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# Factorisation question????????????? Watch

1. how do you factorise : 6a^2 x-abx-12b^2 x
2. I'm not a 100% sure but this is what I would do:

You have: y = 6a^2x - abx - 12b^2x
First thing I would do is take x out:
y = x (6a^2 - ab - 12b^2)

Then you can take two approaches, either factorize it regularily, by taking (a - b) (a - b) and respectfully changing the signs and the number or you could use the formula:

(-b (plus/minus)(square root of b^2 - 4ac)) / 2a = x

In this case you have the variables: a = 6, b = -1 and c = -12, as seen in your equation above.

If you plug in the values you will get:

x = 1 (plus/minus) (root of: (-1)^2 - 4*6*(-12)) / 2*6
x = 1 (plus/minus) (root of: 1 + 288) / 12
x = 1 (plus/minus) (root of: 289) / 12

you can find the root of 289, which is 17 and put it in:

x = 1 (plus/minus) 17 / 12

From this you will get two values:

x = (1 + 17)/ 12 and x = (1 - 17)/ 12

Thus you will get two values for x, being x = 3/2 and x = -4/3

Thus;

y = x (a -3/2b) (a + 4/3b)

or, more neatly:

y = x (2a -3b) (3a + 4b)

If you want to check it would give:

y = x ((2a)(3a)+(2a)(4b)+(-3b)(3a)+(-3b)(4b))
y = x (6a^2 + 8ab - 9ab -12b^2)
y = x (6a^2 - ab -12b^2)

And

y = 6a^2x - abx -12b^2x

I hope this helped.
3. (Original post by emmalav)
how do you factorise : 6a^2 x-abx-12b^2 x
Take the x out and you are left with

Are you familiar with the method 6 times -12 = -72

Then -9 x 8 = -72
And -9 x 8 = -1

So you have

x() = x() = x
4. ^^ cant just take out x, the x is still there, need to write it in
5. (Original post by TenOfThem)
Of course it is still there

I do not believe that I suggested otherwise

I showed a method not an answer
obviously i can see that you knew it but dont leave it out yeh?
6. (Original post by Blake-inator)
...
(Original post by Mr M)
...
Happy now?

7. (Original post by TenOfThem)
Happy now?

For now but make sure you don't leave it out next time yeh?
8. (Original post by Mr M)
For now but make sure you don't leave it out next time yeh?
yeh

I is lurnt me lesson innit
9. (Original post by TenOfThem)
yeh

I is lurnt me lesson innit
leave it out guv!
10. (Original post by Dornishman)
I'm not a 100% sure but this is what I would do:

You have: y = 6a^2x - abx - 12b^2x
First thing I would do is take x out:
y = x (6a^2 - ab - 12b^2)

Then you can take two approaches, either factorize it regularily, by taking (a - b) (a - b) and respectfully changing the signs and the number or you could use the formula:

(-b (plus/minus)(square root of b^2 - 4ac)) / 2a = x

In this case you have the variables: a = 6, b = -1 and c = -12, as seen in your equation above.

If you plug in the values you will get:

x = 1 (plus/minus) (root of: (-1)^2 - 4*6*(-12)) / 2*6
x = 1 (plus/minus) (root of: 1 + 288) / 12
x = 1 (plus/minus) (root of: 289) / 12

you can find the root of 289, which is 17 and put it in:

x = 1 (plus/minus) 17 / 12

From this you will get two values:

x = (1 + 17)/ 12 and x = (1 - 17)/ 12

Thus you will get two values for x, being x = 3/2 and x = -4/3

Thus;

y = x (a -3/2b) (a + 4/3b)

or, more neatly:

y = x (2a -3b) (3a + 4b)

If you want to check it would give:

y = x ((2a)(3a)+(2a)(4b)+(-3b)(3a)+(-3b)(4b))
y = x (6a^2 + 8ab - 9ab -12b^2)
y = x (6a^2 - ab -12b^2)

And

y = 6a^2x - abx -12b^2x

I hope this helped.
thank you

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Updated: April 14, 2013
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