Factorise Question (1 Viewer)

[ejo]

Hey could anyone help me factorise the following:

8x^2 - 30xy + 27y^2

Thanks it would be a great help if you factorised it for me.

 

[Aplus]

(4X - 9Y)(2X - 3Y)

I hope you improve by the time the HSC comes around.

 

[ejo]

I know how to factorise it by looking at it, I was just hoping that there was an easier way then trial and error.

 

[watatank]

quadratic formula.

 

[undalay]

(4X - 9Y)(2X - 3Y)

I hope you improve by the time the HSC comes around.

fkuapluswhybadickfor

 

[YannY]

Hey could anyone help me factorise the following:

8x^2 - 30xy + 27y^2

Thanks it would be a great help if you factorised it for me.

Yes there is a way to factorise this... And when we did it wasnt by trial and error mate.

Okay do you know how to factorise 8x^2-30x+27?

This is factorised by the cross method which you should be familiar with. Or other methods i.e quadratic or changing it to a monic factorisation but never mind that now. When factorised it should be (4x - 9)(2x - 3)

Lets expand this to make sure. (4x - 9)(2x - 3) = 8x^2-12x-18x+27 = 8x^2-30x+27.

Now your question is 8x^2 - 30xy + 27y^2 which only adds a y on the second term and y^2 on the last term. When they do this you just add a y to the second term in the brackets of (4x - 9)(2x - 3). which gives you
(4x - 9y)(2x - 3y).

If you dont get this dont worry. Its not really needed in the hsc that is factorising a quadratic with two variables.

 

[culic]

Yes there is a way to factorise this... And when we did it wasnt by trial and error mate.

Okay do you know how to factorise 8x^2-30x+27?

This is factorised by the cross method which you should be familiar with. Or other methods i.e quadratic or changing it to a monic factorisation but never mind that now. When factorised it should be (4x - 9)(2x - 3)

Lets expand this to make sure. (4x - 9)(2x - 3) = 8x^2-12x-18x+27 = 8x^2-30x+27.

Now your question is 8x^2 - 30xy + 27y^2 which only adds a y on the second term and y^2 on the last term. When they do this you just add a y to the second term in the brackets of (4x - 9)(2x - 3). which gives you
(4x - 9y)(2x - 3y).

If you dont get this dont worry. Its not really needed in the hsc that is factorising a quadratic with two variables.

:O :O :O.... *lost* please explain..in more simpler terms..?

 

[tommykins]

if the coefficients are above 20ish, use the quadratic formula.

 

[ejo]

Okay thanks for all your replies.