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vortish
 3 years ago
subtract the polynomials
(2x^4y^3+4x^3y^4)(5x^4y^36x^3y^4)
vortish
 3 years ago
subtract the polynomials (2x^4y^3+4x^3y^4)(5x^4y^36x^3y^4)

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asnaseer
 3 years ago
Best ResponseYou've already chosen the best response.1which aspect is troubling you here?

asnaseer
 3 years ago
Best ResponseYou've already chosen the best response.1ok, first step is to remove the braces. do you know how to remove the braces? especially the braces with the minus sign outside of it.

asnaseer
 3 years ago
Best ResponseYou've already chosen the best response.1ok  the rule here is that any expression within braces remains "asis" when the braces are removed unless the braces have a minus sign outside of them. In that case all the signs inside the braces change  i.e. all ""'s change to "+" and all "+"'s change to ""

asnaseer
 3 years ago
Best ResponseYou've already chosen the best response.1e.g.: (1+45) + (63)  (2+41) = 1+45 + 63 24+1

asnaseer
 3 years ago
Best ResponseYou've already chosen the best response.1does that make sense so far?

asnaseer
 3 years ago
Best ResponseYou've already chosen the best response.1good, so now try to use these rules to remove the braces in your expression

vortish
 3 years ago
Best ResponseYou've already chosen the best response.0\[2x ^{4}y ^{3}+4x ^{3}y ^{4}5x ^{4}y ^{3}+6x ^{3}y ^{4}\]

asnaseer
 3 years ago
Best ResponseYou've already chosen the best response.1perfect! now you just need to combine "like" terms

asnaseer
 3 years ago
Best ResponseYou've already chosen the best response.1do you know what "like" terms are?

asnaseer
 3 years ago
Best ResponseYou've already chosen the best response.1no  that is not quite right

asnaseer
 3 years ago
Best ResponseYou've already chosen the best response.1the way to think about this might be to imagine each unique combination of products of powers of x and y as a separate entity. e.g. in \(3x^3y^2+2x^2y^3x^3y^2\) we have two unique combinations, namely: \(x^3y^2\) and \(x^2y^3\)

asnaseer
 3 years ago
Best ResponseYou've already chosen the best response.1think of each of these combinations as a new "variable". so, in the example I gave, let \(A=x^3y^2\) and \(B=x^2y^3\). we can then write my expression as: \(3x^3y^2+2x^2y^3x^3y^2=3A+2BA=2A+2B\) make sense?

asnaseer
 3 years ago
Best ResponseYou've already chosen the best response.1is it one specific part here that is unclear or the whole thing?

asnaseer
 3 years ago
Best ResponseYou've already chosen the best response.1ok  so you understand how I picked out the two unique combinations of powers of x and y in my initial expression  correct?

vortish
 3 years ago
Best ResponseYou've already chosen the best response.0that is where i am having the issue

asnaseer
 3 years ago
Best ResponseYou've already chosen the best response.1ok  lets take a look in more detail. we have the expression: \(3x^3y^2+2x^2y^3x^3y^2\) do you agree that this contains terms that involve multiples of \(x^3y^2\) and \(x^2y^3\) only? i.e. there are no other "combinations" of powers of x and y within my expression.

asnaseer
 3 years ago
Best ResponseYou've already chosen the best response.1now, the next step I took was to replace all occurrences of \(x^3y^2\) by a new variable that I called \(A\). this leads to: \(3x^3y^2+2x^2y^3x^3y^2=3A+2x^2y^3A\) make sense so far?

vortish
 3 years ago
Best ResponseYou've already chosen the best response.0ok i think that make sense

asnaseer
 3 years ago
Best ResponseYou've already chosen the best response.1good  the next step was to tackel the remaining unique combination which was \(x^2y^3\)  I replaced all occurrences of this with another new variable that I called \(B\). this leads to: \(3x^3y^2+2x^2y^3x^3y^2=3A+2x^2y^3A=3A+2BA\) ok so far?

vortish
 3 years ago
Best ResponseYou've already chosen the best response.0so i would come up with 3x^4y^3 +7x^3y^4

asnaseer
 3 years ago
Best ResponseYou've already chosen the best response.1good  now we just simplify our new expression by combining all terms involving \(A\) and separately all terms involving \(B\). this leads to: \(3x^3y^2+2x^2y^3x^3y^2=3A+2x^2y^3A=3A+2BA\) = \(3AA+2B\) = \(2A+2B\)

asnaseer
 3 years ago
Best ResponseYou've already chosen the best response.1your expression after removing the braces was:\[2x ^{4}y ^{3}+4x ^{3}y ^{4}5x ^{4}y ^{3}+6x ^{3}y ^{4}\] so if you first replace \(x^4y^3\) by \(A\) and \(x^3y^4\) by \(B\) then you would get:\[2x ^{4}y ^{3}+4x ^{3}y ^{4}5x ^{4}y ^{3}+6x ^{3}y ^{4}=2A+4B5A+6B\] what do you think that simplifies to in terms of A and B?

asnaseer
 3 years ago
Best ResponseYou've already chosen the best response.1correct  now just replace A and B by what they represent

asnaseer
 3 years ago
Best ResponseYou've already chosen the best response.1the "trick" is to recognise these "unique" combinations of terms  that is what is meant by "like" terms in an expression.

vortish
 3 years ago
Best ResponseYou've already chosen the best response.0got it right thanks asnaseer

asnaseer
 3 years ago
Best ResponseYou've already chosen the best response.1it is similar to saying: 2*apples + 3*pears  5*apples = 3*apples + 3*pears i.e. you cannot combine "apples" and "pears"

asnaseer
 3 years ago
Best ResponseYou've already chosen the best response.1ok  glad you got it now
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