## vortish 2 years ago subtract the polynomials (2x^4y^3+4x^3y^4)-(5x^4y^3-6x^3y^4)

1. asnaseer

which aspect is troubling you here?

2. vortish

all of it

3. asnaseer

ok, 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.

4. vortish

no

5. asnaseer

ok - the rule here is that any expression within braces remains "as-is" 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 "-"

6. asnaseer

e.g.: (1+4-5) + (6-3) - (2+4-1) = 1+4-5 + 6-3 -2-4+1

7. asnaseer

does that make sense so far?

8. vortish

yes

9. asnaseer

good, so now try to use these rules to remove the braces in your expression

10. vortish

\[2x ^{4}y ^{3}+4x ^{3}y ^{4}-5x ^{4}y ^{3}+6x ^{3}y ^{4}\]

11. asnaseer

perfect! now you just need to combine "like" terms

12. asnaseer

do you know what "like" terms are?

13. vortish

7x^14y^14

14. asnaseer

no - that is not quite right

15. vortish

fudge

16. asnaseer

the 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 \(3x^3y^2+2x^2y^3-x^3y^2\) we have two unique combinations, namely: \(x^3y^2\) and \(x^2y^3\)

17. asnaseer

think 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: \(3x^3y^2+2x^2y^3-x^3y^2=3A+2B-A=2A+2B\) make sense?

18. vortish

no

19. asnaseer

is it one specific part here that is unclear or the whole thing?

20. vortish

the last explanation

21. asnaseer

ok - so you understand how I picked out the two unique combinations of powers of x and y in my initial expression - correct?

22. vortish

that is where i am having the issue

23. asnaseer

ok - lets take a look in more detail. we have the expression: \(3x^3y^2+2x^2y^3-x^3y^2\) 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.

24. vortish

yes

25. asnaseer

now, 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: \(3x^3y^2+2x^2y^3-x^3y^2=3A+2x^2y^3-A\) make sense so far?

26. vortish

ok i think that make sense

27. asnaseer

good - 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: \(3x^3y^2+2x^2y^3-x^3y^2=3A+2x^2y^3-A=3A+2B-A\) ok so far?

28. vortish

yes

29. vortish

so i would come up with -3x^4y^3 +7x^3y^4

30. asnaseer

good - now we just simplify our new expression by combining all terms involving \(A\) and separately all terms involving \(B\). this leads to: \(3x^3y^2+2x^2y^3-x^3y^2=3A+2x^2y^3-A=3A+2B-A\) = \(3A-A+2B\) = \(2A+2B\)

31. vortish

is t hat right

32. asnaseer

your 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:\[2x ^{4}y ^{3}+4x ^{3}y ^{4}-5x ^{4}y ^{3}+6x ^{3}y ^{4}=2A+4B-5A+6B\] what do you think that simplifies to in terms of A and B?

33. vortish

-3A+10B

34. asnaseer

correct - now just replace A and B by what they represent

35. vortish

ok that makes sense

36. asnaseer

the "trick" is to recognise these "unique" combinations of terms - that is what is meant by "like" terms in an expression.

37. vortish

got it right thanks asnaseer

38. asnaseer

it is similar to saying: 2*apples + 3*pears - 5*apples = -3*apples + 3*pears i.e. you cannot combine "apples" and "pears"

39. asnaseer

ok - glad you got it now