Polynomial factorisation

Robegk

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Joined
Oct 13, 2021
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3
Greetings,

Can anyone explain to me where the -1 comes from in the answer to the attached equation? Thanks in advance...Q2IMG-20211013-WA0002.jpeg
 

lookagain

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Aug 22, 2010
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2,836
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

New member
Joined
Oct 13, 2021
Messages
3
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

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Joined
Apr 22, 2015
Messages
3,500
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.

:)
 
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