ) + 1(± # ) = –7.It's 3(–2) + 1(–1) = –7.So 3x2 – 7x + 2 = (3x –1)(1x – 2)





3(± # ) + 1(± # ) = –7.It's 3(–2) + 1(–1) = –7.So 3x2 – 7x + 2 = (3x –1)(1x – 2)



We need to fill in 1 and 2 as #'s



3(± # ) + 1(± # ) = –7.It's 3(–2) + 1(–1) = –7.So 3x2 – 7x + 2 = (3x –1)(1x – 2)



We need to fill in 1 and 2 as #'s so that

3(± # ) + 1(± # ) = +5.



3(± # ) + 1(± # ) = –7.It's 3(–2) + 1(–1) = –7.So 3x2 – 7x + 2 = (3x –1)(1x – 2)




3(± # ) + 1(± # ) = +5.Since c is negative, they must have opposite signs .



3(± # ) + 1(± # ) = –7.It's 3(–2) + 1(–1) = –7.So 3x2 – 7x + 2 = (3x –1)(1x – 2)




3(± # ) + 1(± # ) = +5.

It is 3(+2) + 1(–1) = +5.Since c is negative, they must have opposite signs .



3(± # ) + 1(± # ) = –7.It's 3(–2) + 1(–1) = –7.So 3x2 – 7x + 2 = (3x –1)(1x – 2)




3(± # ) + 1(± # ) = +5.

It is 3(+2) + 1(–1) = +5.So 3x2 + 5x + 2 = (3x –1)(1x + 2)

Since c is negative, they must have opposite signs .



Example E. Factor 3x2 + 8x + 2.







We need to fill in 1&2 so that

3(± # ) + 1(± # ) = +8.





3(± # ) + 1(± # ) = +8.This is impossible.





3(± # ) + 1(± # ) = +8.This is impossible. Hence the expression is prime.







If both the numbers a and c in ax2 + bx + c have many factors then there are many possibilities to check.







Example F. Factor 3x2 + 11x – 4.
















We start with (3x ± #)(1x ± #). Since 4 = 2(2) = 1(4),





3(± # ) + 1(± # ) = +11.




We start with (3x ± #)(1x ± #). Since 4 = 2(2) = 1(4), we need to fill in 2&2 or 1&4 as #'s so that





3(± # ) + 1(± # ) = +11. It can't be 2&2.





3(± # ) + 1(± # ) = +11.





3(± # ) + 1(± # ) = +11. It can't be 2&2.





Try 1&4, 3(± # ) + 1(± # ) = +11.





3(± # ) + 1(± # ) = +11. It can't be 2&2.





Try 1&4, it is 3(+4) + 1(–1) = +11.

3(± # ) + 1(± # ) = +11.





3(± # ) + 1(± # ) = +11. It can't be 2&2.





Try 1&4, it is 3(+4) + 1(–1) = +11. So 3x2 + 11x – 4 = (3x – 1)(1x + 4).

Factoring Trinomials IIIt's not necessary to always start with ax2. If c is a prime number, we start with c.

Example G. Factor 12x2 – 5x – 3.



Since 3 must be 3(1),



Since 3 must be 3(1), we need to find two numbers such that (#)(#) = 12



Since 3 must be 3(1), we need to find two numbers such that (#)(#) = 12 and that


(± #)(± 3) + (± #)(±1) = – 5.




(± #)(± 3) + (± #)(±1) = – 5. 12 = 1(12) = 2(6) = 3(4)




(± #)(± 3) + (± #)(±1) = – 5. 12 = 1(12) = 2(6) = 3(4) 1&12 and 2&6 can be quickly eliminated.




(± #)(± 3) + (± #)(±1) = – 5. 12 = 1(12) = 2(6) = 3(4) 1&12 and 2&6 can be quickly eliminated. We get (3)(–3) + (4)(+1) = – 5.



So 12x2 – 5x – 3 = (3x + 1)(4x – 3).





So 12x2 – 5x – 3 = (3x + 1)(4x – 3).



Remark:In the above method, finding(#)(± #) + (#)( ± #) = b does not guarantee that the trinomial will factor.



So 12x2 – 5x – 3 = (3x + 1)(4x – 3).



Remark:In the above method, finding(#)(± #) + (#)( ± #) = b does not guarantee that the trinomial will factor. We have to match the sign of c also.

Example H. Factor 3x2 – 7x – 2 .






We start with (3x ± #)(1x ± #). We find that:

3(–2) + 1(–1) = –7.




3(–2) + 1(–1) = –7.But this won't work since (–2)(–1) = 2 = c.




3(–2) + 1(–1) = –7.But this won't work since (–2)(–1) = 2 = c. In fact this trinomial is prime.






There might be multiple matchings for(#)(± #) + (#)( ± #) = b make sure you chose the correct one, if any.





Example I: Factor 1x2 + 5x – 6 .







We have:1(+3) + 1(+2) = +5







We have:1(+3) + 1(+2) = +5 1(+6) + 1(–1) = +5







We have:1(+3) + 1(+2) = +5

The one that works is x2 + 5x – 6 = (x + 6)(x – 1).

1(+6) + 1(–1) = +5


Factoring Trinomials IIFinally, before starting the reverse-FOIL procedure

Factoring Trinomials IIFinally, before starting the reverse-FOIL procedure1. make sure the terms are arranged in order.

Factoring Trinomials IIFinally, before starting the reverse-FOIL procedure1. make sure the terms are arranged in order. 2. if there is any common factor, pull out the GCF first.

Factoring Trinomials IIFinally, before starting the reverse-FOIL procedure1. make sure the terms are arranged in order. 2. if there is any common factor, pull out the GCF first. 3. make sure that x2 is positive, if not, factor out the negative sign first.


Example J. Factor –x3 + 3x + 2x2



–x3 + 3x + 2x2 Arrange the terms in order

= –x3 + 2x2 + 3x




= –x3 + 2x2 + 3x Factor out the GCF

= – x(x2 – 2x – 3)





= – x(x2 – 2x – 3)

= – x(x – 3)(x + 1)





= – x(x2 – 2x – 3)

= – x(x – 3)(x + 1)

Ex. A. Factor the following trinomials. If it’s prime, state so.

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

Education

10 more on factoring trinomials and factoring by formulas