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Factor:

3/6x^2 - 4x minus x - 2/9x - 6

2006-10-16 05:57:35 · 6 answers · asked by sillyboys_trucksare4girls 2 in Science & Mathematics Mathematics

RevKev, yes, you are correct.

2006-10-16 06:08:53 · update #1

6 answers

3/2x(3x-2)-(x-2)/3(3x-2)
LCD=6x(3x-2)
9-2x^2+4x/6x(3x-2)

2006-10-16 06:07:14 · answer #1 · answered by raj 7 · 0 0

Since you say Rev Kev's interpretation of this as 3/6x^2 - 4x - (x - 2/9x - 6) is the correct one:

3/6x² - 4x - (x - 2/9x - 6)
Distribute the negative:
3/6x² - 4x - x + 2/9x + 6
Undistribute the x:
3/6x² + (-4 - 1 + 2/9)x + 6
Simplify:
1/2x² + 43/9x + 6
Since the factorization is not obvious, we will find the roots of this function and then use the fundamental theorem of algebra to find the factors. First we multiply everything by 2:
x² + 86/9x + 12 = 0
Subtract 12 from both sides:
x² + 86/9x = -12
Add 1849/81 to both sides (this is (43/9)²):
x² + 86/9x + 1849/81 = 877/81
Factor:
(x+43/9)²=877/81
Take the square root of both sides:
x+43/9=±√877/9
Subtract 43/9 from both sides:
x=(-43±√877)/9
Now that we have the roots of the equation, we know that (x+(43-√877)/9) and (x+(43+√877)/9 are factors of the original expression. When multiplied by the constant 1/2 (which was the fraction in front of the x² term) they will give the original equation. Therefore the factorixation is:

1/2(x + (43 - √877)/9)(x + (43 + √877)/9)

Check:

1/2(x + (43 - √877)/9)(x + (43 + √877)/9)
1/2(x² + 43/9x + √877/9x + 43/9x + 1849/81 + 43√877/81 -√877/9x - 43√877/81 - 877/81)
1/2(x² + 86/9x + 972/81)
1/2(x² + 86/9x + 12)
1/2x² + 43/9x + 6

Which you will note is the original expression.

2006-10-16 13:35:53 · answer #2 · answered by Pascal 7 · 0 0

Am I right in assuming that you mean:
3/6x^2 - 4x - (x^2 - 2/9x - 6)?

I ask because that is what I would expect from an algebra problem. Yours could very well be different, and my assumption is wrong.

First, I suggest getting rid of those fractions. Let's multiply everything by 18, shall we? There is no law that says we must do this, but it'll help us out. In essence, we are factoring out 1/18. So, we get:
1/18 [9x^2 - 72x - (18x^2 - 4x - 108)] =
1/18 [9x^2 - 72x - 18x^2 + 4x + 108)] =
1/18 (-9x^2 - 68x +108)

That's better. Let's factor out -1. So, you get:
-1/18 (9x^2 + 68x - 108)

This does not foil nicely, so my assumption may be wrong. Hopefully, those steps will help you in getting to where you need to be. In ugly situations like those, I usually get rid of the fractions. You can multiply the -1/18 back in whenever you need to.

Edit: Aha, I totally did not see that as the difference between two fractions at first. Please indulge me and verify for a second time that my original assumption was correct. If it is not, please let us know. Parentheses are very important in algebra. Without knowing where exactly you are in your class, I had to go with one assumption that you were simply missing a letter. The assumption that it's a difference of fractions makes sense, as well, and it tells me how far along you are in your class.

If it is indeed a difference of fractions then the solutions you received of (-2x^2 + 4x + 9) / (6x)(3x - 2) are quite correct. Personally, I would factor out the -1 to get -(2x^2 - 4x - 9) / (6x)(3x - 2) but that's only to have the numerator in an easy-to-use format. I'll confirm with the others that this does not have integer roots, so this cannot be factored beyond this.

At least you have learned a valuable lesson on the importance of parentheses. This would, incidentally, be ideally written as:
[3/(6x^2 - 4x)] – [(x – 2)/(9x – 6)].
Since multiplication and division happen before addition, you don't need the brackets. It does tell the reader that, without a shadow of doubt, you are subtracting two fractions. Most math-savvy folks could read this without the brackets, though. But definitely put parentheses when you need the addition/subtraction done first. There is a huge difference between (x – 2)/(9x – 6) and x – 2/9x – 6.

2006-10-16 13:05:54 · answer #3 · answered by Rev Kev 5 · 0 0

You didn't clarify where things are grouped with parentheses, so we are going to have to guess a little. What makes the most sense is for this to be the subtraction of two fractions. What you replied to RevKev doesn't make sense... I don't see where the additional x² fits in... I'm going to assume the following, like others seem to have done.

Is this correct?
3 / (6x² - 4x) - (x-2) / (9x - 6)

Factor the denominators:
3 / (2x)(3x - 2) - (x-2) / 3(3x - 2)

You have a common 3x-2. You need to multiply the first fraction by 3, and the second equation by 2x to get the same denominator of (6x)(3x - 2)

9 / (6x)(3x - 2) - 2x(x-2) / (6x)(3x - 2)

Now add the numerators to make it a single fraction:
(9 - 2x(x-2)) / (6x)(3x - 2)

Expand out the numerator:
(-2x² + 4x + 9) / (6x)(3x - 2)

That's about as much as you can do to simplify it. I tried factoring the numerator, but it doesn't come up with integer roots.

2006-10-16 13:14:42 · answer #4 · answered by Puzzling 7 · 0 0

(3/(6x^2 - 4x)) - (x-2)/(9x-6)=
(3/(2x(3x-2)) - (x-2)/(3(3x-2)) =
(9/(6x(3x-2)) - 2x(x-2)/(6x(3x-2)) =
(9 - 2x^2 + 2x) / (6x(3x-2))

2006-10-16 13:11:30 · answer #5 · answered by Anonymous · 0 0

3/[6x^2-4x]-x/[9x-6]
={3[9x-6]-2[6x^2-4x]}/[6x^2-4x][9x-6]
={27x-18-12x^2+8x}/2x[3x-2][3][3x-2]
={-18+35x-12x^2}/6x[3x-2]^2
={-18+9x+24x-12x^2}/6x[3x-2]^2
={-9(2-x)+12x(2-x)}/6x[3x-2]^2
=[2-x][12x-9]/6x[3x-2]^2
=3[2-x][4x-3]/6x[3x-2]^2
=[2-x][4x-3]/2x[3x-2]^2

2006-10-16 13:42:46 · answer #6 · answered by openpsychy 6 · 0 0

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