Mixing
How to find $P(x)$ here?
$begingroup$
For such polynomial $P(x)$,
$$P(x^4) = ax^8+(b+1)x^5+(ab)x^4+(a-1)x^2 + 2b-2$$
Find the remainder of the division of $P(x^3-3)$ with $x+1$.
We might have to find $P(x)$ first. My attempt is shown below
$$P(x^4) = a(x^4)^2 + (b+1)(x^4)x+(ab)x^4 + (a-1)dfrac{x^4}{x^2}+2b-2$$
So
$$P(x) = ax^2 +(b+1)x^2 + (ab)x + color{red}{(a-1)x^{-1}}+2b-2$$
However, the degree of a polynomial should be a natural number. Correspondingly I've gone wrong somewhere.
Regards
polynomials
asked Jan 19 at 10:37
EnzoEnzo
19917
$endgroup$
add a comment |
$begingroup$
For such polynomial $P(x)$,
$$P(x^4) = ax^8+(b+1)x^5+(ab)x^4+(a-1)x^2 + 2b-2$$
Find the remainder of the division of $P(x^3-3)$ with $x+1$.
We might have to find $P(x)$ first. My attempt is shown below
$$P(x^4) = a(x^4)^2 + (b+1)(x^4)x+(ab)x^4 + (a-1)dfrac{x^4}{x^2}+2b-2$$
So
$$P(x) = ax^2 +(b+1)x^2 + (ab)x + color{red}{(a-1)x^{-1}}+2b-2$$
However, the degree of a polynomial should be a natural number. Correspondingly I've gone wrong somewhere.
Regards
polynomials
asked Jan 19 at 10:37
EnzoEnzo
19917
$endgroup$
$begingroup$
Just must be $b=-1$ and $a=1$.
$endgroup$
– Michael Rozenberg
Jan 19 at 10:39
$begingroup$
@MichaelRozenberg Why is that?
$endgroup$
– Enzo
Jan 19 at 10:42
$begingroup$
By the definition of the polynomial. $p(x)=a_0x^n+a_1x^{n-1}+...+a_n$, where $a_0neq0$.
$endgroup$
– Michael Rozenberg
Jan 19 at 10:43
add a comment |
$begingroup$
For such polynomial $P(x)$,
$$P(x^4) = ax^8+(b+1)x^5+(ab)x^4+(a-1)x^2 + 2b-2$$
Find the remainder of the division of $P(x^3-3)$ with $x+1$.
We might have to find $P(x)$ first. My attempt is shown below
$$P(x^4) = a(x^4)^2 + (b+1)(x^4)x+(ab)x^4 + (a-1)dfrac{x^4}{x^2}+2b-2$$
So
$$P(x) = ax^2 +(b+1)x^2 + (ab)x + color{red}{(a-1)x^{-1}}+2b-2$$
However, the degree of a polynomial should be a natural number. Correspondingly I've gone wrong somewhere.
Regards
polynomials
asked Jan 19 at 10:37
EnzoEnzo
19917
$endgroup$
For such polynomial $P(x)$,
$$P(x^4) = ax^8+(b+1)x^5+(ab)x^4+(a-1)x^2 + 2b-2$$
Find the remainder of the division of $P(x^3-3)$ with $x+1$.
We might have to find $P(x)$ first. My attempt is shown below
$$P(x^4) = a(x^4)^2 + (b+1)(x^4)x+(ab)x^4 + (a-1)dfrac{x^4}{x^2}+2b-2$$
So
$$P(x) = ax^2 +(b+1)x^2 + (ab)x + color{red}{(a-1)x^{-1}}+2b-2$$
However, the degree of a polynomial should be a natural number. Correspondingly I've gone wrong somewhere.
Regards
polynomials
polynomials
asked Jan 19 at 10:37
EnzoEnzo
19917
asked Jan 19 at 10:37
EnzoEnzo
19917
asked Jan 19 at 10:37
EnzoEnzo
19917
asked Jan 19 at 10:37
EnzoEnzo
19917
asked Jan 19 at 10:37
EnzoEnzo
19917
19917
$begingroup$
Just must be $b=-1$ and $a=1$.
$endgroup$
– Michael Rozenberg
Jan 19 at 10:39
$begingroup$
@MichaelRozenberg Why is that?
$endgroup$
– Enzo
Jan 19 at 10:42
$begingroup$
By the definition of the polynomial. $p(x)=a_0x^n+a_1x^{n-1}+...+a_n$, where $a_0neq0$.
$endgroup$
– Michael Rozenberg
Jan 19 at 10:43
add a comment |
$begingroup$
Just must be $b=-1$ and $a=1$.
$endgroup$
– Michael Rozenberg
Jan 19 at 10:39
$begingroup$
@MichaelRozenberg Why is that?
$endgroup$
– Enzo
Jan 19 at 10:42
$begingroup$
By the definition of the polynomial. $p(x)=a_0x^n+a_1x^{n-1}+...+a_n$, where $a_0neq0$.
$endgroup$
– Michael Rozenberg
Jan 19 at 10:43
$begingroup$
Just must be $b=-1$ and $a=1$.
$endgroup$
– Michael Rozenberg
Jan 19 at 10:39
$begingroup$
Just must be $b=-1$ and $a=1$.
$endgroup$
– Michael Rozenberg
Jan 19 at 10:39
$begingroup$
@MichaelRozenberg Why is that?
$endgroup$
– Enzo
Jan 19 at 10:42
$begingroup$
@MichaelRozenberg Why is that?
$endgroup$
– Enzo
Jan 19 at 10:42
$begingroup$
By the definition of the polynomial. $p(x)=a_0x^n+a_1x^{n-1}+...+a_n$, where $a_0neq0$.
$endgroup$
– Michael Rozenberg
Jan 19 at 10:43
$begingroup$
By the definition of the polynomial. $p(x)=a_0x^n+a_1x^{n-1}+...+a_n$, where $a_0neq0$.
$endgroup$
– Michael Rozenberg
Jan 19 at 10:43
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
From
$$P(t) = at^2+(b+1)t^{5/4}+(ab)t+(a-1)t^{1/2} + 2b-2$$ you deduce that $b=-1$ and $a=1$ for $P$ to indeed be a polynomial.
Then the requested remainder is
$$P((-1)^3-3)=(-4)^2-(-4)-4=16.$$
answered Jan 19 at 10:53
Yves DaoustYves Daoust
129k675227
$endgroup$
$begingroup$
Could you explain this part a bit: $P((-1)^3-3)=(-4)^2-(-4)-4=16.$?
$endgroup$
– Enzo
Jan 19 at 10:54
1
$begingroup$
@Yves Daoust Maybe there is another way to write $p$ like polynomial without square radicals? I think your reasoning just is total wrong.
$endgroup$
– Michael Rozenberg
Jan 19 at 11:01
$begingroup$
@MichaelRozenberg: please show me.
$endgroup$
– Yves Daoust
Jan 19 at 11:55
$begingroup$
@Enzo: review your theory of division by a linear binomial.
$endgroup$
– Yves Daoust
Jan 19 at 11:56
1
$begingroup$
@Yves Daoust You need to show us that it's impossible. Otherwise, your reasoning is nothing.
$endgroup$
– Michael Rozenberg
Jan 19 at 12:28
|
show 1 more comment
$begingroup$
Obviously $P$ must be a quadratic polynomial:
$$ P(x) = cx^2+dx+e implies P(x^4) = cx^8+dx^4+e$$
So $b+1=0$ and $a-1=0$.
answered Jan 19 at 10:42
greedoidgreedoid
45k1157112
$endgroup$
1
$begingroup$
How did we conclude that $p$ must be quadratic polynomial?
$endgroup$
– Enzo
Jan 19 at 10:43
$begingroup$
Can it be third degree or more? What about first or less?
$endgroup$
– greedoid
Jan 19 at 10:43
$begingroup$
Yes, why not? It can be third degree or more.
$endgroup$
– Enzo
Jan 19 at 10:44
$begingroup$
Then I can't help you
$endgroup$
– greedoid
Jan 19 at 10:45
$begingroup$
What do you mean?
$endgroup$
– Enzo
Jan 19 at 10:45
|
show 1 more comment
$begingroup$
Just should be $b=-1$ and $a=1$.
The rest is smooth.
Let $p(x)=a_0x^{n}+a_1x^{n-1}+...+a_n,$ where $a_0neq0.$
Thus, by the given
$$a_0x^{4n}+a_1x^{4n-4}+...+a_n= ax^8+(b+1)x^5+(ab)x^4+(a-1)x^2 + 2b-2.$$
We know that two polynomials are equal iff all coefficients before powers are equal.
Thus, $b+1=0$ and $a-1=0$, which gives
$$p(x)=x^2-x-4.$$
Now, the remainder it's $p(x^3-3),$ where $x=-1$ and since
$$p(-4)=(-4)^2-(-4)-4=16,$$ we are done!
answered Jan 19 at 10:40
Michael RozenbergMichael Rozenberg
106k1893198
$endgroup$
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
From
$$P(t) = at^2+(b+1)t^{5/4}+(ab)t+(a-1)t^{1/2} + 2b-2$$ you deduce that $b=-1$ and $a=1$ for $P$ to indeed be a polynomial.
Then the requested remainder is
$$P((-1)^3-3)=(-4)^2-(-4)-4=16.$$
answered Jan 19 at 10:53
Yves DaoustYves Daoust
129k675227
$endgroup$
$begingroup$
Could you explain this part a bit: $P((-1)^3-3)=(-4)^2-(-4)-4=16.$?
$endgroup$
– Enzo
Jan 19 at 10:54
1
$begingroup$
@Yves Daoust Maybe there is another way to write $p$ like polynomial without square radicals? I think your reasoning just is total wrong.
$endgroup$
– Michael Rozenberg
Jan 19 at 11:01
$begingroup$
@MichaelRozenberg: please show me.
$endgroup$
– Yves Daoust
Jan 19 at 11:55
$begingroup$
@Enzo: review your theory of division by a linear binomial.
$endgroup$
– Yves Daoust
Jan 19 at 11:56
1
$begingroup$
@Yves Daoust You need to show us that it's impossible. Otherwise, your reasoning is nothing.
$endgroup$
– Michael Rozenberg
Jan 19 at 12:28
|
show 1 more comment
$begingroup$
From
$$P(t) = at^2+(b+1)t^{5/4}+(ab)t+(a-1)t^{1/2} + 2b-2$$ you deduce that $b=-1$ and $a=1$ for $P$ to indeed be a polynomial.
Then the requested remainder is
$$P((-1)^3-3)=(-4)^2-(-4)-4=16.$$
answered Jan 19 at 10:53
Yves DaoustYves Daoust
129k675227
$endgroup$
$begingroup$
Could you explain this part a bit: $P((-1)^3-3)=(-4)^2-(-4)-4=16.$?
$endgroup$
– Enzo
Jan 19 at 10:54
1
$begingroup$
@Yves Daoust Maybe there is another way to write $p$ like polynomial without square radicals? I think your reasoning just is total wrong.
$endgroup$
– Michael Rozenberg
Jan 19 at 11:01
$begingroup$
@MichaelRozenberg: please show me.
$endgroup$
– Yves Daoust
Jan 19 at 11:55
$begingroup$
@Enzo: review your theory of division by a linear binomial.
$endgroup$
– Yves Daoust
Jan 19 at 11:56
1
$begingroup$
@Yves Daoust You need to show us that it's impossible. Otherwise, your reasoning is nothing.
$endgroup$
– Michael Rozenberg
Jan 19 at 12:28
|
show 1 more comment
$begingroup$
From
$$P(t) = at^2+(b+1)t^{5/4}+(ab)t+(a-1)t^{1/2} + 2b-2$$ you deduce that $b=-1$ and $a=1$ for $P$ to indeed be a polynomial.
Then the requested remainder is
$$P((-1)^3-3)=(-4)^2-(-4)-4=16.$$
answered Jan 19 at 10:53
Yves DaoustYves Daoust
129k675227
$endgroup$
From
$$P(t) = at^2+(b+1)t^{5/4}+(ab)t+(a-1)t^{1/2} + 2b-2$$ you deduce that $b=-1$ and $a=1$ for $P$ to indeed be a polynomial.
Then the requested remainder is
$$P((-1)^3-3)=(-4)^2-(-4)-4=16.$$
answered Jan 19 at 10:53
Yves DaoustYves Daoust
129k675227
answered Jan 19 at 10:53
Yves DaoustYves Daoust
129k675227
answered Jan 19 at 10:53
Yves DaoustYves Daoust
129k675227
answered Jan 19 at 10:53
Yves DaoustYves Daoust
129k675227
129k675227
$begingroup$
Could you explain this part a bit: $P((-1)^3-3)=(-4)^2-(-4)-4=16.$?
$endgroup$
– Enzo
Jan 19 at 10:54
1
$begingroup$
@Yves Daoust Maybe there is another way to write $p$ like polynomial without square radicals? I think your reasoning just is total wrong.
$endgroup$
– Michael Rozenberg
Jan 19 at 11:01
$begingroup$
@MichaelRozenberg: please show me.
$endgroup$
– Yves Daoust
Jan 19 at 11:55
$begingroup$
@Enzo: review your theory of division by a linear binomial.
$endgroup$
– Yves Daoust
Jan 19 at 11:56
1
$begingroup$
@Yves Daoust You need to show us that it's impossible. Otherwise, your reasoning is nothing.
$endgroup$
– Michael Rozenberg
Jan 19 at 12:28
|
show 1 more comment
$begingroup$
Could you explain this part a bit: $P((-1)^3-3)=(-4)^2-(-4)-4=16.$?
$endgroup$
– Enzo
Jan 19 at 10:54
1
$begingroup$
@Yves Daoust Maybe there is another way to write $p$ like polynomial without square radicals? I think your reasoning just is total wrong.
$endgroup$
– Michael Rozenberg
Jan 19 at 11:01
$begingroup$
@MichaelRozenberg: please show me.
$endgroup$
– Yves Daoust
Jan 19 at 11:55
$begingroup$
@Enzo: review your theory of division by a linear binomial.
$endgroup$
– Yves Daoust
Jan 19 at 11:56
1
$begingroup$
@Yves Daoust You need to show us that it's impossible. Otherwise, your reasoning is nothing.
$endgroup$
– Michael Rozenberg
Jan 19 at 12:28
$begingroup$
Could you explain this part a bit: $P((-1)^3-3)=(-4)^2-(-4)-4=16.$?
$endgroup$
– Enzo
Jan 19 at 10:54
$begingroup$
Could you explain this part a bit: $P((-1)^3-3)=(-4)^2-(-4)-4=16.$?
$endgroup$
– Enzo
Jan 19 at 10:54
1
1
$begingroup$
@Yves Daoust Maybe there is another way to write $p$ like polynomial without square radicals? I think your reasoning just is total wrong.
$endgroup$
– Michael Rozenberg
Jan 19 at 11:01
$begingroup$
@Yves Daoust Maybe there is another way to write $p$ like polynomial without square radicals? I think your reasoning just is total wrong.
$endgroup$
– Michael Rozenberg
Jan 19 at 11:01
$begingroup$
@MichaelRozenberg: please show me.
$endgroup$
– Yves Daoust
Jan 19 at 11:55
$begingroup$
@MichaelRozenberg: please show me.
$endgroup$
– Yves Daoust
Jan 19 at 11:55
$begingroup$
@Enzo: review your theory of division by a linear binomial.
$endgroup$
– Yves Daoust
Jan 19 at 11:56
$begingroup$
@Enzo: review your theory of division by a linear binomial.
$endgroup$
– Yves Daoust
Jan 19 at 11:56
1
1
$begingroup$
@Yves Daoust You need to show us that it's impossible. Otherwise, your reasoning is nothing.
$endgroup$
– Michael Rozenberg
Jan 19 at 12:28
$begingroup$
@Yves Daoust You need to show us that it's impossible. Otherwise, your reasoning is nothing.
$endgroup$
– Michael Rozenberg
Jan 19 at 12:28
|
show 1 more comment
$begingroup$
Obviously $P$ must be a quadratic polynomial:
$$ P(x) = cx^2+dx+e implies P(x^4) = cx^8+dx^4+e$$
So $b+1=0$ and $a-1=0$.
answered Jan 19 at 10:42
greedoidgreedoid
45k1157112
$endgroup$
1
$begingroup$
How did we conclude that $p$ must be quadratic polynomial?
$endgroup$
– Enzo
Jan 19 at 10:43
$begingroup$
Can it be third degree or more? What about first or less?
$endgroup$
– greedoid
Jan 19 at 10:43
$begingroup$
Yes, why not? It can be third degree or more.
$endgroup$
– Enzo
Jan 19 at 10:44
$begingroup$
Then I can't help you
$endgroup$
– greedoid
Jan 19 at 10:45
$begingroup$
What do you mean?
$endgroup$
– Enzo
Jan 19 at 10:45
|
show 1 more comment
$begingroup$
Obviously $P$ must be a quadratic polynomial:
$$ P(x) = cx^2+dx+e implies P(x^4) = cx^8+dx^4+e$$
So $b+1=0$ and $a-1=0$.
answered Jan 19 at 10:42
greedoidgreedoid
45k1157112
$endgroup$
1
$begingroup$
How did we conclude that $p$ must be quadratic polynomial?
$endgroup$
– Enzo
Jan 19 at 10:43
$begingroup$
Can it be third degree or more? What about first or less?
$endgroup$
– greedoid
Jan 19 at 10:43
$begingroup$
Yes, why not? It can be third degree or more.
$endgroup$
– Enzo
Jan 19 at 10:44
$begingroup$
Then I can't help you
$endgroup$
– greedoid
Jan 19 at 10:45
$begingroup$
What do you mean?
$endgroup$
– Enzo
Jan 19 at 10:45
|
show 1 more comment
$begingroup$
Obviously $P$ must be a quadratic polynomial:
$$ P(x) = cx^2+dx+e implies P(x^4) = cx^8+dx^4+e$$
So $b+1=0$ and $a-1=0$.
answered Jan 19 at 10:42
greedoidgreedoid
45k1157112
$endgroup$
Obviously $P$ must be a quadratic polynomial:
$$ P(x) = cx^2+dx+e implies P(x^4) = cx^8+dx^4+e$$
So $b+1=0$ and $a-1=0$.
answered Jan 19 at 10:42
greedoidgreedoid
45k1157112
answered Jan 19 at 10:42
greedoidgreedoid
45k1157112
answered Jan 19 at 10:42
greedoidgreedoid
45k1157112
answered Jan 19 at 10:42
greedoidgreedoid
45k1157112
45k1157112
1
$begingroup$
How did we conclude that $p$ must be quadratic polynomial?
$endgroup$
– Enzo
Jan 19 at 10:43
$begingroup$
Can it be third degree or more? What about first or less?
$endgroup$
– greedoid
Jan 19 at 10:43
$begingroup$
Yes, why not? It can be third degree or more.
$endgroup$
– Enzo
Jan 19 at 10:44
$begingroup$
Then I can't help you
$endgroup$
– greedoid
Jan 19 at 10:45
$begingroup$
What do you mean?
$endgroup$
– Enzo
Jan 19 at 10:45
|
show 1 more comment
1
$begingroup$
How did we conclude that $p$ must be quadratic polynomial?
$endgroup$
– Enzo
Jan 19 at 10:43
$begingroup$
Can it be third degree or more? What about first or less?
$endgroup$
– greedoid
Jan 19 at 10:43
$begingroup$
Yes, why not? It can be third degree or more.
$endgroup$
– Enzo
Jan 19 at 10:44
$begingroup$
Then I can't help you
$endgroup$
– greedoid
Jan 19 at 10:45
$begingroup$
What do you mean?
$endgroup$
– Enzo
Jan 19 at 10:45
1
1
$begingroup$
How did we conclude that $p$ must be quadratic polynomial?
$endgroup$
– Enzo
Jan 19 at 10:43
$begingroup$
How did we conclude that $p$ must be quadratic polynomial?
$endgroup$
– Enzo
Jan 19 at 10:43
$begingroup$
Can it be third degree or more? What about first or less?
$endgroup$
– greedoid
Jan 19 at 10:43
$begingroup$
Can it be third degree or more? What about first or less?
$endgroup$
– greedoid
Jan 19 at 10:43
$begingroup$
Yes, why not? It can be third degree or more.
$endgroup$
– Enzo
Jan 19 at 10:44
$begingroup$
Yes, why not? It can be third degree or more.
$endgroup$
– Enzo
Jan 19 at 10:44
$begingroup$
Then I can't help you
$endgroup$
– greedoid
Jan 19 at 10:45
$begingroup$
Then I can't help you
$endgroup$
– greedoid
Jan 19 at 10:45
$begingroup$
What do you mean?
$endgroup$
– Enzo
Jan 19 at 10:45
$begingroup$
What do you mean?
$endgroup$
– Enzo
Jan 19 at 10:45
|
show 1 more comment
$begingroup$
Just should be $b=-1$ and $a=1$.
The rest is smooth.
Let $p(x)=a_0x^{n}+a_1x^{n-1}+...+a_n,$ where $a_0neq0.$
Thus, by the given
$$a_0x^{4n}+a_1x^{4n-4}+...+a_n= ax^8+(b+1)x^5+(ab)x^4+(a-1)x^2 + 2b-2.$$
We know that two polynomials are equal iff all coefficients before powers are equal.
Thus, $b+1=0$ and $a-1=0$, which gives
$$p(x)=x^2-x-4.$$
Now, the remainder it's $p(x^3-3),$ where $x=-1$ and since
$$p(-4)=(-4)^2-(-4)-4=16,$$ we are done!
answered Jan 19 at 10:40
Michael RozenbergMichael Rozenberg
106k1893198
$endgroup$
add a comment |
$begingroup$
Just should be $b=-1$ and $a=1$.
The rest is smooth.
Let $p(x)=a_0x^{n}+a_1x^{n-1}+...+a_n,$ where $a_0neq0.$
Thus, by the given
$$a_0x^{4n}+a_1x^{4n-4}+...+a_n= ax^8+(b+1)x^5+(ab)x^4+(a-1)x^2 + 2b-2.$$
We know that two polynomials are equal iff all coefficients before powers are equal.
Thus, $b+1=0$ and $a-1=0$, which gives
$$p(x)=x^2-x-4.$$
Now, the remainder it's $p(x^3-3),$ where $x=-1$ and since
$$p(-4)=(-4)^2-(-4)-4=16,$$ we are done!
answered Jan 19 at 10:40
Michael RozenbergMichael Rozenberg
106k1893198
$endgroup$
add a comment |
$begingroup$
Just should be $b=-1$ and $a=1$.
The rest is smooth.
Let $p(x)=a_0x^{n}+a_1x^{n-1}+...+a_n,$ where $a_0neq0.$
Thus, by the given
$$a_0x^{4n}+a_1x^{4n-4}+...+a_n= ax^8+(b+1)x^5+(ab)x^4+(a-1)x^2 + 2b-2.$$
We know that two polynomials are equal iff all coefficients before powers are equal.
Thus, $b+1=0$ and $a-1=0$, which gives
$$p(x)=x^2-x-4.$$
Now, the remainder it's $p(x^3-3),$ where $x=-1$ and since
$$p(-4)=(-4)^2-(-4)-4=16,$$ we are done!
answered Jan 19 at 10:40
Michael RozenbergMichael Rozenberg
106k1893198
$endgroup$
Just should be $b=-1$ and $a=1$.
The rest is smooth.
Let $p(x)=a_0x^{n}+a_1x^{n-1}+...+a_n,$ where $a_0neq0.$
Thus, by the given
$$a_0x^{4n}+a_1x^{4n-4}+...+a_n= ax^8+(b+1)x^5+(ab)x^4+(a-1)x^2 + 2b-2.$$
We know that two polynomials are equal iff all coefficients before powers are equal.
Thus, $b+1=0$ and $a-1=0$, which gives
$$p(x)=x^2-x-4.$$
Now, the remainder it's $p(x^3-3),$ where $x=-1$ and since
$$p(-4)=(-4)^2-(-4)-4=16,$$ we are done!
answered Jan 19 at 10:40
Michael RozenbergMichael Rozenberg
106k1893198
edited Jan 19 at 10:57
answered Jan 19 at 10:40
Michael RozenbergMichael Rozenberg
106k1893198
answered Jan 19 at 10:40
Michael RozenbergMichael Rozenberg
106k1893198
answered Jan 19 at 10:40
Michael RozenbergMichael Rozenberg
106k1893198
106k1893198
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$begingroup$
Just must be $b=-1$ and $a=1$.
$endgroup$
– Michael Rozenberg
Jan 19 at 10:39
$begingroup$
@MichaelRozenberg Why is that?
$endgroup$
– Enzo
Jan 19 at 10:42
$begingroup$
By the definition of the polynomial. $p(x)=a_0x^n+a_1x^{n-1}+...+a_n$, where $a_0neq0$.
$endgroup$
– Michael Rozenberg
Jan 19 at 10:43