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Just trying to make sure I understand this correctly... (the step in the red box).

 

 

So if I'm understanding this correctly, bx^2 - 2x^2 = bx^2 (as per the general polynomial expression). Therefore as these 2 expressions can be equated, you can match up the bx^2 values of both sides to equal each other?

 

This can also be done for any other variable too, and as c = -2c on the LHS and +30 on the RHS, c = -15.

 

The book I'm going through has never explained this... hence my first road bump heh. Just wanted to make sure I understood this correctly.

 

Thanks!

 Jul 8, 2016
edited by Guest  Jul 8, 2016
edited by Guest  Jul 8, 2016
edited by Guest  Jul 8, 2016
edited by Guest  Jul 8, 2016
 #1
avatar+118608 
+1

Hi Guest,

I answered this question in some detail yesterday.

I do understand that you are just asking for clarification but it would be polite for you to comment and to say thanks on my original response.

It would also be better for you to become a member because it is much easier to track your questions.  (They are automatically put on your watch list) and you can get email alerts if someone answers you.

PLUS we get to know you and members, especially conscientious and polite members, get more attention from answerers.  I usually answer member questions before I answer other questions :)

 

Ok now the lecture is over.

 

Yes, when you have one polynomial = another polynomial

the coefficients of like term must be equal.

so for instance, if

 

\(a_1x^7 +a_2x^6 +a_3x^5 +a_4x^4 +a_5x^3 +a_6x^2 +a_7x^1 +a_8= b_1x^7 +b_2x^6 +b_3x^5 +b_4x^4 +b_5x^3 +b_6x^2 +b_7x^1 +b_8\\ then\\ a_1=b_1, \quad a_2=b_2, \quad a_3=b_3, \quad a_4=b_4, \quad a_5=b_5, \quad a_6=b_6, \quad a_7=b_7, \quad a_8=b_8 \quad \)

 

When I answered your question before I did not expand the whole algebraic multiplication.

I did not need to because I was only interestest in the coefficients of the x^2 terms.

 

So yes I think you have a full understanding of how this works :)

 Jul 8, 2016
 #2
avatar+108 
+5

Thanks heaps!

 

Sorry I didn't mean to be rude, just wasn't sure if you would see a comment that I posted on the other thread :D.

 

I definately appreciate your help!

 

I'll make an account for next time I have a problem :p. 

 Jul 8, 2016
 #3
avatar+118608 
0

Hi Kreyn,

Welcome to the Web2.0calc forum :))

 

Yes, you are right, old questions and responses to them can get lost very easily here.  I think I do see most responses - I have learned a few tricks I guess, but i admit that I would miss some.  :(

 

If you respond and you are not sure I will see it then send me a private message and include the address ofthe question so I can just  hyperlink to it.

 Jul 8, 2016

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