Help factoring?

HI,
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.

3x² + 2x - 1 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.
(3x.......)(3x..........) First sign goes in first parentheses.
(3x..+....)(3x.........) Product of signs goes in 2nd parentheses.
(3x..+....)(3x...-.....) <== 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 1 = 3 So, out to the side list pairs of factors of 3.

3
------
1, 3

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.)

(3x...+.....)(3x..-....) Your signs are different, so you want to subtract factors to get 2. Those factors are 1 and 3. When you put the numbers in the parentheses, the bigger number is pushy - it always goes first. So your factors now are:
(3x.+..3)(3x.-.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 3 but the second parentheses does not reduce.

(3x.+..3)(3x.-.1)
----------
.....3
This reduces to your final factors of

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

x² + 6x + 9 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.
(x.....)(x........) First sign goes in first parentheses.
(x..+...)(x......) Product of signs goes in 2nd parentheses.
(x..+...)(x..+..) <== plus is because pos x pos = positive

Now multiply your first and third numbers together. Ignore their signs - you've already done them. 1 x 9 = 9 So, out to the side list pairs of factors of 9.

9
------
1, 9
3, 3

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.)

(x..+....)(x..+.....) Your signs are the same, so you want to add factors to get 6. Those factors are 3 and 3. When you put the numbers in the parentheses, the bigger number is pushy - it always goes first. So your factors now are:
(x.+.3)(x.+.3)

These factors have no number in front of the x, so they don't reduce and you're done.

(x + 3)(x + 3) is the answer.

Here's another example.

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 that helps!! :-)

You have to look at the coefficients of the first and last terms and their integer factors. Then you need to be able to add some combination of those factors to get the coefficient of the middle term.

For the first one, 3 factors into 3 and 1; -1 factors into -1 and 1. To get 2 from those factors you can take 3 -1. So you have 3x - x = 2x. That's the way I think you about them.

The second one is (x+3)(x+3). 9 factors into 9 and 1 or 3 and 3. Well 3+3 = 6, so that makes that a promising route to explore. With practice you'll begin to see the patterns automatically.

consider this:
(x+a)(x+b)=x^2+x(a+b)+ab



(x+3)^2

x^2 + 6x + 9
= x^2 + 3x + 3x + 9
= (x^2 + 3x) + (3x + 9)
= x(x + 3) + 3(x + 3)
= (x + 3)(x + 3)
= (x + 3)^2

3x² + 2x - 1

(3x-----)(x-------)
(3x-----1)(x-----1)
(3x - 1)(x + 1) = 0
x = 1/3 , x = - 1


(x + 3)² is answer for factorising next question.

(3x-1) (x+1)=0
you distribute: (3x sq. - 1)=0
3x sq. = 1
get the square-root of: 3x sq.
and then 1 will be divided to the ans.

3x^2 + 2x -1 =0

find 2 numbers that when you multiply them you get the 3rd part which is -1 and the same numbers when you add them you get 2
3x -1
x +1
try to multiply 3x by x and 1
and -1 by x and 1
when you add them all you'll get the original equation
---------------------------------------

x^2 + 6x +9

(x+3)(x+3) =0
x= -3