Polynomial factorisation

[Robegk]

Greetings,

Can anyone explain to me where the -1 comes from in the answer to the attached equation? Thanks in advance...Q2

 

[lookagain]

Robegk, I got this:

\(\displaystyle (x - y)^2 \ + \ 3(y - x)^3 \ = \)

\(\displaystyle (x - y)^2 \ - \ 3(x - y)^3 \ = \)

\(\displaystyle (x - y)^2[1 - 3(x - y)] \ =\)

\(\displaystyle (x - y)^2(1 - 3x + 3y) \ = \)

\(\displaystyle (x - y)^2(-3x + 3y + 1)\)

___________________________

What did you get?

 

[Robegk]

Robegk, I got this:

\(\displaystyle (x - y)^2 \ + \ 3(y - x)^3 \ = \)

\(\displaystyle (x - y)^2 \ - \ 3(x - y)^3 \ = \)

\(\displaystyle (x - y)^2[1 - 3(x - y)] \ =\)

\(\displaystyle (x - y)^2(1 - 3x + 3y) \ = \)

\(\displaystyle (x - y)^2(-3x + 3y + 1)\)

___________________________

What did you get?

Thank you for the detailed reply, can I ask you on step 3 of your workings where the 1 comes from, I can't understand it, sorry

 

[Otis]

where the 1 comes from

Hi Robegk. He factored out the expression (x-y)^2. Like this pattern:

z^2 - 3z^3 = z^2(1 - 3z)

If you're still puzzling over it, then use the distributive property to multiply out (expand) the right-hand side. Do that twice, with the 1 and without the 1, and see what happens.

 

[Robegk]

Now I understand! Thank you very much for your help people, much appreciated