Factor x(2) + 4x - 12, how?

[itgl72]

I have not taken algebra in 20 years, and have an algebra class left to complete my associates. Ive been reviewing like crazy, but much of this is still stumping me and I feel behind so I'll be coming in here a bit to see if I can get help, and find a tutor at the college. Im making progress, but its SLOW, and PAINFUL, and time-consuming. Have to do this when my kids are at school in order to have peace to study.

Anyway, stuck on something here, I'll give you part of the problem below:

Code: Select all

3               1                4
---      +   ---      =     -----
X+6          x-2           x(2) + 4x - 12


The section x(2) + 4x - 12 in the book is show factored to (x+6)(x-2)

This is done to also find the LCD.

I dont understand how it got to (x+6)(x-2)??

[little_dragon]

The section x(2) + 4x - 12 in the book is show factored to (x+6)(x-2)

I dont understand how it got to (x+6)(x-2)??

they show U here: http://www.purplemath.com/modules/factquad.htm
its 12=(+6)(-2) & +6-2=+4 so x^2+4x-12=(x+6)(x-2)

[itgl72]

I'm a litte confused over the box method. So far its been helful.


http://www.purplemath.com/modules/factquad2.htm

Down in the box area:

4x2 – 19x + 12 = (x – 4)(4x – 3)

To the left of the box there is an x and below that a -4



Same thing with this 2x2 + x – 6 = (2x – 3)(x + 2).:



How do you get that x and +2 in the front of the box.

Sorry, I am terrible at algebra

[little_dragon]

How do you get that x and +2 in the front of the box.

U take x out of 2x^2 & -3x
U take +2 out of +4x & -6
they show how here: http://www.purplemath.com/modules/simpfact.htm

[MrAlgebra]

I prefer the "magic number" way of factoring trinomials since the people I've worked with have seemed to understood it better that way.

If your trinomial is in the form of ax² + bx + c, then you find two numbers that multiply to give you the product ac and that add to give you b.

With your expression x² + 4x - 12, you have a = 1, b = 4 and c = -12. The product ac is 1 * -12 = -12, so you'll want to find two numbers that multiply to give you -12 and add to give you 4. Those numbers are 6 and -2. You use these two numbers to break up the middle part of the expression like so:

x² + 4x - 12
x² + 6x - 2x - 12

Then you pull everything you can out of the first two terms:

x² + 6x - 2x - 12
x(x + 6) - 2x - 12

Then you do the same thing with the last two terms:

x(x + 6) - 2x - 12
x(x + 6) - 2(x + 6)

Finally, you pull out whatever is in the parentheses to get your final answer:

x(x + 6) - 2(x + 6)
(x + 6)(x - 2)

Hope that helps.

[itgl72]

I prefer the "magic number" way of factoring trinomials since the people I've worked with have seemed to understood it better that way.

If your trinomial is in the form of ax² + bx + c, then you find two numbers that multiply to give you the product ac and that add to give you b.

With your expression x² + 4x - 12, you have a = 1, b = 4 and c = -12. The product ac is 1 * -12 = -12, so you'll want to find two numbers that multiply to give you -12 and add to give you 4. Those numbers are 6 and -2. You use these two numbers to break up the middle part of the expression like so:

x² + 4x - 12
x² + 6x - 2x - 12

Then you pull everything you can out of the first two terms:

x² + 6x - 2x - 12
x(x + 6) - 2x - 12

Then you do the same thing with the last two terms:

x(x + 6) - 2x - 12
x(x + 6) - 2(x + 6)

Finally, you pull out whatever is in the parentheses to get your final answer:

x(x + 6) - 2(x + 6)
(x + 6)(x - 2)

Hope that helps.

Im going to try this.

Ive been doing it lately with the box version listed above, but I still get hung up on BIG numbers.

[theshadow]

Ive been doing it lately with the box version listed above, but I still get hung up on BIG numbers.

For big numbers try factor pairs. When a*c is large write out a list of factors using your calculator to do divisions by 1, 2, 3, 4, &tc whatever goes in evenly until you get a pair that works.

[MrAlgebra]

It's not as difficult as it might sound to list out all of the possible factors to see what they add up to. Suppose we have:

8x² + 24x + 10

So you're looking for two numbers that will multiply to give you 80 and that add to give you 24. Let's look at the factors of 80 by just starting at 1 and seeing if 80 is divisible by different numbers.

1 80, sum = 81
2 40, sum = 42
3 nope
4 20, sum = 24 (bingo)

Let's list out the rest of them just to show how fast it can be.

5 16, sum = 21
6 nope
7 nope
8 10, sum = 18
9 nope

And with 10 we're back to 8, so we can stop there since we've gotten all of the pairs of factors that matter.