Maths Help : Algebra 'Obtaining factors using the grouping method' (1 Viewer)

[BLIT2014]

Could you please explain what the grouping method is too?



Please provide necessary working out if needed too


Thank you...

Q 11.
3x²-9xy+3y²


Q 13. x(a-b)+ y(b-a)

 

[Kurosaki]

Is that supposed to be 3x^2 in q11?

 

[BLIT2014]

Question is now correct / do able

 

[kazemagic]

Could you please explain what the grouping method is too?



Please provide necessary working out if needed too


Thank you...

Q 11.
3x²-9xy+3y²


Q 13. x(a-b)+ y(b-a)

Q13.
(b-a) = -(a-b)
Therfore equation can now be x(a-b) - y(a-b)
Factorising (a-b)
and you should get (x-y)(a-b)

 

[stanton_liu]

Could you please explain what the grouping method is too?



Please provide necessary working out if needed too


Thank you...

Q 11.
3x²-9xy+3y²


Q 13. x(a-b)+ y(b-a)

Question 13: x(a-b)+ y(b-a) = x(a-b) - y(a-b) , by taking out the negative of (b-a)
= (x-y) (a-b) , using the grouping method

So the grouping method is basically, as follows:
If you're given a(c+d) + b(c+d) , it can be "grouped" (factorised is a better word) into (a+b)(c+d), so that is
a(c+d) + b(c+d) = (a+b)(c+d)

If you don't believe that result, just expand both sides.

As for question 11, does it have to be fully factorised in terms of x and y in each bracket or just taking out the factor of 3?
If it's the former, we get an answer with a squroot (xy) term. :S

 

[stanton_liu]

Q13.
(b-a) = -(a-b)
Therfore equation can now be x(a-b) - y(a-b)
Factorising (a-b)
and you should get (x-y)(a-b)

Dammit got in quicker than me. I blame my laggy computer, which made my browser freeze for 5 minutes

 

[girlworld_club]

the cute days when math was easy *sigh*

:-(