Mixing
Polynomial Long Division Confusion (simplifying $frac{x^{5}}{x^{2}+1}$)
$begingroup$
I need to simplify begin{equation}
frac{x^{5}}{x^{2}+1}
end{equation}
by long division in order to solve an integral.
However, I keep getting an infinite series:
begin{equation}
x^{3}+x+frac{1}{x}-frac{1}{x^{3}}+...
end{equation}
algebra-precalculus polynomials
edited Jan 9 at 19:02
robjohn♦
267k27308631
asked Jan 9 at 7:38
MrDerDartMrDerDart
485
$endgroup$
add a comment |
$begingroup$
I need to simplify begin{equation}
frac{x^{5}}{x^{2}+1}
end{equation}
by long division in order to solve an integral.
However, I keep getting an infinite series:
begin{equation}
x^{3}+x+frac{1}{x}-frac{1}{x^{3}}+...
end{equation}
algebra-precalculus polynomials
edited Jan 9 at 19:02
robjohn♦
267k27308631
asked Jan 9 at 7:38
MrDerDartMrDerDart
485
$endgroup$
$begingroup$
In polynomial long division, we need to divide until the degree of remainder became less than degree of divisor, if we continue it further then quotient will no longer be a polynomial.
$endgroup$
– user629353
Jan 9 at 8:39
add a comment |
$begingroup$
I need to simplify begin{equation}
frac{x^{5}}{x^{2}+1}
end{equation}
by long division in order to solve an integral.
However, I keep getting an infinite series:
begin{equation}
x^{3}+x+frac{1}{x}-frac{1}{x^{3}}+...
end{equation}
algebra-precalculus polynomials
edited Jan 9 at 19:02
robjohn♦
267k27308631
asked Jan 9 at 7:38
MrDerDartMrDerDart
485
$endgroup$
I need to simplify begin{equation}
frac{x^{5}}{x^{2}+1}
end{equation}
by long division in order to solve an integral.
However, I keep getting an infinite series:
begin{equation}
x^{3}+x+frac{1}{x}-frac{1}{x^{3}}+...
end{equation}
algebra-precalculus polynomials
algebra-precalculus polynomials
edited Jan 9 at 19:02
robjohn♦
267k27308631
asked Jan 9 at 7:38
MrDerDartMrDerDart
485
edited Jan 9 at 19:02
robjohn♦
267k27308631
asked Jan 9 at 7:38
MrDerDartMrDerDart
485
edited Jan 9 at 19:02
robjohn♦
267k27308631
edited Jan 9 at 19:02
robjohn♦
267k27308631
edited Jan 9 at 19:02
robjohn♦
267k27308631
267k27308631
asked Jan 9 at 7:38
MrDerDartMrDerDart
485
asked Jan 9 at 7:38
MrDerDartMrDerDart
485
asked Jan 9 at 7:38
MrDerDartMrDerDart
485
485
$begingroup$
In polynomial long division, we need to divide until the degree of remainder became less than degree of divisor, if we continue it further then quotient will no longer be a polynomial.
$endgroup$
– user629353
Jan 9 at 8:39
add a comment |
$begingroup$
In polynomial long division, we need to divide until the degree of remainder became less than degree of divisor, if we continue it further then quotient will no longer be a polynomial.
$endgroup$
– user629353
Jan 9 at 8:39
$begingroup$
In polynomial long division, we need to divide until the degree of remainder became less than degree of divisor, if we continue it further then quotient will no longer be a polynomial.
$endgroup$
– user629353
Jan 9 at 8:39
$begingroup$
In polynomial long division, we need to divide until the degree of remainder became less than degree of divisor, if we continue it further then quotient will no longer be a polynomial.
$endgroup$
– user629353
Jan 9 at 8:39
add a comment |
6 Answers
6
active
oldest
votes
$begingroup$
When dividing one polynomial by another, you expect a polynomial quotient. For example, when you divide $x^3+x+1$ by $x+1$,
First multiply $x+1$ by $x^2$ and subtract the result, $x^3+x^2$, from $x^3+x+1$, giving the remainder $-x^2+x+1$;
Now, multiply $x+1$ by $-x$ and subtract the result, $-x^2-x$, from $-x^2+x+1$, giving the remainder $2x+1$;
As a final step, multiply $x+1$ by $2$ and subtract the result, $2x+2$, from $2x+1$, giving you the remainder $-1$. Thus,$$x^3+x+1=(x+1)(x^2-x+2)-1$$
Note that you don't go beyond the last step, because the degree of the remainder, $2,(0)$ is smaller than the degree of the divisor, $x+1, (1)$. To go any further, you will have to multiply $x+1$ by terms containing negative powers of $x$, which would mean the quotient will no longer be a polynomial in $x$. Moreover, there is no guarantee that this procedure will terminate, as you have seen.
The essence of performing long-division to solve the integral $$intfrac{x^5}{x^2+1}dx$$ is to express $x^5=(x^2+1)Q(x)+R(x)$, where $Q(x),R(x)$ are polynomials in $x$ with degree of $R(x)<2$, which is the degree of the divisor, $x^2+1$.
This gives $$frac{x^5}{x^2+1}=Q(x)+frac{R(x)}{x^2+1}$$where $Q(x)$ can be easily integrated because it is a polynomial, and $displaystylefrac{R(x)}{x^2+1}$ can be integrated using partial fractions or similar techniques.
As other answers have pointed out, $$intfrac{x^5}{x^2+1}dx=int x^3-x+frac x{x^2+1}dx=frac{x^4}4-frac{x^2}2+frac12ln(x^2+1)+C$$
answered Jan 9 at 8:07
Shubham JohriShubham Johri
5,122717
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add a comment |
$begingroup$
The idea of polynomial division is like integer division. With integer division of $frac nd$, we want integer $q,r$ so that $n=qd+r$ and $rlt d$. With polynomial division of $frac nd$, we want polynomial $q,r$ so that $n=qd+r$ and $deg(r)lt deg(d)$.
$$
require{enclose}
begin{array}{rl}
&phantom{),}color{#C00}{x^3}color{#090}{-x}\[-4pt]
x^2+1!!!!!&enclose{longdiv}{x^5qquad}\[-4pt]
&phantom{),}underline{color{#C00}{x^5+x^3}}\[-2pt]
&phantom{),x^5}{}-x^3\[-4pt]
&phantom{),x^5}underline{color{#090}{{}-x^3-x}}\[-4pt]
&phantom{),x^5{}-x^3-{}}x\[-4pt]
end{array}
$$
So we get a quotient of $x^3-x$ and a remainder of $x$, which allows us to write both
$$
overbrace{quad,x^5quad,}^n=overbrace{left(x^3-xright)}^qoverbrace{left(x^2+1right)}^d+overbrace{vphantom{x^5}quad;xquad;}^r
$$
and
$$
frac{x^5}{x^2+1}=x^3-x+frac{x}{x^2+1}
$$
answered Jan 9 at 8:26
robjohn♦robjohn
267k27308631
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add a comment |
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Why wouldn't you get an infinite series?
If $x^2 + 1$ doesn't divide evenly into $x^5$ (which it doesn't) you will get a remainder. Just like with numbers with remainders if you try to continue you will get a decimal. Here if try to divide into the remainder you will get an expression with a negative power.
The thing is when you get a remainder that can be divided into any further.... you stop. And you let it be simply a remainder.
$frac {x^5}{x^2+ 1} = frac {x^5 + x^3}{x^2 + 1} -frac {x^3}{x^2 + 1}$
$= x^3 - frac {x^3 + x}{x^2 + 1} + frac {x}{x^2+ 1} =$
$x^3 - x + frac {x}{x^2 + 1}$.
Now we can't divide any further as the degree of the denominator ($x^2 + 1$) is $2$ and that is larger than the degree of that numerator ($x$). So we are done.
$frac {x^5}{x^2 +1} = x^3 - x +frac {x}{x^2 +1}$.
Put another way: $x^5 = (x^3 - x)(x^2 + 1) + x$.
$x$ is .... just a remainder you cant do any thing with.
It is exactly like.
$frac {249}{7} = frac {210 + 39}{7} = frac {210}7 + frac {39}7=$
$30 + frac {35 + 4}{7} = 30 + frac {35}7 + frac 47=$
$30 + 5 + frac 47 = 35frac 47$.
We've divided as far as we can go.
If you tried to go further we would get decimals:
$30 + 5 + frac {40}{7*10} = 30 + 5 + {35 + 5}{70} = $
$30 + 5 + frac 5{10} + frac 5{70} =30 + 5 + frac 5{10} + frac {50}{700} =$
$30 + 5 + frac 5{10} + frac 7{100} + frac 1{1000} + ......$.
$= 35.571428571428571428571428571429.....$
But we weren't asked to go that for and as we aren't masochists.... we stopped at $35frac 47$.
$
answered Jan 9 at 8:19
fleabloodfleablood
69.7k22685
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add a comment |
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$frac {x^{5}} {x^{2}+1}= x^{3}-x+frac x {x^{2}+1}$.
answered Jan 9 at 7:42
Kavi Rama MurthyKavi Rama Murthy
57.5k42160
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add a comment |
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Since $x^5 = left(x^2 + 1right)left(x^3 - xright) + x$, we have that
$$cfrac{x^5}{x^2 + 1} = x^3 - x + cfrac{x}{x^2 + 1} tag{1}label{eq1} $$
To do this in general, first note that $x^2$ divides into $x^5$ a total of $x^3$ times. However, this gives $x^3left(x^2 + 1right) = x^5 + x^3$, so it's too big by $x^3$. As such, you need to subtract an appropriate value, with it being $x$ here due to $x times x^2 = x^3$. However, $xleft(x^2 + 1right) = x^3 + x$, so subtracting this means that you are now $x$ too small, so you need to add that $x$ back. However, as the degree of $x$ is only $1$, which is less than the degree of $2$ in $x^2 + 1$, you leave that final $x$ as the remainder. This, in total, gives eqref{eq1}.
answered Jan 9 at 7:42
John OmielanJohn Omielan
1,961210
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If the integral is well defined, you can write $${x^5over 1+x^2}dx={1over 2}{(x^2)^2over 1+x^2}dx^2={1over 2}{u^2over 1+u}du$$
answered Jan 9 at 8:11
Mostafa AyazMostafa Ayaz
15.4k3939
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6 Answers
6
active
oldest
votes
6 Answers
6
active
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active
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active
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$begingroup$
When dividing one polynomial by another, you expect a polynomial quotient. For example, when you divide $x^3+x+1$ by $x+1$,
First multiply $x+1$ by $x^2$ and subtract the result, $x^3+x^2$, from $x^3+x+1$, giving the remainder $-x^2+x+1$;
Now, multiply $x+1$ by $-x$ and subtract the result, $-x^2-x$, from $-x^2+x+1$, giving the remainder $2x+1$;
As a final step, multiply $x+1$ by $2$ and subtract the result, $2x+2$, from $2x+1$, giving you the remainder $-1$. Thus,$$x^3+x+1=(x+1)(x^2-x+2)-1$$
Note that you don't go beyond the last step, because the degree of the remainder, $2,(0)$ is smaller than the degree of the divisor, $x+1, (1)$. To go any further, you will have to multiply $x+1$ by terms containing negative powers of $x$, which would mean the quotient will no longer be a polynomial in $x$. Moreover, there is no guarantee that this procedure will terminate, as you have seen.
The essence of performing long-division to solve the integral $$intfrac{x^5}{x^2+1}dx$$ is to express $x^5=(x^2+1)Q(x)+R(x)$, where $Q(x),R(x)$ are polynomials in $x$ with degree of $R(x)<2$, which is the degree of the divisor, $x^2+1$.
This gives $$frac{x^5}{x^2+1}=Q(x)+frac{R(x)}{x^2+1}$$where $Q(x)$ can be easily integrated because it is a polynomial, and $displaystylefrac{R(x)}{x^2+1}$ can be integrated using partial fractions or similar techniques.
As other answers have pointed out, $$intfrac{x^5}{x^2+1}dx=int x^3-x+frac x{x^2+1}dx=frac{x^4}4-frac{x^2}2+frac12ln(x^2+1)+C$$
answered Jan 9 at 8:07
Shubham JohriShubham Johri
5,122717
$endgroup$
add a comment |
$begingroup$
When dividing one polynomial by another, you expect a polynomial quotient. For example, when you divide $x^3+x+1$ by $x+1$,
First multiply $x+1$ by $x^2$ and subtract the result, $x^3+x^2$, from $x^3+x+1$, giving the remainder $-x^2+x+1$;
Now, multiply $x+1$ by $-x$ and subtract the result, $-x^2-x$, from $-x^2+x+1$, giving the remainder $2x+1$;
As a final step, multiply $x+1$ by $2$ and subtract the result, $2x+2$, from $2x+1$, giving you the remainder $-1$. Thus,$$x^3+x+1=(x+1)(x^2-x+2)-1$$
Note that you don't go beyond the last step, because the degree of the remainder, $2,(0)$ is smaller than the degree of the divisor, $x+1, (1)$. To go any further, you will have to multiply $x+1$ by terms containing negative powers of $x$, which would mean the quotient will no longer be a polynomial in $x$. Moreover, there is no guarantee that this procedure will terminate, as you have seen.
The essence of performing long-division to solve the integral $$intfrac{x^5}{x^2+1}dx$$ is to express $x^5=(x^2+1)Q(x)+R(x)$, where $Q(x),R(x)$ are polynomials in $x$ with degree of $R(x)<2$, which is the degree of the divisor, $x^2+1$.
This gives $$frac{x^5}{x^2+1}=Q(x)+frac{R(x)}{x^2+1}$$where $Q(x)$ can be easily integrated because it is a polynomial, and $displaystylefrac{R(x)}{x^2+1}$ can be integrated using partial fractions or similar techniques.
As other answers have pointed out, $$intfrac{x^5}{x^2+1}dx=int x^3-x+frac x{x^2+1}dx=frac{x^4}4-frac{x^2}2+frac12ln(x^2+1)+C$$
answered Jan 9 at 8:07
Shubham JohriShubham Johri
5,122717
$endgroup$
add a comment |
$begingroup$
When dividing one polynomial by another, you expect a polynomial quotient. For example, when you divide $x^3+x+1$ by $x+1$,
First multiply $x+1$ by $x^2$ and subtract the result, $x^3+x^2$, from $x^3+x+1$, giving the remainder $-x^2+x+1$;
Now, multiply $x+1$ by $-x$ and subtract the result, $-x^2-x$, from $-x^2+x+1$, giving the remainder $2x+1$;
As a final step, multiply $x+1$ by $2$ and subtract the result, $2x+2$, from $2x+1$, giving you the remainder $-1$. Thus,$$x^3+x+1=(x+1)(x^2-x+2)-1$$
Note that you don't go beyond the last step, because the degree of the remainder, $2,(0)$ is smaller than the degree of the divisor, $x+1, (1)$. To go any further, you will have to multiply $x+1$ by terms containing negative powers of $x$, which would mean the quotient will no longer be a polynomial in $x$. Moreover, there is no guarantee that this procedure will terminate, as you have seen.
The essence of performing long-division to solve the integral $$intfrac{x^5}{x^2+1}dx$$ is to express $x^5=(x^2+1)Q(x)+R(x)$, where $Q(x),R(x)$ are polynomials in $x$ with degree of $R(x)<2$, which is the degree of the divisor, $x^2+1$.
This gives $$frac{x^5}{x^2+1}=Q(x)+frac{R(x)}{x^2+1}$$where $Q(x)$ can be easily integrated because it is a polynomial, and $displaystylefrac{R(x)}{x^2+1}$ can be integrated using partial fractions or similar techniques.
As other answers have pointed out, $$intfrac{x^5}{x^2+1}dx=int x^3-x+frac x{x^2+1}dx=frac{x^4}4-frac{x^2}2+frac12ln(x^2+1)+C$$
answered Jan 9 at 8:07
Shubham JohriShubham Johri
5,122717
$endgroup$
When dividing one polynomial by another, you expect a polynomial quotient. For example, when you divide $x^3+x+1$ by $x+1$,
First multiply $x+1$ by $x^2$ and subtract the result, $x^3+x^2$, from $x^3+x+1$, giving the remainder $-x^2+x+1$;
Now, multiply $x+1$ by $-x$ and subtract the result, $-x^2-x$, from $-x^2+x+1$, giving the remainder $2x+1$;
As a final step, multiply $x+1$ by $2$ and subtract the result, $2x+2$, from $2x+1$, giving you the remainder $-1$. Thus,$$x^3+x+1=(x+1)(x^2-x+2)-1$$
Note that you don't go beyond the last step, because the degree of the remainder, $2,(0)$ is smaller than the degree of the divisor, $x+1, (1)$. To go any further, you will have to multiply $x+1$ by terms containing negative powers of $x$, which would mean the quotient will no longer be a polynomial in $x$. Moreover, there is no guarantee that this procedure will terminate, as you have seen.
The essence of performing long-division to solve the integral $$intfrac{x^5}{x^2+1}dx$$ is to express $x^5=(x^2+1)Q(x)+R(x)$, where $Q(x),R(x)$ are polynomials in $x$ with degree of $R(x)<2$, which is the degree of the divisor, $x^2+1$.
This gives $$frac{x^5}{x^2+1}=Q(x)+frac{R(x)}{x^2+1}$$where $Q(x)$ can be easily integrated because it is a polynomial, and $displaystylefrac{R(x)}{x^2+1}$ can be integrated using partial fractions or similar techniques.
As other answers have pointed out, $$intfrac{x^5}{x^2+1}dx=int x^3-x+frac x{x^2+1}dx=frac{x^4}4-frac{x^2}2+frac12ln(x^2+1)+C$$
answered Jan 9 at 8:07
Shubham JohriShubham Johri
5,122717
answered Jan 9 at 8:07
Shubham JohriShubham Johri
5,122717
answered Jan 9 at 8:07
Shubham JohriShubham Johri
5,122717
answered Jan 9 at 8:07
Shubham JohriShubham Johri
5,122717
5,122717
add a comment |
add a comment |
$begingroup$
The idea of polynomial division is like integer division. With integer division of $frac nd$, we want integer $q,r$ so that $n=qd+r$ and $rlt d$. With polynomial division of $frac nd$, we want polynomial $q,r$ so that $n=qd+r$ and $deg(r)lt deg(d)$.
$$
require{enclose}
begin{array}{rl}
&phantom{),}color{#C00}{x^3}color{#090}{-x}\[-4pt]
x^2+1!!!!!&enclose{longdiv}{x^5qquad}\[-4pt]
&phantom{),}underline{color{#C00}{x^5+x^3}}\[-2pt]
&phantom{),x^5}{}-x^3\[-4pt]
&phantom{),x^5}underline{color{#090}{{}-x^3-x}}\[-4pt]
&phantom{),x^5{}-x^3-{}}x\[-4pt]
end{array}
$$
So we get a quotient of $x^3-x$ and a remainder of $x$, which allows us to write both
$$
overbrace{quad,x^5quad,}^n=overbrace{left(x^3-xright)}^qoverbrace{left(x^2+1right)}^d+overbrace{vphantom{x^5}quad;xquad;}^r
$$
and
$$
frac{x^5}{x^2+1}=x^3-x+frac{x}{x^2+1}
$$
answered Jan 9 at 8:26
robjohn♦robjohn
267k27308631
$endgroup$
add a comment |
$begingroup$
The idea of polynomial division is like integer division. With integer division of $frac nd$, we want integer $q,r$ so that $n=qd+r$ and $rlt d$. With polynomial division of $frac nd$, we want polynomial $q,r$ so that $n=qd+r$ and $deg(r)lt deg(d)$.
$$
require{enclose}
begin{array}{rl}
&phantom{),}color{#C00}{x^3}color{#090}{-x}\[-4pt]
x^2+1!!!!!&enclose{longdiv}{x^5qquad}\[-4pt]
&phantom{),}underline{color{#C00}{x^5+x^3}}\[-2pt]
&phantom{),x^5}{}-x^3\[-4pt]
&phantom{),x^5}underline{color{#090}{{}-x^3-x}}\[-4pt]
&phantom{),x^5{}-x^3-{}}x\[-4pt]
end{array}
$$
So we get a quotient of $x^3-x$ and a remainder of $x$, which allows us to write both
$$
overbrace{quad,x^5quad,}^n=overbrace{left(x^3-xright)}^qoverbrace{left(x^2+1right)}^d+overbrace{vphantom{x^5}quad;xquad;}^r
$$
and
$$
frac{x^5}{x^2+1}=x^3-x+frac{x}{x^2+1}
$$
answered Jan 9 at 8:26
robjohn♦robjohn
267k27308631
$endgroup$
add a comment |
$begingroup$
The idea of polynomial division is like integer division. With integer division of $frac nd$, we want integer $q,r$ so that $n=qd+r$ and $rlt d$. With polynomial division of $frac nd$, we want polynomial $q,r$ so that $n=qd+r$ and $deg(r)lt deg(d)$.
$$
require{enclose}
begin{array}{rl}
&phantom{),}color{#C00}{x^3}color{#090}{-x}\[-4pt]
x^2+1!!!!!&enclose{longdiv}{x^5qquad}\[-4pt]
&phantom{),}underline{color{#C00}{x^5+x^3}}\[-2pt]
&phantom{),x^5}{}-x^3\[-4pt]
&phantom{),x^5}underline{color{#090}{{}-x^3-x}}\[-4pt]
&phantom{),x^5{}-x^3-{}}x\[-4pt]
end{array}
$$
So we get a quotient of $x^3-x$ and a remainder of $x$, which allows us to write both
$$
overbrace{quad,x^5quad,}^n=overbrace{left(x^3-xright)}^qoverbrace{left(x^2+1right)}^d+overbrace{vphantom{x^5}quad;xquad;}^r
$$
and
$$
frac{x^5}{x^2+1}=x^3-x+frac{x}{x^2+1}
$$
answered Jan 9 at 8:26
robjohn♦robjohn
267k27308631
$endgroup$
The idea of polynomial division is like integer division. With integer division of $frac nd$, we want integer $q,r$ so that $n=qd+r$ and $rlt d$. With polynomial division of $frac nd$, we want polynomial $q,r$ so that $n=qd+r$ and $deg(r)lt deg(d)$.
$$
require{enclose}
begin{array}{rl}
&phantom{),}color{#C00}{x^3}color{#090}{-x}\[-4pt]
x^2+1!!!!!&enclose{longdiv}{x^5qquad}\[-4pt]
&phantom{),}underline{color{#C00}{x^5+x^3}}\[-2pt]
&phantom{),x^5}{}-x^3\[-4pt]
&phantom{),x^5}underline{color{#090}{{}-x^3-x}}\[-4pt]
&phantom{),x^5{}-x^3-{}}x\[-4pt]
end{array}
$$
So we get a quotient of $x^3-x$ and a remainder of $x$, which allows us to write both
$$
overbrace{quad,x^5quad,}^n=overbrace{left(x^3-xright)}^qoverbrace{left(x^2+1right)}^d+overbrace{vphantom{x^5}quad;xquad;}^r
$$
and
$$
frac{x^5}{x^2+1}=x^3-x+frac{x}{x^2+1}
$$
answered Jan 9 at 8:26
robjohn♦robjohn
267k27308631
edited Jan 9 at 8:48
answered Jan 9 at 8:26
robjohn♦robjohn
267k27308631
answered Jan 9 at 8:26
robjohn♦robjohn
267k27308631
answered Jan 9 at 8:26
robjohn♦robjohn
267k27308631
267k27308631
add a comment |
add a comment |
$begingroup$
Why wouldn't you get an infinite series?
If $x^2 + 1$ doesn't divide evenly into $x^5$ (which it doesn't) you will get a remainder. Just like with numbers with remainders if you try to continue you will get a decimal. Here if try to divide into the remainder you will get an expression with a negative power.
The thing is when you get a remainder that can be divided into any further.... you stop. And you let it be simply a remainder.
$frac {x^5}{x^2+ 1} = frac {x^5 + x^3}{x^2 + 1} -frac {x^3}{x^2 + 1}$
$= x^3 - frac {x^3 + x}{x^2 + 1} + frac {x}{x^2+ 1} =$
$x^3 - x + frac {x}{x^2 + 1}$.
Now we can't divide any further as the degree of the denominator ($x^2 + 1$) is $2$ and that is larger than the degree of that numerator ($x$). So we are done.
$frac {x^5}{x^2 +1} = x^3 - x +frac {x}{x^2 +1}$.
Put another way: $x^5 = (x^3 - x)(x^2 + 1) + x$.
$x$ is .... just a remainder you cant do any thing with.
It is exactly like.
$frac {249}{7} = frac {210 + 39}{7} = frac {210}7 + frac {39}7=$
$30 + frac {35 + 4}{7} = 30 + frac {35}7 + frac 47=$
$30 + 5 + frac 47 = 35frac 47$.
We've divided as far as we can go.
If you tried to go further we would get decimals:
$30 + 5 + frac {40}{7*10} = 30 + 5 + {35 + 5}{70} = $
$30 + 5 + frac 5{10} + frac 5{70} =30 + 5 + frac 5{10} + frac {50}{700} =$
$30 + 5 + frac 5{10} + frac 7{100} + frac 1{1000} + ......$.
$= 35.571428571428571428571428571429.....$
But we weren't asked to go that for and as we aren't masochists.... we stopped at $35frac 47$.
$
answered Jan 9 at 8:19
fleabloodfleablood
69.7k22685
$endgroup$
add a comment |
$begingroup$
Why wouldn't you get an infinite series?
If $x^2 + 1$ doesn't divide evenly into $x^5$ (which it doesn't) you will get a remainder. Just like with numbers with remainders if you try to continue you will get a decimal. Here if try to divide into the remainder you will get an expression with a negative power.
The thing is when you get a remainder that can be divided into any further.... you stop. And you let it be simply a remainder.
$frac {x^5}{x^2+ 1} = frac {x^5 + x^3}{x^2 + 1} -frac {x^3}{x^2 + 1}$
$= x^3 - frac {x^3 + x}{x^2 + 1} + frac {x}{x^2+ 1} =$
$x^3 - x + frac {x}{x^2 + 1}$.
Now we can't divide any further as the degree of the denominator ($x^2 + 1$) is $2$ and that is larger than the degree of that numerator ($x$). So we are done.
$frac {x^5}{x^2 +1} = x^3 - x +frac {x}{x^2 +1}$.
Put another way: $x^5 = (x^3 - x)(x^2 + 1) + x$.
$x$ is .... just a remainder you cant do any thing with.
It is exactly like.
$frac {249}{7} = frac {210 + 39}{7} = frac {210}7 + frac {39}7=$
$30 + frac {35 + 4}{7} = 30 + frac {35}7 + frac 47=$
$30 + 5 + frac 47 = 35frac 47$.
We've divided as far as we can go.
If you tried to go further we would get decimals:
$30 + 5 + frac {40}{7*10} = 30 + 5 + {35 + 5}{70} = $
$30 + 5 + frac 5{10} + frac 5{70} =30 + 5 + frac 5{10} + frac {50}{700} =$
$30 + 5 + frac 5{10} + frac 7{100} + frac 1{1000} + ......$.
$= 35.571428571428571428571428571429.....$
But we weren't asked to go that for and as we aren't masochists.... we stopped at $35frac 47$.
$
answered Jan 9 at 8:19
fleabloodfleablood
69.7k22685
$endgroup$
add a comment |
$begingroup$
Why wouldn't you get an infinite series?
If $x^2 + 1$ doesn't divide evenly into $x^5$ (which it doesn't) you will get a remainder. Just like with numbers with remainders if you try to continue you will get a decimal. Here if try to divide into the remainder you will get an expression with a negative power.
The thing is when you get a remainder that can be divided into any further.... you stop. And you let it be simply a remainder.
$frac {x^5}{x^2+ 1} = frac {x^5 + x^3}{x^2 + 1} -frac {x^3}{x^2 + 1}$
$= x^3 - frac {x^3 + x}{x^2 + 1} + frac {x}{x^2+ 1} =$
$x^3 - x + frac {x}{x^2 + 1}$.
Now we can't divide any further as the degree of the denominator ($x^2 + 1$) is $2$ and that is larger than the degree of that numerator ($x$). So we are done.
$frac {x^5}{x^2 +1} = x^3 - x +frac {x}{x^2 +1}$.
Put another way: $x^5 = (x^3 - x)(x^2 + 1) + x$.
$x$ is .... just a remainder you cant do any thing with.
It is exactly like.
$frac {249}{7} = frac {210 + 39}{7} = frac {210}7 + frac {39}7=$
$30 + frac {35 + 4}{7} = 30 + frac {35}7 + frac 47=$
$30 + 5 + frac 47 = 35frac 47$.
We've divided as far as we can go.
If you tried to go further we would get decimals:
$30 + 5 + frac {40}{7*10} = 30 + 5 + {35 + 5}{70} = $
$30 + 5 + frac 5{10} + frac 5{70} =30 + 5 + frac 5{10} + frac {50}{700} =$
$30 + 5 + frac 5{10} + frac 7{100} + frac 1{1000} + ......$.
$= 35.571428571428571428571428571429.....$
But we weren't asked to go that for and as we aren't masochists.... we stopped at $35frac 47$.
$
answered Jan 9 at 8:19
fleabloodfleablood
69.7k22685
$endgroup$
Why wouldn't you get an infinite series?
If $x^2 + 1$ doesn't divide evenly into $x^5$ (which it doesn't) you will get a remainder. Just like with numbers with remainders if you try to continue you will get a decimal. Here if try to divide into the remainder you will get an expression with a negative power.
The thing is when you get a remainder that can be divided into any further.... you stop. And you let it be simply a remainder.
$frac {x^5}{x^2+ 1} = frac {x^5 + x^3}{x^2 + 1} -frac {x^3}{x^2 + 1}$
$= x^3 - frac {x^3 + x}{x^2 + 1} + frac {x}{x^2+ 1} =$
$x^3 - x + frac {x}{x^2 + 1}$.
Now we can't divide any further as the degree of the denominator ($x^2 + 1$) is $2$ and that is larger than the degree of that numerator ($x$). So we are done.
$frac {x^5}{x^2 +1} = x^3 - x +frac {x}{x^2 +1}$.
Put another way: $x^5 = (x^3 - x)(x^2 + 1) + x$.
$x$ is .... just a remainder you cant do any thing with.
It is exactly like.
$frac {249}{7} = frac {210 + 39}{7} = frac {210}7 + frac {39}7=$
$30 + frac {35 + 4}{7} = 30 + frac {35}7 + frac 47=$
$30 + 5 + frac 47 = 35frac 47$.
We've divided as far as we can go.
If you tried to go further we would get decimals:
$30 + 5 + frac {40}{7*10} = 30 + 5 + {35 + 5}{70} = $
$30 + 5 + frac 5{10} + frac 5{70} =30 + 5 + frac 5{10} + frac {50}{700} =$
$30 + 5 + frac 5{10} + frac 7{100} + frac 1{1000} + ......$.
$= 35.571428571428571428571428571429.....$
But we weren't asked to go that for and as we aren't masochists.... we stopped at $35frac 47$.
$
answered Jan 9 at 8:19
fleabloodfleablood
69.7k22685
answered Jan 9 at 8:19
fleabloodfleablood
69.7k22685
answered Jan 9 at 8:19
fleabloodfleablood
69.7k22685
answered Jan 9 at 8:19
fleabloodfleablood
69.7k22685
69.7k22685
add a comment |
add a comment |
$begingroup$
$frac {x^{5}} {x^{2}+1}= x^{3}-x+frac x {x^{2}+1}$.
answered Jan 9 at 7:42
Kavi Rama MurthyKavi Rama Murthy
57.5k42160
$endgroup$
add a comment |
$begingroup$
$frac {x^{5}} {x^{2}+1}= x^{3}-x+frac x {x^{2}+1}$.
answered Jan 9 at 7:42
Kavi Rama MurthyKavi Rama Murthy
57.5k42160
$endgroup$
add a comment |
$begingroup$
$frac {x^{5}} {x^{2}+1}= x^{3}-x+frac x {x^{2}+1}$.
answered Jan 9 at 7:42
Kavi Rama MurthyKavi Rama Murthy
57.5k42160
$endgroup$
$frac {x^{5}} {x^{2}+1}= x^{3}-x+frac x {x^{2}+1}$.
answered Jan 9 at 7:42
Kavi Rama MurthyKavi Rama Murthy
57.5k42160
answered Jan 9 at 7:42
Kavi Rama MurthyKavi Rama Murthy
57.5k42160
answered Jan 9 at 7:42
Kavi Rama MurthyKavi Rama Murthy
57.5k42160
answered Jan 9 at 7:42
Kavi Rama MurthyKavi Rama Murthy
57.5k42160
57.5k42160
add a comment |
add a comment |
$begingroup$
Since $x^5 = left(x^2 + 1right)left(x^3 - xright) + x$, we have that
$$cfrac{x^5}{x^2 + 1} = x^3 - x + cfrac{x}{x^2 + 1} tag{1}label{eq1} $$
To do this in general, first note that $x^2$ divides into $x^5$ a total of $x^3$ times. However, this gives $x^3left(x^2 + 1right) = x^5 + x^3$, so it's too big by $x^3$. As such, you need to subtract an appropriate value, with it being $x$ here due to $x times x^2 = x^3$. However, $xleft(x^2 + 1right) = x^3 + x$, so subtracting this means that you are now $x$ too small, so you need to add that $x$ back. However, as the degree of $x$ is only $1$, which is less than the degree of $2$ in $x^2 + 1$, you leave that final $x$ as the remainder. This, in total, gives eqref{eq1}.
answered Jan 9 at 7:42
John OmielanJohn Omielan
1,961210
$endgroup$
add a comment |
$begingroup$
Since $x^5 = left(x^2 + 1right)left(x^3 - xright) + x$, we have that
$$cfrac{x^5}{x^2 + 1} = x^3 - x + cfrac{x}{x^2 + 1} tag{1}label{eq1} $$
To do this in general, first note that $x^2$ divides into $x^5$ a total of $x^3$ times. However, this gives $x^3left(x^2 + 1right) = x^5 + x^3$, so it's too big by $x^3$. As such, you need to subtract an appropriate value, with it being $x$ here due to $x times x^2 = x^3$. However, $xleft(x^2 + 1right) = x^3 + x$, so subtracting this means that you are now $x$ too small, so you need to add that $x$ back. However, as the degree of $x$ is only $1$, which is less than the degree of $2$ in $x^2 + 1$, you leave that final $x$ as the remainder. This, in total, gives eqref{eq1}.
answered Jan 9 at 7:42
John OmielanJohn Omielan
1,961210
$endgroup$
add a comment |
$begingroup$
Since $x^5 = left(x^2 + 1right)left(x^3 - xright) + x$, we have that
$$cfrac{x^5}{x^2 + 1} = x^3 - x + cfrac{x}{x^2 + 1} tag{1}label{eq1} $$
To do this in general, first note that $x^2$ divides into $x^5$ a total of $x^3$ times. However, this gives $x^3left(x^2 + 1right) = x^5 + x^3$, so it's too big by $x^3$. As such, you need to subtract an appropriate value, with it being $x$ here due to $x times x^2 = x^3$. However, $xleft(x^2 + 1right) = x^3 + x$, so subtracting this means that you are now $x$ too small, so you need to add that $x$ back. However, as the degree of $x$ is only $1$, which is less than the degree of $2$ in $x^2 + 1$, you leave that final $x$ as the remainder. This, in total, gives eqref{eq1}.
answered Jan 9 at 7:42
John OmielanJohn Omielan
1,961210
$endgroup$
Since $x^5 = left(x^2 + 1right)left(x^3 - xright) + x$, we have that
$$cfrac{x^5}{x^2 + 1} = x^3 - x + cfrac{x}{x^2 + 1} tag{1}label{eq1} $$
To do this in general, first note that $x^2$ divides into $x^5$ a total of $x^3$ times. However, this gives $x^3left(x^2 + 1right) = x^5 + x^3$, so it's too big by $x^3$. As such, you need to subtract an appropriate value, with it being $x$ here due to $x times x^2 = x^3$. However, $xleft(x^2 + 1right) = x^3 + x$, so subtracting this means that you are now $x$ too small, so you need to add that $x$ back. However, as the degree of $x$ is only $1$, which is less than the degree of $2$ in $x^2 + 1$, you leave that final $x$ as the remainder. This, in total, gives eqref{eq1}.
answered Jan 9 at 7:42
John OmielanJohn Omielan
1,961210
edited Jan 9 at 7:49
answered Jan 9 at 7:42
John OmielanJohn Omielan
1,961210
answered Jan 9 at 7:42
John OmielanJohn Omielan
1,961210
answered Jan 9 at 7:42
John OmielanJohn Omielan
1,961210
1,961210
add a comment |
add a comment |
$begingroup$
If the integral is well defined, you can write $${x^5over 1+x^2}dx={1over 2}{(x^2)^2over 1+x^2}dx^2={1over 2}{u^2over 1+u}du$$
answered Jan 9 at 8:11
Mostafa AyazMostafa Ayaz
15.4k3939
$endgroup$
add a comment |
$begingroup$
If the integral is well defined, you can write $${x^5over 1+x^2}dx={1over 2}{(x^2)^2over 1+x^2}dx^2={1over 2}{u^2over 1+u}du$$
answered Jan 9 at 8:11
Mostafa AyazMostafa Ayaz
15.4k3939
$endgroup$
add a comment |
$begingroup$
If the integral is well defined, you can write $${x^5over 1+x^2}dx={1over 2}{(x^2)^2over 1+x^2}dx^2={1over 2}{u^2over 1+u}du$$
answered Jan 9 at 8:11
Mostafa AyazMostafa Ayaz
15.4k3939
$endgroup$
If the integral is well defined, you can write $${x^5over 1+x^2}dx={1over 2}{(x^2)^2over 1+x^2}dx^2={1over 2}{u^2over 1+u}du$$
answered Jan 9 at 8:11
Mostafa AyazMostafa Ayaz
15.4k3939
answered Jan 9 at 8:11
Mostafa AyazMostafa Ayaz
15.4k3939
answered Jan 9 at 8:11
Mostafa AyazMostafa Ayaz
15.4k3939
answered Jan 9 at 8:11
Mostafa AyazMostafa Ayaz
15.4k3939
15.4k3939
add a comment |
add a comment |
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In polynomial long division, we need to divide until the degree of remainder became less than degree of divisor, if we continue it further then quotient will no longer be a polynomial.
$endgroup$
– user629353
Jan 9 at 8:39