In polynomial division, the remainder's degree is always less than that of the divisor, but when dividing...
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In polynomial division, the remainder's degree is always less than that of the divisor, but when dividing $x^3+y^3$ by $x+5$, it isn't
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$begingroup$
I'm just a 9th grader trying to self-study, so if the question sounds silly to you please excuse me.
$P(x) = x^3 + y^3$ is divided by something like $g(x) = x + 5;$ the degree of the polynomial is $3$. Doesn't the degree of the remainder always be less than that of the divisor. Can someone explain what is happening?
algebra-precalculus polynomials
edited Mar 25 at 9:00
Blue
49.7k870158
asked Feb 20 at 7:11
Aditya BharadwajAditya Bharadwaj
1261
$endgroup$
add a comment |
$begingroup$
I'm just a 9th grader trying to self-study, so if the question sounds silly to you please excuse me.
$P(x) = x^3 + y^3$ is divided by something like $g(x) = x + 5;$ the degree of the polynomial is $3$. Doesn't the degree of the remainder always be less than that of the divisor. Can someone explain what is happening?
algebra-precalculus polynomials
edited Mar 25 at 9:00
Blue
49.7k870158
asked Feb 20 at 7:11
Aditya BharadwajAditya Bharadwaj
1261
$endgroup$
1
$begingroup$
I believe the problem is simply that you're dealing with a polynomial in two variables, $x,y$ for $P$. That the remainder has a lesser degree than the divisor only really holds in the case of polynomials in one variable, I believe.
$endgroup$
– Eevee Trainer
Feb 20 at 7:17
2
$begingroup$
To rewrite my previous (now deleted) comment because I'm dumb and it's been a while since I've dealt with this. Basically I noticed you have $P(x)$, but $P(x,y)$ would be more appropriate if $y$ was a variable. If $y$ is not a variable, i.e. some constant, then your remainder (which I got to be $y^3 - 125$) would be of degree zero as intended in the one-variable case. Of course if we have multiple variables a lot of nice stuff we have simply flies out the window. Such is math.
$endgroup$
– Eevee Trainer
Feb 20 at 7:23
3
$begingroup$
The degree of that result viewed as a polynomial in x is less than 3.
$endgroup$
– William Elliot
Feb 20 at 7:53
$begingroup$
So, y is not a variable but just a constant that we do not know the value of.Since y is a constant its degree will be 0 and not 3.
$endgroup$
– Aditya Bharadwaj
Feb 20 at 8:12
$begingroup$
If you wish to learn how to extend the division algorithm to multivariate polynomails then search on "grobner basis".
$endgroup$
– Bill Dubuque
Feb 20 at 21:52
add a comment |
$begingroup$
I'm just a 9th grader trying to self-study, so if the question sounds silly to you please excuse me.
$P(x) = x^3 + y^3$ is divided by something like $g(x) = x + 5;$ the degree of the polynomial is $3$. Doesn't the degree of the remainder always be less than that of the divisor. Can someone explain what is happening?
algebra-precalculus polynomials
edited Mar 25 at 9:00
Blue
49.7k870158
asked Feb 20 at 7:11
Aditya BharadwajAditya Bharadwaj
1261
$endgroup$
I'm just a 9th grader trying to self-study, so if the question sounds silly to you please excuse me.
$P(x) = x^3 + y^3$ is divided by something like $g(x) = x + 5;$ the degree of the polynomial is $3$. Doesn't the degree of the remainder always be less than that of the divisor. Can someone explain what is happening?
algebra-precalculus polynomials
algebra-precalculus polynomials
edited Mar 25 at 9:00
Blue
49.7k870158
asked Feb 20 at 7:11
Aditya BharadwajAditya Bharadwaj
1261
edited Mar 25 at 9:00
Blue
49.7k870158
asked Feb 20 at 7:11
Aditya BharadwajAditya Bharadwaj
1261
edited Mar 25 at 9:00
Blue
49.7k870158
edited Mar 25 at 9:00
Blue
49.7k870158
edited Mar 25 at 9:00
Blue
49.7k870158
49.7k870158
asked Feb 20 at 7:11
Aditya BharadwajAditya Bharadwaj
1261
asked Feb 20 at 7:11
Aditya BharadwajAditya Bharadwaj
1261
asked Feb 20 at 7:11
Aditya BharadwajAditya Bharadwaj
1261
1261
1
$begingroup$
I believe the problem is simply that you're dealing with a polynomial in two variables, $x,y$ for $P$. That the remainder has a lesser degree than the divisor only really holds in the case of polynomials in one variable, I believe.
$endgroup$
– Eevee Trainer
Feb 20 at 7:17
2
$begingroup$
To rewrite my previous (now deleted) comment because I'm dumb and it's been a while since I've dealt with this. Basically I noticed you have $P(x)$, but $P(x,y)$ would be more appropriate if $y$ was a variable. If $y$ is not a variable, i.e. some constant, then your remainder (which I got to be $y^3 - 125$) would be of degree zero as intended in the one-variable case. Of course if we have multiple variables a lot of nice stuff we have simply flies out the window. Such is math.
$endgroup$
– Eevee Trainer
Feb 20 at 7:23
3
$begingroup$
The degree of that result viewed as a polynomial in x is less than 3.
$endgroup$
– William Elliot
Feb 20 at 7:53
$begingroup$
So, y is not a variable but just a constant that we do not know the value of.Since y is a constant its degree will be 0 and not 3.
$endgroup$
– Aditya Bharadwaj
Feb 20 at 8:12
$begingroup$
If you wish to learn how to extend the division algorithm to multivariate polynomails then search on "grobner basis".
$endgroup$
– Bill Dubuque
Feb 20 at 21:52
add a comment |
1
$begingroup$
I believe the problem is simply that you're dealing with a polynomial in two variables, $x,y$ for $P$. That the remainder has a lesser degree than the divisor only really holds in the case of polynomials in one variable, I believe.
$endgroup$
– Eevee Trainer
Feb 20 at 7:17
2
$begingroup$
To rewrite my previous (now deleted) comment because I'm dumb and it's been a while since I've dealt with this. Basically I noticed you have $P(x)$, but $P(x,y)$ would be more appropriate if $y$ was a variable. If $y$ is not a variable, i.e. some constant, then your remainder (which I got to be $y^3 - 125$) would be of degree zero as intended in the one-variable case. Of course if we have multiple variables a lot of nice stuff we have simply flies out the window. Such is math.
$endgroup$
– Eevee Trainer
Feb 20 at 7:23
3
$begingroup$
The degree of that result viewed as a polynomial in x is less than 3.
$endgroup$
– William Elliot
Feb 20 at 7:53
$begingroup$
So, y is not a variable but just a constant that we do not know the value of.Since y is a constant its degree will be 0 and not 3.
$endgroup$
– Aditya Bharadwaj
Feb 20 at 8:12
$begingroup$
If you wish to learn how to extend the division algorithm to multivariate polynomails then search on "grobner basis".
$endgroup$
– Bill Dubuque
Feb 20 at 21:52
1
1
$begingroup$
I believe the problem is simply that you're dealing with a polynomial in two variables, $x,y$ for $P$. That the remainder has a lesser degree than the divisor only really holds in the case of polynomials in one variable, I believe.
$endgroup$
– Eevee Trainer
Feb 20 at 7:17
$begingroup$
I believe the problem is simply that you're dealing with a polynomial in two variables, $x,y$ for $P$. That the remainder has a lesser degree than the divisor only really holds in the case of polynomials in one variable, I believe.
$endgroup$
– Eevee Trainer
Feb 20 at 7:17
2
2
$begingroup$
To rewrite my previous (now deleted) comment because I'm dumb and it's been a while since I've dealt with this. Basically I noticed you have $P(x)$, but $P(x,y)$ would be more appropriate if $y$ was a variable. If $y$ is not a variable, i.e. some constant, then your remainder (which I got to be $y^3 - 125$) would be of degree zero as intended in the one-variable case. Of course if we have multiple variables a lot of nice stuff we have simply flies out the window. Such is math.
$endgroup$
– Eevee Trainer
Feb 20 at 7:23
$begingroup$
To rewrite my previous (now deleted) comment because I'm dumb and it's been a while since I've dealt with this. Basically I noticed you have $P(x)$, but $P(x,y)$ would be more appropriate if $y$ was a variable. If $y$ is not a variable, i.e. some constant, then your remainder (which I got to be $y^3 - 125$) would be of degree zero as intended in the one-variable case. Of course if we have multiple variables a lot of nice stuff we have simply flies out the window. Such is math.
$endgroup$
– Eevee Trainer
Feb 20 at 7:23
3
3
$begingroup$
The degree of that result viewed as a polynomial in x is less than 3.
$endgroup$
– William Elliot
Feb 20 at 7:53
$begingroup$
The degree of that result viewed as a polynomial in x is less than 3.
$endgroup$
– William Elliot
Feb 20 at 7:53
$begingroup$
So, y is not a variable but just a constant that we do not know the value of.Since y is a constant its degree will be 0 and not 3.
$endgroup$
– Aditya Bharadwaj
Feb 20 at 8:12
$begingroup$
So, y is not a variable but just a constant that we do not know the value of.Since y is a constant its degree will be 0 and not 3.
$endgroup$
– Aditya Bharadwaj
Feb 20 at 8:12
$begingroup$
If you wish to learn how to extend the division algorithm to multivariate polynomails then search on "grobner basis".
$endgroup$
– Bill Dubuque
Feb 20 at 21:52
$begingroup$
If you wish to learn how to extend the division algorithm to multivariate polynomails then search on "grobner basis".
$endgroup$
– Bill Dubuque
Feb 20 at 21:52
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
As clarified by the OP in the comments that $y$ is a constant. I would rewrite $y$ as $a$ so that it looks more constant-like for the sake of convenience.
Now $P(x)=x^3+a^3$ and $g(x)=x+5$. By Euclid's Division Lemma for Polynomials you do have $text{deg} r(x)=0$ or $text{deg }r(x)lttext{deg }g(x)$.
You may confirm this by long division method which gives you the following result: $$dfrac{x^3+a^3}{x+5}=x^2+5x-25+dfrac{125+a^3}{x+5}$$ or $r(x)=125+a^3$ and $q(x)=x^2+5x-25$ which clearly satisfies $text{deg }g(x)=1gttext{deg }r(x)=0$. So there is no discrepancy.
answered Mar 25 at 7:53
Paras KhoslaParas Khosla
3,2851627
$endgroup$
add a comment |
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$begingroup$
As clarified by the OP in the comments that $y$ is a constant. I would rewrite $y$ as $a$ so that it looks more constant-like for the sake of convenience.
Now $P(x)=x^3+a^3$ and $g(x)=x+5$. By Euclid's Division Lemma for Polynomials you do have $text{deg} r(x)=0$ or $text{deg }r(x)lttext{deg }g(x)$.
You may confirm this by long division method which gives you the following result: $$dfrac{x^3+a^3}{x+5}=x^2+5x-25+dfrac{125+a^3}{x+5}$$ or $r(x)=125+a^3$ and $q(x)=x^2+5x-25$ which clearly satisfies $text{deg }g(x)=1gttext{deg }r(x)=0$. So there is no discrepancy.
answered Mar 25 at 7:53
Paras KhoslaParas Khosla
3,2851627
$endgroup$
add a comment |
$begingroup$
As clarified by the OP in the comments that $y$ is a constant. I would rewrite $y$ as $a$ so that it looks more constant-like for the sake of convenience.
Now $P(x)=x^3+a^3$ and $g(x)=x+5$. By Euclid's Division Lemma for Polynomials you do have $text{deg} r(x)=0$ or $text{deg }r(x)lttext{deg }g(x)$.
You may confirm this by long division method which gives you the following result: $$dfrac{x^3+a^3}{x+5}=x^2+5x-25+dfrac{125+a^3}{x+5}$$ or $r(x)=125+a^3$ and $q(x)=x^2+5x-25$ which clearly satisfies $text{deg }g(x)=1gttext{deg }r(x)=0$. So there is no discrepancy.
answered Mar 25 at 7:53
Paras KhoslaParas Khosla
3,2851627
$endgroup$
add a comment |
$begingroup$
As clarified by the OP in the comments that $y$ is a constant. I would rewrite $y$ as $a$ so that it looks more constant-like for the sake of convenience.
Now $P(x)=x^3+a^3$ and $g(x)=x+5$. By Euclid's Division Lemma for Polynomials you do have $text{deg} r(x)=0$ or $text{deg }r(x)lttext{deg }g(x)$.
You may confirm this by long division method which gives you the following result: $$dfrac{x^3+a^3}{x+5}=x^2+5x-25+dfrac{125+a^3}{x+5}$$ or $r(x)=125+a^3$ and $q(x)=x^2+5x-25$ which clearly satisfies $text{deg }g(x)=1gttext{deg }r(x)=0$. So there is no discrepancy.
answered Mar 25 at 7:53
Paras KhoslaParas Khosla
3,2851627
$endgroup$
As clarified by the OP in the comments that $y$ is a constant. I would rewrite $y$ as $a$ so that it looks more constant-like for the sake of convenience.
Now $P(x)=x^3+a^3$ and $g(x)=x+5$. By Euclid's Division Lemma for Polynomials you do have $text{deg} r(x)=0$ or $text{deg }r(x)lttext{deg }g(x)$.
You may confirm this by long division method which gives you the following result: $$dfrac{x^3+a^3}{x+5}=x^2+5x-25+dfrac{125+a^3}{x+5}$$ or $r(x)=125+a^3$ and $q(x)=x^2+5x-25$ which clearly satisfies $text{deg }g(x)=1gttext{deg }r(x)=0$. So there is no discrepancy.
answered Mar 25 at 7:53
Paras KhoslaParas Khosla
3,2851627
answered Mar 25 at 7:53
Paras KhoslaParas Khosla
3,2851627
answered Mar 25 at 7:53
Paras KhoslaParas Khosla
3,2851627
answered Mar 25 at 7:53
Paras KhoslaParas Khosla
3,2851627
3,2851627
add a comment |
add a comment |
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$begingroup$
I believe the problem is simply that you're dealing with a polynomial in two variables, $x,y$ for $P$. That the remainder has a lesser degree than the divisor only really holds in the case of polynomials in one variable, I believe.
$endgroup$
– Eevee Trainer
Feb 20 at 7:17
2
$begingroup$
To rewrite my previous (now deleted) comment because I'm dumb and it's been a while since I've dealt with this. Basically I noticed you have $P(x)$, but $P(x,y)$ would be more appropriate if $y$ was a variable. If $y$ is not a variable, i.e. some constant, then your remainder (which I got to be $y^3 - 125$) would be of degree zero as intended in the one-variable case. Of course if we have multiple variables a lot of nice stuff we have simply flies out the window. Such is math.
$endgroup$
– Eevee Trainer
Feb 20 at 7:23
3
$begingroup$
The degree of that result viewed as a polynomial in x is less than 3.
$endgroup$
– William Elliot
Feb 20 at 7:53
$begingroup$
So, y is not a variable but just a constant that we do not know the value of.Since y is a constant its degree will be 0 and not 3.
$endgroup$
– Aditya Bharadwaj
Feb 20 at 8:12
$begingroup$
If you wish to learn how to extend the division algorithm to multivariate polynomails then search on "grobner basis".
$endgroup$
– Bill Dubuque
Feb 20 at 21:52