Factoring.....I"m confused

[math38]

How do you factor this:
2x^2 + 9xy-35y^2


Please show me how to setup the problem too...

 

[Denis]

How do you factor this: 2x^2 + 9xy-35y^2

Pretend the "y" is not there:
2x^2 + 9x - 35
Factor that: (2x - 5)(x + 7)
Put the "y" back in:
(2x - 5y)(x + 7y)

That's MY way of doing it; find it less confusing.

 

[chrisr]

Or....

2x[sup:c7gywsjx]2[/sup:c7gywsjx]+9xy-35y[sup:c7gywsjx]2[/sup:c7gywsjx] = y[sup:c7gywsjx]2[/sup:c7gywsjx]{2(x[sup:c7gywsjx]2[/sup:c7gywsjx]/y[sup:c7gywsjx]2[/sup:c7gywsjx]) +9(x/y) -35}
to form a quadratic in x/y

=y[sup:c7gywsjx]2[/sup:c7gywsjx](2a[sup:c7gywsjx]2[/sup:c7gywsjx]+9a-35} where a=x/y

=y[sup:c7gywsjx]2[/sup:c7gywsjx](2a-5)(a+7) = y(2a-5)y(a+7)

=(2ay-5y)(ay+7y) = (2x-5y)(x+7y)

 

[Mrspi]

Or....

2x[sup:37l3snbv]2[/sup:37l3snbv]+9xy-35y[sup:37l3snbv]2[/sup:37l3snbv] = y[sup:37l3snbv]2[/sup:37l3snbv]{2(x[sup:37l3snbv]2[/sup:37l3snbv]/y[sup:37l3snbv]2[/sup:37l3snbv]) +9(x/y) -35}
to form a quadratic in x/y

=y[sup:37l3snbv]2[/sup:37l3snbv](2a[sup:37l3snbv]2[/sup:37l3snbv]+9a-35} where a=x/y

=y[sup:37l3snbv]2[/sup:37l3snbv](2a-5)(a+7) = y(2a-5)y(a+7)

=(2ay-5y)(ay+5y) = (2x-5y)(x+7y)

I fail to see why on earth anyone would choose this approach.

 

[chrisr]

The question was "how do you factor this".

There are ways to do so.

The x[sup:uhl3vasg]2[/sup:uhl3vasg] can also be taken outside, that's another way to get the same result.