How do I should that $x-1$ a factor of a positive degree polynomial?
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I'm supposed to show that $x-1$ is a factor of a polynomial P of positive degree if and only if the sum of the coefficients of P is zero. How do I do that exactly?
polynomials
asked Aug 26 at 21:53
Math Love
183
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up vote
1
down vote
favorite
I'm supposed to show that $x-1$ is a factor of a polynomial P of positive degree if and only if the sum of the coefficients of P is zero. How do I do that exactly?
polynomials
asked Aug 26 at 21:53
Math Love
183
4
If $x-1$ is a factor, then what does that tell us about $P(1)$?
– Trevor Norton
Aug 26 at 21:54
That the polynomial would be equal to zero. But what does it mean to be a positive degree? Does it mean that $x^n$ where n starts with a positive number like $x^2$, $x^4$ and such?
– Math Love
Aug 26 at 21:58
Yes, positive degree means it is a polynomial of degree at least $1$, so that degree $0$ (constant) polynomials are not considered.
– Jair Taylor
Aug 26 at 21:59
Got it! Thank you
– Math Love
Aug 26 at 22:01
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
I'm supposed to show that $x-1$ is a factor of a polynomial P of positive degree if and only if the sum of the coefficients of P is zero. How do I do that exactly?
polynomials
asked Aug 26 at 21:53
Math Love
183
I'm supposed to show that $x-1$ is a factor of a polynomial P of positive degree if and only if the sum of the coefficients of P is zero. How do I do that exactly?
polynomials
asked Aug 26 at 21:53
Math Love
183
asked Aug 26 at 21:53
Math Love
183
asked Aug 26 at 21:53
Math Love
183
asked Aug 26 at 21:53
Math Love
183
183
4
If $x-1$ is a factor, then what does that tell us about $P(1)$?
– Trevor Norton
Aug 26 at 21:54
That the polynomial would be equal to zero. But what does it mean to be a positive degree? Does it mean that $x^n$ where n starts with a positive number like $x^2$, $x^4$ and such?
– Math Love
Aug 26 at 21:58
Yes, positive degree means it is a polynomial of degree at least $1$, so that degree $0$ (constant) polynomials are not considered.
– Jair Taylor
Aug 26 at 21:59
Got it! Thank you
– Math Love
Aug 26 at 22:01
add a comment |Â
4
If $x-1$ is a factor, then what does that tell us about $P(1)$?
– Trevor Norton
Aug 26 at 21:54
That the polynomial would be equal to zero. But what does it mean to be a positive degree? Does it mean that $x^n$ where n starts with a positive number like $x^2$, $x^4$ and such?
– Math Love
Aug 26 at 21:58
Yes, positive degree means it is a polynomial of degree at least $1$, so that degree $0$ (constant) polynomials are not considered.
– Jair Taylor
Aug 26 at 21:59
Got it! Thank you
– Math Love
Aug 26 at 22:01
4
4
If $x-1$ is a factor, then what does that tell us about $P(1)$?
– Trevor Norton
Aug 26 at 21:54
If $x-1$ is a factor, then what does that tell us about $P(1)$?
– Trevor Norton
Aug 26 at 21:54
That the polynomial would be equal to zero. But what does it mean to be a positive degree? Does it mean that $x^n$ where n starts with a positive number like $x^2$, $x^4$ and such?
– Math Love
Aug 26 at 21:58
That the polynomial would be equal to zero. But what does it mean to be a positive degree? Does it mean that $x^n$ where n starts with a positive number like $x^2$, $x^4$ and such?
– Math Love
Aug 26 at 21:58
Yes, positive degree means it is a polynomial of degree at least $1$, so that degree $0$ (constant) polynomials are not considered.
– Jair Taylor
Aug 26 at 21:59
Yes, positive degree means it is a polynomial of degree at least $1$, so that degree $0$ (constant) polynomials are not considered.
– Jair Taylor
Aug 26 at 21:59
Got it! Thank you
– Math Love
Aug 26 at 22:01
Got it! Thank you
– Math Love
Aug 26 at 22:01
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
3
down vote
accepted
If $x-1$ is a factor then you have $p(x) = (x-1)q(x)$ where q(x) $ is another polynomial.
Let $x=1$, we get $$p(1) = (1-1)q(1)=0$$
That is if you let $x=1$ in your polynomial you will get $p(1)=0$
Note that when you find $p(1)$ you let $x=1$ , so you are just adding the coefficients of $P(x)$ together.
Thus if $x=1$ is a factor, the sum of coefficients of $p(x)$ is $0$.
answered Aug 26 at 22:02
Mohammad Riazi-Kermani
30.5k41852
add a comment |Â
up vote
0
down vote
The remainder of the division of a polynomial $p(x)$ by $x-a$ is $p(a)$, so
$$p(x);text divisible by ;x-aiff p(a)=0.$$
On the other hand $p(1)$ is just the sum of the coefficients of $p(x)$, whence the conclusion.
answered Aug 26 at 22:39
Bernard
111k635102
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
accepted
If $x-1$ is a factor then you have $p(x) = (x-1)q(x)$ where q(x) $ is another polynomial.
Let $x=1$, we get $$p(1) = (1-1)q(1)=0$$
That is if you let $x=1$ in your polynomial you will get $p(1)=0$
Note that when you find $p(1)$ you let $x=1$ , so you are just adding the coefficients of $P(x)$ together.
Thus if $x=1$ is a factor, the sum of coefficients of $p(x)$ is $0$.
answered Aug 26 at 22:02
Mohammad Riazi-Kermani
30.5k41852
add a comment |Â
up vote
3
down vote
accepted
If $x-1$ is a factor then you have $p(x) = (x-1)q(x)$ where q(x) $ is another polynomial.
Let $x=1$, we get $$p(1) = (1-1)q(1)=0$$
That is if you let $x=1$ in your polynomial you will get $p(1)=0$
Note that when you find $p(1)$ you let $x=1$ , so you are just adding the coefficients of $P(x)$ together.
Thus if $x=1$ is a factor, the sum of coefficients of $p(x)$ is $0$.
answered Aug 26 at 22:02
Mohammad Riazi-Kermani
30.5k41852
add a comment |Â
up vote
3
down vote
accepted
up vote
3
down vote
accepted
If $x-1$ is a factor then you have $p(x) = (x-1)q(x)$ where q(x) $ is another polynomial.
Let $x=1$, we get $$p(1) = (1-1)q(1)=0$$
That is if you let $x=1$ in your polynomial you will get $p(1)=0$
Note that when you find $p(1)$ you let $x=1$ , so you are just adding the coefficients of $P(x)$ together.
Thus if $x=1$ is a factor, the sum of coefficients of $p(x)$ is $0$.
answered Aug 26 at 22:02
Mohammad Riazi-Kermani
30.5k41852
If $x-1$ is a factor then you have $p(x) = (x-1)q(x)$ where q(x) $ is another polynomial.
Let $x=1$, we get $$p(1) = (1-1)q(1)=0$$
That is if you let $x=1$ in your polynomial you will get $p(1)=0$
Note that when you find $p(1)$ you let $x=1$ , so you are just adding the coefficients of $P(x)$ together.
Thus if $x=1$ is a factor, the sum of coefficients of $p(x)$ is $0$.
answered Aug 26 at 22:02
Mohammad Riazi-Kermani
30.5k41852
answered Aug 26 at 22:02
Mohammad Riazi-Kermani
30.5k41852
answered Aug 26 at 22:02
Mohammad Riazi-Kermani
30.5k41852
answered Aug 26 at 22:02
Mohammad Riazi-Kermani
30.5k41852
30.5k41852
add a comment |Â
add a comment |Â
up vote
0
down vote
The remainder of the division of a polynomial $p(x)$ by $x-a$ is $p(a)$, so
$$p(x);text divisible by ;x-aiff p(a)=0.$$
On the other hand $p(1)$ is just the sum of the coefficients of $p(x)$, whence the conclusion.
answered Aug 26 at 22:39
Bernard
111k635102
add a comment |Â
up vote
0
down vote
The remainder of the division of a polynomial $p(x)$ by $x-a$ is $p(a)$, so
$$p(x);text divisible by ;x-aiff p(a)=0.$$
On the other hand $p(1)$ is just the sum of the coefficients of $p(x)$, whence the conclusion.
answered Aug 26 at 22:39
Bernard
111k635102
add a comment |Â
up vote
0
down vote
up vote
0
down vote
The remainder of the division of a polynomial $p(x)$ by $x-a$ is $p(a)$, so
$$p(x);text divisible by ;x-aiff p(a)=0.$$
On the other hand $p(1)$ is just the sum of the coefficients of $p(x)$, whence the conclusion.
answered Aug 26 at 22:39
Bernard
111k635102
The remainder of the division of a polynomial $p(x)$ by $x-a$ is $p(a)$, so
$$p(x);text divisible by ;x-aiff p(a)=0.$$
On the other hand $p(1)$ is just the sum of the coefficients of $p(x)$, whence the conclusion.
answered Aug 26 at 22:39
Bernard
111k635102
answered Aug 26 at 22:39
Bernard
111k635102
answered Aug 26 at 22:39
Bernard
111k635102
answered Aug 26 at 22:39
Bernard
111k635102
111k635102
add a comment |Â
add a comment |Â
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4
If $x-1$ is a factor, then what does that tell us about $P(1)$?
– Trevor Norton
Aug 26 at 21:54
That the polynomial would be equal to zero. But what does it mean to be a positive degree? Does it mean that $x^n$ where n starts with a positive number like $x^2$, $x^4$ and such?
– Math Love
Aug 26 at 21:58
Yes, positive degree means it is a polynomial of degree at least $1$, so that degree $0$ (constant) polynomials are not considered.
– Jair Taylor
Aug 26 at 21:59
Got it! Thank you
– Math Love
Aug 26 at 22:01