Is this a typo or am I missing something? (polynomial equations)

  • Context: Undergrad 
  • Thread starter Thread starter bigmike94
  • Start date Start date
Click For Summary
SUMMARY

The discussion centers on a confusion regarding the expression of a polynomial in terms of three other polynomials, specifically the term "3z" versus "z" in the context of constants with t^0. The participant identifies a discrepancy between their answer, which includes "5x+z," and the provided solution, which states "5x+3z." Despite using a second method that omits "3z" yet yields the same solution, the participant remains perplexed about the origin of the "3" in the first equality. Ultimately, the conclusion is that the book contains an error, and the participant can proceed without concern for missing information.

PREREQUISITES
  • Understanding of polynomial expressions and their components
  • Familiarity with the concept of t^0 in polynomial terms
  • Basic knowledge of algebraic manipulation
  • Experience with verifying polynomial identities
NEXT STEPS
  • Review polynomial identities and their properties
  • Study methods for expressing polynomials in terms of other polynomials
  • Learn about common errors in polynomial equations and how to identify them
  • Explore alternative methods for solving polynomial equations
USEFUL FOR

Students and educators in algebra, particularly those dealing with polynomial equations, as well as anyone involved in mathematical problem-solving and verification processes.

bigmike94
Messages
99
Reaction score
61
TL;DR
i cannot figure out where the 3z comes from
The question is asking to express a polynomial in terms of 3 other given polynomials, which in this case should be very straight forward but i am having issues, with the end terms on the second line they have grouped the constants that have a t^0. They have put 5x+3z, my answer says 5x+z, because they're the only terms with t^0 attached. Every other part of the answer matches mine apart from that one term, which ultimately changes the whole final answer. what am I missing?

1668597132994.png
 
Physics news on Phys.org
The solution presented uses 3z, so I would assume that there is a 3 missing in the last term of the first equality.
 
bigmike94 said:
TL;DR Summary: i cannot figure out where the 3z comes from

The question is asking to express a polynomial in terms of 3 other given polynomials, which in this case should be very straight forward but i am having issues, with the end terms on the second line they have grouped the constants that have a t^0. They have put 5x+3z, my answer says 5x+z, because they're the only terms with t^0 attached. Every other part of the answer matches mine apart from that one term, which ultimately changes the whole final answer. what am I missing?

View attachment 317242

i will add that they used a second method which acknowledges the fact that there is no 3z, yet it has the same solution as using 3z, very confused

1668598063678.png
 
DrClaude said:
The solution presented uses 3z, so I would assume that there is a 3 missing in the last term of the first equality.
But where would that come from, this is the question and i cannot see anywhere where that term comes from, i have also posted their second method that doesn't use the term 3z yet has the same solution

1668598160596.png
 
bigmike94 said:
i will add that they used a second method which acknowledges the fact that there is no 3z, yet it has the same solution as using 3z, very confused

View attachment 317243
You can check that setting ##x=-3## and ##z=4## doesn't solve ##-3 = 5x+z##.

bigmike94 said:
But where would that come from, this is the question and i cannot see anywhere where that term comes from, i have also posted their second method that doesn't use the term 3z yet has the same solution

View attachment 317244
Then the book needs to be revised.
 
thank you for the help.

So can i move on with no worries about missing anything?
 
bigmike94 said:
thank you for the help.

So can i move on with no worries about missing anything?
I guess so. Good for you to see that there was a problem there!
 
  • Like
Likes   Reactions: bigmike94

Similar threads

  • · Replies 0 ·
Replies
0
Views
2K
  • · Replies 2 ·
Replies
2
Views
1K
  • · Replies 8 ·
Replies
8
Views
2K
  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 26 ·
Replies
26
Views
966
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 18 ·
Replies
18
Views
5K
  • · Replies 4 ·
Replies
4
Views
1K
  • · Replies 3 ·
Replies
3
Views
3K