Mixing
The number of distinct terms in a trinomial expansion
$begingroup$
$$(x^3 + frac1{x^3} +1)^{200}$$ is the given expression. How many distinct terms are in this expression when expanded. I know that there are a total of $3^{200}$ terms before combining the terms but I'm struggling to find the distinct terms. Please provide a clear explanation of the answer. Thanks in advance.
binomial-theorem
edited Jan 10 at 16:59
Shubham Johri
5,102717
asked Jan 10 at 16:58
ChaseChase
32
$endgroup$
add a comment |
$begingroup$
$$(x^3 + frac1{x^3} +1)^{200}$$ is the given expression. How many distinct terms are in this expression when expanded. I know that there are a total of $3^{200}$ terms before combining the terms but I'm struggling to find the distinct terms. Please provide a clear explanation of the answer. Thanks in advance.
binomial-theorem
edited Jan 10 at 16:59
Shubham Johri
5,102717
asked Jan 10 at 16:58
ChaseChase
32
$endgroup$
$begingroup$
By distinct you mean “different powers of x”?
$endgroup$
– b00n heT
Jan 10 at 17:04
$begingroup$
This depends on the definition of "distinct terms". I would assume you consider the term distinct, if they differ by the exponent of $x$.
$endgroup$
– user
Jan 10 at 17:06
$begingroup$
This question is almost certainly talking about how many terms will be in the expanded equation after like terms have been collected.
$endgroup$
– Rhys Hughes
Jan 10 at 17:12
$begingroup$
On one hand I had the same interpretation, on the other I feel like the knowing how many terms make up each of the coefficients is also interesting.
$endgroup$
– b00n heT
Jan 10 at 17:21
add a comment |
$begingroup$
$$(x^3 + frac1{x^3} +1)^{200}$$ is the given expression. How many distinct terms are in this expression when expanded. I know that there are a total of $3^{200}$ terms before combining the terms but I'm struggling to find the distinct terms. Please provide a clear explanation of the answer. Thanks in advance.
binomial-theorem
edited Jan 10 at 16:59
Shubham Johri
5,102717
asked Jan 10 at 16:58
ChaseChase
32
$endgroup$
$$(x^3 + frac1{x^3} +1)^{200}$$ is the given expression. How many distinct terms are in this expression when expanded. I know that there are a total of $3^{200}$ terms before combining the terms but I'm struggling to find the distinct terms. Please provide a clear explanation of the answer. Thanks in advance.
binomial-theorem
binomial-theorem
edited Jan 10 at 16:59
Shubham Johri
5,102717
asked Jan 10 at 16:58
ChaseChase
32
edited Jan 10 at 16:59
Shubham Johri
5,102717
asked Jan 10 at 16:58
ChaseChase
32
edited Jan 10 at 16:59
Shubham Johri
5,102717
edited Jan 10 at 16:59
Shubham Johri
5,102717
edited Jan 10 at 16:59
Shubham Johri
5,102717
5,102717
asked Jan 10 at 16:58
ChaseChase
32
asked Jan 10 at 16:58
ChaseChase
32
asked Jan 10 at 16:58
ChaseChase
32
32
$begingroup$
By distinct you mean “different powers of x”?
$endgroup$
– b00n heT
Jan 10 at 17:04
$begingroup$
This depends on the definition of "distinct terms". I would assume you consider the term distinct, if they differ by the exponent of $x$.
$endgroup$
– user
Jan 10 at 17:06
$begingroup$
This question is almost certainly talking about how many terms will be in the expanded equation after like terms have been collected.
$endgroup$
– Rhys Hughes
Jan 10 at 17:12
$begingroup$
On one hand I had the same interpretation, on the other I feel like the knowing how many terms make up each of the coefficients is also interesting.
$endgroup$
– b00n heT
Jan 10 at 17:21
add a comment |
$begingroup$
By distinct you mean “different powers of x”?
$endgroup$
– b00n heT
Jan 10 at 17:04
$begingroup$
This depends on the definition of "distinct terms". I would assume you consider the term distinct, if they differ by the exponent of $x$.
$endgroup$
– user
Jan 10 at 17:06
$begingroup$
This question is almost certainly talking about how many terms will be in the expanded equation after like terms have been collected.
$endgroup$
– Rhys Hughes
Jan 10 at 17:12
$begingroup$
On one hand I had the same interpretation, on the other I feel like the knowing how many terms make up each of the coefficients is also interesting.
$endgroup$
– b00n heT
Jan 10 at 17:21
$begingroup$
By distinct you mean “different powers of x”?
$endgroup$
– b00n heT
Jan 10 at 17:04
$begingroup$
By distinct you mean “different powers of x”?
$endgroup$
– b00n heT
Jan 10 at 17:04
$begingroup$
This depends on the definition of "distinct terms". I would assume you consider the term distinct, if they differ by the exponent of $x$.
$endgroup$
– user
Jan 10 at 17:06
$begingroup$
This depends on the definition of "distinct terms". I would assume you consider the term distinct, if they differ by the exponent of $x$.
$endgroup$
– user
Jan 10 at 17:06
$begingroup$
This question is almost certainly talking about how many terms will be in the expanded equation after like terms have been collected.
$endgroup$
– Rhys Hughes
Jan 10 at 17:12
$begingroup$
This question is almost certainly talking about how many terms will be in the expanded equation after like terms have been collected.
$endgroup$
– Rhys Hughes
Jan 10 at 17:12
$begingroup$
On one hand I had the same interpretation, on the other I feel like the knowing how many terms make up each of the coefficients is also interesting.
$endgroup$
– b00n heT
Jan 10 at 17:21
$begingroup$
On one hand I had the same interpretation, on the other I feel like the knowing how many terms make up each of the coefficients is also interesting.
$endgroup$
– b00n heT
Jan 10 at 17:21
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
The answer should be $401$: in the complete expansion of the polynomial above, there are no negative coefficients, there will be no cancellation, so you just need to count how many powers of $x$ appear. Substituting $x^3$ with $y$, it is clear that every power of $y$ between $y^{200}$ and $y^{-200}$ appears. Hence, the answer is $401$.
answered Jan 10 at 17:07
Leo163Leo163
1,660512
$endgroup$
$begingroup$
Sorry, but can you explain the Substituting $(x)^{3}$ with $y$ bit. Slightly confused on what is y used as here.
$endgroup$
– Chase
Jan 11 at 3:01
$begingroup$
What I was trying to say is the following: let's rewrite your expression as $(y+1+y^{-1})^{200}$, where we put $x^3=y$ (this is not necessary in any way, I just wanted to make my argument more straightforward). At this point, we count how many powers of $y$ appear, and the answer is $401$ as explained above. Now, we substitute back to $x$, which does not change the number of terms of the expression. Hence the answer is $401$.
$endgroup$
– Leo163
Jan 11 at 10:57
add a comment |
$begingroup$
I'm going to simplify the problem a bit to explain my method, then you can apply it to this one:
Suppose instead we had:
$$(x+1)^{200}$$
This is $(x+1)(x+1)(x+1)...$
Suppose we wish to find any term (from the $2^{200}$ in the expanded equation, before collecting like terms). We could do this by going through each of the brackets one at a time and selecting either $x$ or $1$. Once we go through every bracket selecting one of the terms, the product of the terms we picked out is one of the terms of the expansion.
Let's suppose we pick $x$ a total of $t$ times, and therefore we pick $1$ a total of $(200-t)$ times. Then the power, $p$, of a given term is:
$$p=(1)(t)+0(200-t)=t$$
Applying this to your equation, can you see we have:
$$p=(3)(t)+(-3)(u)+(0)=3(t-u)$$
This makes it obvious that the powers are only multiples of $3$, so you just need to figure out how many are in your range.
answered Jan 10 at 17:13
Rhys HughesRhys Hughes
5,9981529
$endgroup$
add a comment |
$begingroup$
Does this work?
You can take a product of $ k_1 $ terms in $ x^3 $, $ k_2 $ terms in $ x^{-3} $ and $ k_3 $ terms in $ x^0 $, where the same power of $ x $ can be obtained in multiple ways. In doing this we require $ k_1+k_2 +k_3 = 200 $ and $ 0 leq k_1, k_2, k_3 leq 200 $. Consolidating terms of the same power of $ x $ then results in gathering terms of the form $ x^{3(k_1-k_2)} $. This yields all distinct powers of $ x^3 $ from $ -200 $ to $+200 $, including zero, implying 401 distinct terms in the expansion. We note that none cancels since the individual coefficients are all 1.
answered Jan 10 at 17:17
WA DonWA Don
261
$endgroup$
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3068910%2fthe-number-of-distinct-terms-in-a-trinomial-expansion%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
The answer should be $401$: in the complete expansion of the polynomial above, there are no negative coefficients, there will be no cancellation, so you just need to count how many powers of $x$ appear. Substituting $x^3$ with $y$, it is clear that every power of $y$ between $y^{200}$ and $y^{-200}$ appears. Hence, the answer is $401$.
answered Jan 10 at 17:07
Leo163Leo163
1,660512
$endgroup$
$begingroup$
Sorry, but can you explain the Substituting $(x)^{3}$ with $y$ bit. Slightly confused on what is y used as here.
$endgroup$
– Chase
Jan 11 at 3:01
$begingroup$
What I was trying to say is the following: let's rewrite your expression as $(y+1+y^{-1})^{200}$, where we put $x^3=y$ (this is not necessary in any way, I just wanted to make my argument more straightforward). At this point, we count how many powers of $y$ appear, and the answer is $401$ as explained above. Now, we substitute back to $x$, which does not change the number of terms of the expression. Hence the answer is $401$.
$endgroup$
– Leo163
Jan 11 at 10:57
add a comment |
$begingroup$
The answer should be $401$: in the complete expansion of the polynomial above, there are no negative coefficients, there will be no cancellation, so you just need to count how many powers of $x$ appear. Substituting $x^3$ with $y$, it is clear that every power of $y$ between $y^{200}$ and $y^{-200}$ appears. Hence, the answer is $401$.
answered Jan 10 at 17:07
Leo163Leo163
1,660512
$endgroup$
$begingroup$
Sorry, but can you explain the Substituting $(x)^{3}$ with $y$ bit. Slightly confused on what is y used as here.
$endgroup$
– Chase
Jan 11 at 3:01
$begingroup$
What I was trying to say is the following: let's rewrite your expression as $(y+1+y^{-1})^{200}$, where we put $x^3=y$ (this is not necessary in any way, I just wanted to make my argument more straightforward). At this point, we count how many powers of $y$ appear, and the answer is $401$ as explained above. Now, we substitute back to $x$, which does not change the number of terms of the expression. Hence the answer is $401$.
$endgroup$
– Leo163
Jan 11 at 10:57
add a comment |
$begingroup$
The answer should be $401$: in the complete expansion of the polynomial above, there are no negative coefficients, there will be no cancellation, so you just need to count how many powers of $x$ appear. Substituting $x^3$ with $y$, it is clear that every power of $y$ between $y^{200}$ and $y^{-200}$ appears. Hence, the answer is $401$.
answered Jan 10 at 17:07
Leo163Leo163
1,660512
$endgroup$
The answer should be $401$: in the complete expansion of the polynomial above, there are no negative coefficients, there will be no cancellation, so you just need to count how many powers of $x$ appear. Substituting $x^3$ with $y$, it is clear that every power of $y$ between $y^{200}$ and $y^{-200}$ appears. Hence, the answer is $401$.
answered Jan 10 at 17:07
Leo163Leo163
1,660512
answered Jan 10 at 17:07
Leo163Leo163
1,660512
answered Jan 10 at 17:07
Leo163Leo163
1,660512
answered Jan 10 at 17:07
Leo163Leo163
1,660512
1,660512
$begingroup$
Sorry, but can you explain the Substituting $(x)^{3}$ with $y$ bit. Slightly confused on what is y used as here.
$endgroup$
– Chase
Jan 11 at 3:01
$begingroup$
What I was trying to say is the following: let's rewrite your expression as $(y+1+y^{-1})^{200}$, where we put $x^3=y$ (this is not necessary in any way, I just wanted to make my argument more straightforward). At this point, we count how many powers of $y$ appear, and the answer is $401$ as explained above. Now, we substitute back to $x$, which does not change the number of terms of the expression. Hence the answer is $401$.
$endgroup$
– Leo163
Jan 11 at 10:57
add a comment |
$begingroup$
Sorry, but can you explain the Substituting $(x)^{3}$ with $y$ bit. Slightly confused on what is y used as here.
$endgroup$
– Chase
Jan 11 at 3:01
$begingroup$
What I was trying to say is the following: let's rewrite your expression as $(y+1+y^{-1})^{200}$, where we put $x^3=y$ (this is not necessary in any way, I just wanted to make my argument more straightforward). At this point, we count how many powers of $y$ appear, and the answer is $401$ as explained above. Now, we substitute back to $x$, which does not change the number of terms of the expression. Hence the answer is $401$.
$endgroup$
– Leo163
Jan 11 at 10:57
$begingroup$
Sorry, but can you explain the Substituting $(x)^{3}$ with $y$ bit. Slightly confused on what is y used as here.
$endgroup$
– Chase
Jan 11 at 3:01
$begingroup$
Sorry, but can you explain the Substituting $(x)^{3}$ with $y$ bit. Slightly confused on what is y used as here.
$endgroup$
– Chase
Jan 11 at 3:01
$begingroup$
What I was trying to say is the following: let's rewrite your expression as $(y+1+y^{-1})^{200}$, where we put $x^3=y$ (this is not necessary in any way, I just wanted to make my argument more straightforward). At this point, we count how many powers of $y$ appear, and the answer is $401$ as explained above. Now, we substitute back to $x$, which does not change the number of terms of the expression. Hence the answer is $401$.
$endgroup$
– Leo163
Jan 11 at 10:57
$begingroup$
What I was trying to say is the following: let's rewrite your expression as $(y+1+y^{-1})^{200}$, where we put $x^3=y$ (this is not necessary in any way, I just wanted to make my argument more straightforward). At this point, we count how many powers of $y$ appear, and the answer is $401$ as explained above. Now, we substitute back to $x$, which does not change the number of terms of the expression. Hence the answer is $401$.
$endgroup$
– Leo163
Jan 11 at 10:57
add a comment |
$begingroup$
I'm going to simplify the problem a bit to explain my method, then you can apply it to this one:
Suppose instead we had:
$$(x+1)^{200}$$
This is $(x+1)(x+1)(x+1)...$
Suppose we wish to find any term (from the $2^{200}$ in the expanded equation, before collecting like terms). We could do this by going through each of the brackets one at a time and selecting either $x$ or $1$. Once we go through every bracket selecting one of the terms, the product of the terms we picked out is one of the terms of the expansion.
Let's suppose we pick $x$ a total of $t$ times, and therefore we pick $1$ a total of $(200-t)$ times. Then the power, $p$, of a given term is:
$$p=(1)(t)+0(200-t)=t$$
Applying this to your equation, can you see we have:
$$p=(3)(t)+(-3)(u)+(0)=3(t-u)$$
This makes it obvious that the powers are only multiples of $3$, so you just need to figure out how many are in your range.
answered Jan 10 at 17:13
Rhys HughesRhys Hughes
5,9981529
$endgroup$
add a comment |
$begingroup$
I'm going to simplify the problem a bit to explain my method, then you can apply it to this one:
Suppose instead we had:
$$(x+1)^{200}$$
This is $(x+1)(x+1)(x+1)...$
Suppose we wish to find any term (from the $2^{200}$ in the expanded equation, before collecting like terms). We could do this by going through each of the brackets one at a time and selecting either $x$ or $1$. Once we go through every bracket selecting one of the terms, the product of the terms we picked out is one of the terms of the expansion.
Let's suppose we pick $x$ a total of $t$ times, and therefore we pick $1$ a total of $(200-t)$ times. Then the power, $p$, of a given term is:
$$p=(1)(t)+0(200-t)=t$$
Applying this to your equation, can you see we have:
$$p=(3)(t)+(-3)(u)+(0)=3(t-u)$$
This makes it obvious that the powers are only multiples of $3$, so you just need to figure out how many are in your range.
answered Jan 10 at 17:13
Rhys HughesRhys Hughes
5,9981529
$endgroup$
add a comment |
$begingroup$
I'm going to simplify the problem a bit to explain my method, then you can apply it to this one:
Suppose instead we had:
$$(x+1)^{200}$$
This is $(x+1)(x+1)(x+1)...$
Suppose we wish to find any term (from the $2^{200}$ in the expanded equation, before collecting like terms). We could do this by going through each of the brackets one at a time and selecting either $x$ or $1$. Once we go through every bracket selecting one of the terms, the product of the terms we picked out is one of the terms of the expansion.
Let's suppose we pick $x$ a total of $t$ times, and therefore we pick $1$ a total of $(200-t)$ times. Then the power, $p$, of a given term is:
$$p=(1)(t)+0(200-t)=t$$
Applying this to your equation, can you see we have:
$$p=(3)(t)+(-3)(u)+(0)=3(t-u)$$
This makes it obvious that the powers are only multiples of $3$, so you just need to figure out how many are in your range.
answered Jan 10 at 17:13
Rhys HughesRhys Hughes
5,9981529
$endgroup$
I'm going to simplify the problem a bit to explain my method, then you can apply it to this one:
Suppose instead we had:
$$(x+1)^{200}$$
This is $(x+1)(x+1)(x+1)...$
Suppose we wish to find any term (from the $2^{200}$ in the expanded equation, before collecting like terms). We could do this by going through each of the brackets one at a time and selecting either $x$ or $1$. Once we go through every bracket selecting one of the terms, the product of the terms we picked out is one of the terms of the expansion.
Let's suppose we pick $x$ a total of $t$ times, and therefore we pick $1$ a total of $(200-t)$ times. Then the power, $p$, of a given term is:
$$p=(1)(t)+0(200-t)=t$$
Applying this to your equation, can you see we have:
$$p=(3)(t)+(-3)(u)+(0)=3(t-u)$$
This makes it obvious that the powers are only multiples of $3$, so you just need to figure out how many are in your range.
answered Jan 10 at 17:13
Rhys HughesRhys Hughes
5,9981529
answered Jan 10 at 17:13
Rhys HughesRhys Hughes
5,9981529
answered Jan 10 at 17:13
Rhys HughesRhys Hughes
5,9981529
answered Jan 10 at 17:13
Rhys HughesRhys Hughes
5,9981529
5,9981529
add a comment |
add a comment |
$begingroup$
Does this work?
You can take a product of $ k_1 $ terms in $ x^3 $, $ k_2 $ terms in $ x^{-3} $ and $ k_3 $ terms in $ x^0 $, where the same power of $ x $ can be obtained in multiple ways. In doing this we require $ k_1+k_2 +k_3 = 200 $ and $ 0 leq k_1, k_2, k_3 leq 200 $. Consolidating terms of the same power of $ x $ then results in gathering terms of the form $ x^{3(k_1-k_2)} $. This yields all distinct powers of $ x^3 $ from $ -200 $ to $+200 $, including zero, implying 401 distinct terms in the expansion. We note that none cancels since the individual coefficients are all 1.
answered Jan 10 at 17:17
WA DonWA Don
261
$endgroup$
add a comment |
$begingroup$
Does this work?
You can take a product of $ k_1 $ terms in $ x^3 $, $ k_2 $ terms in $ x^{-3} $ and $ k_3 $ terms in $ x^0 $, where the same power of $ x $ can be obtained in multiple ways. In doing this we require $ k_1+k_2 +k_3 = 200 $ and $ 0 leq k_1, k_2, k_3 leq 200 $. Consolidating terms of the same power of $ x $ then results in gathering terms of the form $ x^{3(k_1-k_2)} $. This yields all distinct powers of $ x^3 $ from $ -200 $ to $+200 $, including zero, implying 401 distinct terms in the expansion. We note that none cancels since the individual coefficients are all 1.
answered Jan 10 at 17:17
WA DonWA Don
261
$endgroup$
add a comment |
$begingroup$
Does this work?
You can take a product of $ k_1 $ terms in $ x^3 $, $ k_2 $ terms in $ x^{-3} $ and $ k_3 $ terms in $ x^0 $, where the same power of $ x $ can be obtained in multiple ways. In doing this we require $ k_1+k_2 +k_3 = 200 $ and $ 0 leq k_1, k_2, k_3 leq 200 $. Consolidating terms of the same power of $ x $ then results in gathering terms of the form $ x^{3(k_1-k_2)} $. This yields all distinct powers of $ x^3 $ from $ -200 $ to $+200 $, including zero, implying 401 distinct terms in the expansion. We note that none cancels since the individual coefficients are all 1.
answered Jan 10 at 17:17
WA DonWA Don
261
$endgroup$
Does this work?
You can take a product of $ k_1 $ terms in $ x^3 $, $ k_2 $ terms in $ x^{-3} $ and $ k_3 $ terms in $ x^0 $, where the same power of $ x $ can be obtained in multiple ways. In doing this we require $ k_1+k_2 +k_3 = 200 $ and $ 0 leq k_1, k_2, k_3 leq 200 $. Consolidating terms of the same power of $ x $ then results in gathering terms of the form $ x^{3(k_1-k_2)} $. This yields all distinct powers of $ x^3 $ from $ -200 $ to $+200 $, including zero, implying 401 distinct terms in the expansion. We note that none cancels since the individual coefficients are all 1.
answered Jan 10 at 17:17
WA DonWA Don
261
answered Jan 10 at 17:17
WA DonWA Don
261
answered Jan 10 at 17:17
WA DonWA Don
261
answered Jan 10 at 17:17
WA DonWA Don
261
261
add a comment |
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3068910%2fthe-number-of-distinct-terms-in-a-trinomial-expansion%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
$begingroup$
By distinct you mean “different powers of x”?
$endgroup$
– b00n heT
Jan 10 at 17:04
$begingroup$
This depends on the definition of "distinct terms". I would assume you consider the term distinct, if they differ by the exponent of $x$.
$endgroup$
– user
Jan 10 at 17:06
$begingroup$
This question is almost certainly talking about how many terms will be in the expanded equation after like terms have been collected.
$endgroup$
– Rhys Hughes
Jan 10 at 17:12
$begingroup$
On one hand I had the same interpretation, on the other I feel like the knowing how many terms make up each of the coefficients is also interesting.
$endgroup$
– b00n heT
Jan 10 at 17:21