Undoing the FOIL method - 10 pts?

Question

Gosh Algebra was so long ago. I just saw a problem and figured it out but cant figure out how to get binomials to a problem.

How do you take 2x^2 - x - 6 and get it into (2x + 3) (x-2)

I know how to FOIL and get it into 2x^2 -x -6, but I cant reverse.

Thanks guys... no rude comments please!

Answer 1

Hi,

Here is my explanation of how to factor trinomials. I did your problem at the beginning.

On every type of factoring problem ALWAYS look first for a GCF, a greatest common factor. There won't always be one, but if there is, you should divide it out first. A GCF will carry along through the problem and be the first thing in an answer with a GCF.

I' m going to explain factoring trinomials the way that I teach it, which is probably different from what you've done. Try to follow the steps. This method will work on any trinomial.

2x² - x - 6 Look for a GCF None this time. If there was one, factor it out. Then temporarily start both parentheses with the first number and variable.
(2x.......)(2x..........) First sign goes in first parentheses.
(2x..-....)(2x..........) Product of signs goes in 2nd parentheses.
(2x..-....)(2x...+.....) <== plus is because neg x neg = positive

Now multiply your first and third numbers together. Ignore their signs - you've already done them. 2 x 6 = 12 So, out to the side list pairs of factors of 12.

12
------
1, 12
2, 6
3, 4

Now you want to pick which factors go in your parentheses, using these rules:

If the signs you put on your parentheses are the SAME, find the factors that ADD up to the middle number in the problem.
If the signs you put on your parentheses are the DIFFERENT, find the factors that SUBTRACT to the middle number in the problem. (Note you look at signs in the parentheses, not in the problem.)

(2x..-....)(2x...+.....) Your signs are different, so you want to subtract factors to get 1. Those factors are 3 and 4. When you put the numbers in the parentheses, the bigger number is pushy - it always goes first. So your factors now are:
(2x..-.4)(2x.+.3)

Now you have to reduce either or both parentheses by dividing each parentheses' terms by the largest possible divisor. In our problem, the first parentheses is divisible by 2 but the second parentheses does not reduce.

(2x..-.4)(2x.+.3)
----------
......2..... This reduces to your final factors of

(x - 2)(2x + 3) <== YOUR ANSWER

3y² - 13y - 10 Look for a GCF None this time. If there was one, factor it out. Then temporarily start both parentheses with the first number and variable.
(3y.......)(3y..........) First sign goes in first parentheses.
(3y..-....)(3y..........) Product of signs goes in 2nd parentheses.
(3y..-....)(3y...+.....) <== plus is because neg x neg = positive

Now multiply your first and third numbers together. Ignore their signs - you've already done them. 3 x 10 = 30 So, out to the side list pairs of factors of 30.

30
------
1, 30
2, 15
3, 10
5, 6

Now you want to pick which factors go in your parentheses, using these rules:

If the signs you put on your parentheses are the SAME, find the factors that ADD up to the middle number in the problem.
If the signs you put on your parentheses are the DIFFERENT, find the factors that SUBTRACT to the middle number in the problem. (Note you look at signs in the parentheses, not in the problem.)

(3y..-....)(3y...+.....) Your signs are different, so you want to subtract factors to get 13. Those factors are 2 and 15. ( Notice that 3 and 10 would have added to 13, so you HAD to know to subtract this time.) When you put the numbers in the parentheses, the bigger number is pushy - it always goes first. So your factors now are:
(3y..-..15)(3y.+.2.)

Now you have to reduce either or both parentheses by dividing each parentheses' terms by the largest possible divisor. In our problem, the first parentheses is divisible by 3 but the second parentheses does not reduce.

(3y..-..15)(3y.+.2.)
-------------
.......3 This reduces to your final factors of

(y - 5)(3y + 2)

NEXT PROBLEM !!

2x² +15x + 7 Look for a GCF None this time. If there was one, factor it out. Then temporarily start both parentheses
............................. with the first number and variable.
(2x.......)(2x..........) First sign goes in first parentheses.
(2x..+....)(2x..........) Product of signs goes in 2nd parentheses.
(2x..+....)(2x...+.....) <== plus is because pos x pos = positive

Now multiply your first and third numbers together. Ignore their signs - you've already done them. 2 x 7 = 14 So, out to the side list pairs of factors of 14.

14
------
1, 14
2, 7

Now you want to pick which factors go in your parentheses, using these rules:

If the signs you put on your parentheses are the SAME, find the factors that ADD up to the middle number in the problem.
If the signs you put on your parentheses are the DIFFERENT, find the factors that SUBTRACT to the middle number in the problem. (Note you look at signs in the parentheses, not in the problem.)

(2x..+....)(2x...+.....) Your signs are the same, so you want to add factors to get 15. Those factors are 1 and 14. When you put the numbers in the parentheses, the bigger number is pushy - it always goes first. So your factors now are:
(2x.+.14)(2x.+.1)

Now you have to reduce either or both parentheses by dividing each parentheses' terms by the largest possible divisor. In our problem, the first parentheses is divisible by 2 but the second parentheses does not reduce.

(2x.+.14)(2x.+.1)
-------------
.......2 This reduces to your final factors of

(x + 7)(2x + 1)

This takes care of trinomials!!

I hope this helps you!! :-)

Answer 2

follow the following steps each time you are doing this
1) you are looking for two numbers, their sum is = -1 .and their product is equal = (2 * -6) = - 12
( -1 is the co-eff. of x, 2 is the co-eff. of x^2, and -6 is the last term.
2) -12 = - 4 * 3, because if you add them you get the sum = -1
3) replace the term with x , by two terms with these two numbers,
2x^2 -4x +3x - 6
3) now, find the common factor for each two terms,they call this factoring by grouping,
2x ( x - 2) +3( x - 2), in all such problems, you will find that the two brackets are exactly the same(otherwise you should have mistaken),
3) take this bracket as a common factor=
( x- 2) ( 2x +3)

Answer 3

(9x + 2)(9x + 2) = 81x^2 + 18x + 18x + 4 = 81x^2 + 36x + 4. (2x + 5)^2 is (2x + 5)(2x + 5) = 4x^2 + 10x + 10x + 25 = 4x^2 + 20x + 25. (3x - 4)^2 is (3x - 4)(3x - 4) = 9x^2 - 12x - 12x + sixteen = 9x^2 - 24x + sixteen. (3x + 4)^2 is (3x + 4)(3x + 4) = 9x^2 + 12x + 12x + sixteen = 9x^2 + 24x + sixteen. think of of your self as having purely gained a present day. i do no longer do homework!

Answer 4

2x² - x - 6

Step 1
Factors for 2x² are 2x and x
(2x____)(x______)

Step 2
Require factors for 6
Thes could be 3 and 2 or 6 and 1
Try 3 and 2

Step 3
(2x___3)(x____2)

Step 4
Affix signs to obtain - 6 and - x
(2x + 3)(x - 2)

Answer is (2x + 3)(x - 2)

Answer 5

Long time ago for me as well,
I think you just have to look at the first and last numbers to see what happens with they are multiplied.