Remainder of the polynomial: What is wrong with this approach?
$begingroup$
Find the remainder when $P(x)=5x^6 + x^5 - 2x^3 - x^2 + 1$ is divided
by $x^2+1$.
$$
P(x)=(x^2+1)cdot Q(x)+R(x)
$$
When $x^2=-1$,
$$
P(x)=-5+x+2x+1+1 = 3x-3
$$
Exactly, why the method above, does not work for the following question?
Find the remainder when $P(x)=x^{81}+x^{49}+x^{25}+x^9 + x$ is divided
by $x^3-x$.
$$
P(x)=(x^3-x)cdot Q(x)+R(x)=x(x+1)(x-1)cdot Q(x) + R(x)
$$
$P(1)=5$, but $P(-1)=-5$, and $P(0)=0$
However, if I factor $P(x)$ by $x$,
$$
P(x)=x(x^{80}+x^{48}+x^{24}+x^8 + 1)
$$
and consider $P(1)$ and $P(-1)$ without the factored $x$, then I get $5x$ in both cases. (The correct answer.)
Question: I can solve the question by slightly different methods where there is no confusion or ambiguity involved. I just like to know the above method does not work with this problem. (And by “exactly why” I mean to ask, how am I supposed to foresee that the above method should not be applied directly.)
polynomials
asked Dec 30 '18 at 9:21
blackenedblackened
400313
$endgroup$
add a comment |
$begingroup$
Find the remainder when $P(x)=5x^6 + x^5 - 2x^3 - x^2 + 1$ is divided
by $x^2+1$.
$$
P(x)=(x^2+1)cdot Q(x)+R(x)
$$
When $x^2=-1$,
$$
P(x)=-5+x+2x+1+1 = 3x-3
$$
Exactly, why the method above, does not work for the following question?
Find the remainder when $P(x)=x^{81}+x^{49}+x^{25}+x^9 + x$ is divided
by $x^3-x$.
$$
P(x)=(x^3-x)cdot Q(x)+R(x)=x(x+1)(x-1)cdot Q(x) + R(x)
$$
$P(1)=5$, but $P(-1)=-5$, and $P(0)=0$
However, if I factor $P(x)$ by $x$,
$$
P(x)=x(x^{80}+x^{48}+x^{24}+x^8 + 1)
$$
and consider $P(1)$ and $P(-1)$ without the factored $x$, then I get $5x$ in both cases. (The correct answer.)
Question: I can solve the question by slightly different methods where there is no confusion or ambiguity involved. I just like to know the above method does not work with this problem. (And by “exactly why” I mean to ask, how am I supposed to foresee that the above method should not be applied directly.)
polynomials
asked Dec 30 '18 at 9:21
blackenedblackened
400313
$endgroup$
1
$begingroup$
You got $R(1)=P(1)=5$, $R(-1)=P(-1)=-5$, $R(0)=P(0)=0$. Aren't these all compatible with $R(x)=5x$?? In other words, I think the problem was in the step when you tried to extract $R(x)$ from those three data points. The Chinese remainder theorem is usually called upon at that point.
$endgroup$
– Jyrki Lahtonen
Dec 30 '18 at 10:00
add a comment |
$begingroup$
Find the remainder when $P(x)=5x^6 + x^5 - 2x^3 - x^2 + 1$ is divided
by $x^2+1$.
$$
P(x)=(x^2+1)cdot Q(x)+R(x)
$$
When $x^2=-1$,
$$
P(x)=-5+x+2x+1+1 = 3x-3
$$
Exactly, why the method above, does not work for the following question?
Find the remainder when $P(x)=x^{81}+x^{49}+x^{25}+x^9 + x$ is divided
by $x^3-x$.
$$
P(x)=(x^3-x)cdot Q(x)+R(x)=x(x+1)(x-1)cdot Q(x) + R(x)
$$
$P(1)=5$, but $P(-1)=-5$, and $P(0)=0$
However, if I factor $P(x)$ by $x$,
$$
P(x)=x(x^{80}+x^{48}+x^{24}+x^8 + 1)
$$
and consider $P(1)$ and $P(-1)$ without the factored $x$, then I get $5x$ in both cases. (The correct answer.)
Question: I can solve the question by slightly different methods where there is no confusion or ambiguity involved. I just like to know the above method does not work with this problem. (And by “exactly why” I mean to ask, how am I supposed to foresee that the above method should not be applied directly.)
polynomials
asked Dec 30 '18 at 9:21
blackenedblackened
400313
$endgroup$
Find the remainder when $P(x)=5x^6 + x^5 - 2x^3 - x^2 + 1$ is divided
by $x^2+1$.
$$
P(x)=(x^2+1)cdot Q(x)+R(x)
$$
When $x^2=-1$,
$$
P(x)=-5+x+2x+1+1 = 3x-3
$$
Exactly, why the method above, does not work for the following question?
Find the remainder when $P(x)=x^{81}+x^{49}+x^{25}+x^9 + x$ is divided
by $x^3-x$.
$$
P(x)=(x^3-x)cdot Q(x)+R(x)=x(x+1)(x-1)cdot Q(x) + R(x)
$$
$P(1)=5$, but $P(-1)=-5$, and $P(0)=0$
However, if I factor $P(x)$ by $x$,
$$
P(x)=x(x^{80}+x^{48}+x^{24}+x^8 + 1)
$$
and consider $P(1)$ and $P(-1)$ without the factored $x$, then I get $5x$ in both cases. (The correct answer.)
Question: I can solve the question by slightly different methods where there is no confusion or ambiguity involved. I just like to know the above method does not work with this problem. (And by “exactly why” I mean to ask, how am I supposed to foresee that the above method should not be applied directly.)
polynomials
polynomials
asked Dec 30 '18 at 9:21
blackenedblackened
400313
asked Dec 30 '18 at 9:21
blackenedblackened
400313
edited Dec 30 '18 at 9:40
blackened
asked Dec 30 '18 at 9:21
blackenedblackened
400313
asked Dec 30 '18 at 9:21
blackenedblackened
400313
asked Dec 30 '18 at 9:21
blackenedblackened
400313
400313
1
$begingroup$
You got $R(1)=P(1)=5$, $R(-1)=P(-1)=-5$, $R(0)=P(0)=0$. Aren't these all compatible with $R(x)=5x$?? In other words, I think the problem was in the step when you tried to extract $R(x)$ from those three data points. The Chinese remainder theorem is usually called upon at that point.
$endgroup$
– Jyrki Lahtonen
Dec 30 '18 at 10:00
add a comment |
1
$begingroup$
You got $R(1)=P(1)=5$, $R(-1)=P(-1)=-5$, $R(0)=P(0)=0$. Aren't these all compatible with $R(x)=5x$?? In other words, I think the problem was in the step when you tried to extract $R(x)$ from those three data points. The Chinese remainder theorem is usually called upon at that point.
$endgroup$
– Jyrki Lahtonen
Dec 30 '18 at 10:00
1
1
$begingroup$
You got $R(1)=P(1)=5$, $R(-1)=P(-1)=-5$, $R(0)=P(0)=0$. Aren't these all compatible with $R(x)=5x$?? In other words, I think the problem was in the step when you tried to extract $R(x)$ from those three data points. The Chinese remainder theorem is usually called upon at that point.
$endgroup$
– Jyrki Lahtonen
Dec 30 '18 at 10:00
$begingroup$
You got $R(1)=P(1)=5$, $R(-1)=P(-1)=-5$, $R(0)=P(0)=0$. Aren't these all compatible with $R(x)=5x$?? In other words, I think the problem was in the step when you tried to extract $R(x)$ from those three data points. The Chinese remainder theorem is usually called upon at that point.
$endgroup$
– Jyrki Lahtonen
Dec 30 '18 at 10:00
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
I think the problem is, with P(1), P(-1) and P(0), you are actually getting the value of r(1), r(-1) and r(0), instead of r(x).
r(x) is linear in this case, so you are able to get the correct answer by factoring P(x) by x
answered Dec 30 '18 at 9:38
Mengqi ChenMengqi Chen
165
$endgroup$
add a comment |
$begingroup$
The first method does work:
When $x^3=x$, we have $P(x)=x^{81}+x^{49}+x^{25}+x^9 + x= x+x+x+x+x = 5x$.
When $x^3=x$, we have $P(x)=x(x^{80}+x^{48}+x^{24}+x^8 + 1)= x(x^2+x^2+x^2+x^2+1) = x(4x^2+1)=4x^3+x=5x$.
answered Dec 30 '18 at 9:53
lhflhf
165k10171396
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
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$begingroup$
I think the problem is, with P(1), P(-1) and P(0), you are actually getting the value of r(1), r(-1) and r(0), instead of r(x).
r(x) is linear in this case, so you are able to get the correct answer by factoring P(x) by x
answered Dec 30 '18 at 9:38
Mengqi ChenMengqi Chen
165
$endgroup$
add a comment |
$begingroup$
I think the problem is, with P(1), P(-1) and P(0), you are actually getting the value of r(1), r(-1) and r(0), instead of r(x).
r(x) is linear in this case, so you are able to get the correct answer by factoring P(x) by x
answered Dec 30 '18 at 9:38
Mengqi ChenMengqi Chen
165
$endgroup$
add a comment |
$begingroup$
I think the problem is, with P(1), P(-1) and P(0), you are actually getting the value of r(1), r(-1) and r(0), instead of r(x).
r(x) is linear in this case, so you are able to get the correct answer by factoring P(x) by x
answered Dec 30 '18 at 9:38
Mengqi ChenMengqi Chen
165
$endgroup$
I think the problem is, with P(1), P(-1) and P(0), you are actually getting the value of r(1), r(-1) and r(0), instead of r(x).
r(x) is linear in this case, so you are able to get the correct answer by factoring P(x) by x
answered Dec 30 '18 at 9:38
Mengqi ChenMengqi Chen
165
answered Dec 30 '18 at 9:38
Mengqi ChenMengqi Chen
165
answered Dec 30 '18 at 9:38
Mengqi ChenMengqi Chen
165
answered Dec 30 '18 at 9:38
Mengqi ChenMengqi Chen
165
165
add a comment |
add a comment |
$begingroup$
The first method does work:
When $x^3=x$, we have $P(x)=x^{81}+x^{49}+x^{25}+x^9 + x= x+x+x+x+x = 5x$.
When $x^3=x$, we have $P(x)=x(x^{80}+x^{48}+x^{24}+x^8 + 1)= x(x^2+x^2+x^2+x^2+1) = x(4x^2+1)=4x^3+x=5x$.
answered Dec 30 '18 at 9:53
lhflhf
165k10171396
$endgroup$
add a comment |
$begingroup$
The first method does work:
When $x^3=x$, we have $P(x)=x^{81}+x^{49}+x^{25}+x^9 + x= x+x+x+x+x = 5x$.
When $x^3=x$, we have $P(x)=x(x^{80}+x^{48}+x^{24}+x^8 + 1)= x(x^2+x^2+x^2+x^2+1) = x(4x^2+1)=4x^3+x=5x$.
answered Dec 30 '18 at 9:53
lhflhf
165k10171396
$endgroup$
add a comment |
$begingroup$
The first method does work:
When $x^3=x$, we have $P(x)=x^{81}+x^{49}+x^{25}+x^9 + x= x+x+x+x+x = 5x$.
When $x^3=x$, we have $P(x)=x(x^{80}+x^{48}+x^{24}+x^8 + 1)= x(x^2+x^2+x^2+x^2+1) = x(4x^2+1)=4x^3+x=5x$.
answered Dec 30 '18 at 9:53
lhflhf
165k10171396
$endgroup$
The first method does work:
When $x^3=x$, we have $P(x)=x^{81}+x^{49}+x^{25}+x^9 + x= x+x+x+x+x = 5x$.
When $x^3=x$, we have $P(x)=x(x^{80}+x^{48}+x^{24}+x^8 + 1)= x(x^2+x^2+x^2+x^2+1) = x(4x^2+1)=4x^3+x=5x$.
answered Dec 30 '18 at 9:53
lhflhf
165k10171396
answered Dec 30 '18 at 9:53
lhflhf
165k10171396
answered Dec 30 '18 at 9:53
lhflhf
165k10171396
answered Dec 30 '18 at 9:53
lhflhf
165k10171396
165k10171396
add a comment |
add a comment |
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1
$begingroup$
You got $R(1)=P(1)=5$, $R(-1)=P(-1)=-5$, $R(0)=P(0)=0$. Aren't these all compatible with $R(x)=5x$?? In other words, I think the problem was in the step when you tried to extract $R(x)$ from those three data points. The Chinese remainder theorem is usually called upon at that point.
$endgroup$
– Jyrki Lahtonen
Dec 30 '18 at 10:00