Factorisation question?????????????

how do you factorise : 6a^2 x-abx-12b^2 x

Of course it is still there

I do not believe that I suggested otherwise

I showed a method not an answer

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Happy now?

For now but make sure you don't leave it out next time yeh?

yeh

I is lurnt me lesson innit

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.