Please help, Algebra question?

6x² + 4x - 2 I like to factor this by grouping. To do so, we want to expand the middle term + 4x. Take the 6 from the x^2 term and multiply it by the -2 at the end. This gives you -12. Now, can we find two factors of -12 with a sum of +4 ? Yes, with +6 and -2, we get -12 as a product and +4 as a sum. So expand the +4x using these numbers.

6x² + 6x - 2x - 2

Now group them two at a time:
(6x² + 6x) + (- 2x - 2)
Now factor what is common from each:
6x(x + 1) -2(x + 1)
Now factor the common factor: (x + 1)(6x - 2)

Your answer is: ( x + 1 )( 6x - 2 )

For your second problem, you must recognize it as a difference of two squares. Use the formula: A² - B² =( A - B)(A + B)

In the case of x² - 36, your A = x and your B = 6
So, x² - 6² = (x - 6)(x + 6)

This third one doesn't fall into the category of difference of cubes, but in general, if you have something of the form A³ - B³ this is known as the difference of cubes and factors as follows:
A³ - B³= (A - B)(A² + AB + B²)

Hope this helped

Edit. In problem 2, the final answer is not correct. I failed to factor out a 2. So, your answer should be:
2( x + 1 )( 3x - 1 )

Also, 3) doesn't factor using rational numbers. It will factor if you use radical expressions(square roots). Note also that the person above who did factor it did so incorrectly.

use BODMAS rule. That is solve brackets then Off then Division then Multiplication then Addition and finally Subtraction

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

Use the FOIL method. You can use logic by saying that in 1)
3x times 2x is 6x^2. 3x times 2 is 6x, and -1 times 2x is -2x (6x-2x=4x). Then -1 times 2 is -2. This leaves 6x^2+4x-2x

6x^2+4x-2
First you need to figure out what factors give you 6. They would be either 2x3 or 6x1. Then do the same for 2. They would be simply 2x1. Then you need to set up 2 quantities and almost play with it. Start with the 2 and the 3 to make x^2.
(2x +/- _)(3x +/- _). Now when you insert the factors for the 2 (2 and 1) you need check, that when you multiply it out will you get 4x left over. Remember to play with the + and - signs. Since the 2 is negative, each quantity will have a different one to give you the negative.

This answer for the 1st one is (6x - 2)(1x + 1)
Do a quick FOIL check.
F: 6x and 1x make 6x^2
O: 6x and 1 make 6x
I: -2 and 1x make -2x

NOTE that O and I go together to make 4x

and then L: -2 and 1 make -2.

The trick to getting it is to make the 4x. Think how to arrange the factor of both 6x^2 and -2 to make the 4.


x^2-36
These problems are more memorization than anything. Any problem like this follows this patten

(A-B)= ( sqrt(A) + sqrt(B) )( sqrt(A) - sqrt(B) )

Making this problem (x+6)(x-6)

NOTE, this is only when there is a minus sign between them. That is very important.

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

hope i helped

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

2)
x^2 - 36
= (x + 6)(x - 6)

3)
x^3 - 36
= cannot be factored

First of all: Don't Stress Out. Mostly, the numbers are guessing as well as trial and error.

1. 6x^2 + 4x - 2: So, lets try 2 and 3 as the factors of 6. We can also figure out that one of the second number in the parenthases is a negative and the other a positive because they have to multiply to equal -2. So lets try (3x - 2) (2x + 1), which would become 6x^2 -x -2 by FOIL. So this doesn't work. Let's switch the 2 and the 1. So lets try (3x - 1) (2x + 2) which would become 6x^2 + 4x - 2 by FOIL. It worked, so that would be your answer: (3x - 1) (2x + 2).

2. So we know that that the numbers multiply to 36, and that one is even and the other odd. We also know that when added together they cancel each other out, because there is no number inbetween the x^2 and the -36. So it would have to be (x - 6) (x + 6).

3. This time we do it the same way, except now we know that the x^2 is now x^3. So instead of x * x, we do x^2 * x. The answer would then be: (x^2 - 6) (x + 6)

First step to factoring is removing common terms.

So for the first one:

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

Now we have to determine the factors of the trinomial.

To do this you have to think about whats going to happen when the factors are multiplied to give this answer. We know that the first terms must give 3x^2 therefore the first terms of the factors are 3x and x. Since the last number is negative they have different signs:

(3x-?)(x+?)
Now think of two numbers that when multiplied give one. The only answer of course is one therefore the factors are:

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

For the nest two, its the exact same logic, except that these are perfect squares meaning the middle term cancels out.

therefore the factors of the second one is:

(x-6)(x+6)

As for the last one it's not a perfect sube so i dont think its factorable...

6x^2+4x-2
6x^2+6x-2x-2
6x(x+1)-2(x+1)
(x+1)(6x-2)
Find 2 numbers such that their sum is 4 and product is -12. The numbers are 6 and -2. You write 4x as 6x-2x

x^2-36
a^2-b^2=(a+b)(a-b)
(x+6)(x-6)

1. 6x^2+4x-2 factor out 2
2(3x^2+2x-1) factor binomial in parentheses
2(3x-1)(x+1)

2. x^2-36
this is a difference of squares a^2-b^2=(a-b)(a+b)
(x-6)(x+6)