Trinomial Factoring?

Question

I've been trying to factor 6x^2 + x - 2. I've been trying to use the "what multiplies to -2 and adds to to equal x" method but I can't get it to work. I think it's -1 and 2. But when I foil it it doesn't work. maybe I'm completely missing something. If anyone has an easier way to do this then please answer!!

Answer 1

Hi,

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.

6x² + x - 2 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.
(6x.......)(6x..........) First sign goes in first parentheses.
(6x..+....)(6x........) Product of signs goes in 2nd parentheses.
(6x..+....)(6x..-....) <== neg is because pos x neg = negative

Now multiply your first and third numbers together. Ignore their signs - you've already done them. 6 x 2 = 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.)

(6x..+...)(6x..-.....) 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:
(6x..+..4)(6x..-.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 and the second parentheses is divisible by 3.

(6x..+..4)(6x..-.3)
-----------..---------
.......2...........3.......These reduce to your final factors of

(3x.+.2)(2x.-.1)

Here's another problem.
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)


I hope this helped!! :-)

Answer 2

Step 1) Multiply first and last term ( 6x-2=-12) Step 2) Find multiples of -12 that add up to the middles term ( 1 ). It comes out to be -3 and 4, because -3+4=1. Step 3) Write extended formula(6x^2-3x+4x-2). Step 4) Insert parenthesis [ (6x^2-3x)+(4x-2) ]. Step 5) Take out common factors [ 3x(2x-1)+2(2x-1) . Step 6) Take out common factor. [ (3x+2)(2x-1)

Answer 3

no...u have to multiply -2 and 6. that would be -12. now find something that multiplies to -12 and adds to 1.....you guessed it, 4 and -3. now you'll have 6x^2-3x+4x-2. factor each half to get 3x(2x-1)+2(2x-1). you can now write the answer as (3x+2)(2x-1). If you want to go further and find the zeros of the equation, set it equal to zero and you'll get -2/3 and 1/2.

Answer 4

6x^2 + x - 2
6x^2 + 4x - 3x - 2
2x(3x + 2) - 1(3x + 2)
(2x - 1)(3x + 2)

If ax^2 + bx + c is the general eq...just split 'b' into two numbers that add upto 'b' and multiply to give the product of 'a' and 'c'.
Here, +4 and -3 add to give 1 and muliply to give -12

Answer 5

(2x-1)(3x+2)

What multiplies to -2 is -1 and +2 or +1 and -2.
What multiplies to 6x^2 is 3x and 2x or 6x and 1x.

Answer 6

(2x - 1)(3x + 2)

Answer 7

(3X-1)(2X+1)