Fundamental theorem of algebra question: Why/how does q(x) = a_n ?

[Aion]

A polynomial of degree $\displaystyle n$ has exactly $\displaystyle n$ complex roots if you count multiplicity.

Let $\displaystyle p(x) = a_nx^n + ... +a_1x + a_0$ be a polynomial of degree $\displaystyle n$ and let $\displaystyle c_1, ..., c_m$ be roots of multiplicity $\displaystyle k_1, ..., k_m$ such that $\displaystyle k_1 + ... + k_m = n$.

Then we can factor $\displaystyle p$ as

$\displaystyle p(x) = (x-c_1)^{k_1} ... (x-c_m)^{k_m}q(x)$, for some polynomial $\displaystyle q$ which lacks roots. But the degree of $\displaystyle p$ is $\displaystyle n$ and the degree of the product $\displaystyle (x-c_1)^{k_1}...(x-c_m)^{k_m}$ is also $\displaystyle n$.


deg $\displaystyle p =$ deg $\displaystyle (x-c_1)^{k_1} + ... +$ deg $\displaystyle (x-c_m)^{k_m}+$ deg $\displaystyle q$

$\displaystyle n = k_1 + ... + k_m +$ deg $\displaystyle q$
$\displaystyle n = n +$ deg $\displaystyle q$

Therefore $\displaystyle q$ much have degree $\displaystyle 0$ hence be a constant. If we look at the $\displaystyle x^n$ terms on both sides we can see that $\displaystyle q(x) = a_n$ i.e. the highest coefficient in $\displaystyle p$, thus

$\displaystyle p(x) = a_n(x-c_1)^{k_1}...(x-c_m)^{k_m}$


My question is​:

Why/how does $\displaystyle q(x) = a_n$ ?

 

[Deleted member 4993]

A polynomial of degree $\displaystyle n$ has exactly $\displaystyle n$ complex roots if you count multiplicity.

Let $\displaystyle p(x) = a_nx^n + ... +a_1x + a_0$ be a polynomial of degree $\displaystyle n$ and let $\displaystyle c_1, ..., c_m$ be roots of multiplicity $\displaystyle k_1, ..., k_m$ such that $\displaystyle k_1 + ... + k_m = n$.

Then we can factor $\displaystyle p$ as

$\displaystyle p(x) = (x-c_1)^{k_1} ... (x-c_m)^{k_m}q(x)$, for some polynomial $\displaystyle q$ which lacks roots. But the degree of $\displaystyle p$ is $\displaystyle n$ and the degree of the product $\displaystyle (x-c_1)^{k_1}...(x-c_m)^{k_m}$ is also $\displaystyle n$.


deg $\displaystyle p =$ deg $\displaystyle (x-c_1)^{k_1} + ... +$ deg $\displaystyle (x-c_m)^{k_m}+$ deg $\displaystyle q$

$\displaystyle n = k_1 + ... + k_m +$ deg $\displaystyle q$
$\displaystyle n = n +$ deg $\displaystyle q$

Therefore $\displaystyle q$ much have degree $\displaystyle 0$ hence be a constant. If we look at the $\displaystyle x^n$ terms on both sides we can see that $\displaystyle q(x) = a_n$ i.e. the highest coefficient in $\displaystyle p$, thus

$\displaystyle p(x) = a_n(x-c_1)^{k_1}...(x-c_m)^{k_m}$


My question is​:

Why/how does $\displaystyle q(x) = a_n$ ?

You did answer it yourself!!!

 

[Aion]

You did answer it yourself!!!

No, I'm quoting my math book. I don't fully understand it myself.

But for this to actually work according to what I've stated above

$\displaystyle q(x) = a_nx^0$

But this makes no sense to me since the constant term should be $\displaystyle a_0$ hence $\displaystyle q(x) = a_0x^0$

Clearly, I'm lacking some fundamental insight...

 

[HallsofIvy]

A polynomial of degree $\displaystyle n$ has exactly $\displaystyle n$ complex roots if you count multiplicity.

Let $\displaystyle p(x) = a_nx^n + ... +a_1x + a_0$ be a polynomial of degree $\displaystyle n$ and let $\displaystyle c_1, ..., c_m$ be roots of multiplicity $\displaystyle k_1, ..., k_m$ such that $\displaystyle k_1 + ... + k_m = n$.

Then we can factor $\displaystyle p$ as

$\displaystyle p(x) = (x-c_1)^{k_1} ... (x-c_m)^{k_m}q(x)$, for some polynomial $\displaystyle q$ which lacks roots. But the degree of $\displaystyle p$ is $\displaystyle n$ and the degree of the product $\displaystyle (x-c_1)^{k_1}...(x-c_m)^{k_m}$ is also $\displaystyle n$.

And so it follows that $\displaystyle q(x)$ is a constant.

deg $\displaystyle p =$ deg $\displaystyle (x-c_1)^{k_1} + ... +$ deg $\displaystyle (x-c_m)^{k_m}+$ deg $\displaystyle q$

$\displaystyle n = k_1 + ... + k_m +$ deg $\displaystyle q$
$\displaystyle n = n +$ deg $\displaystyle q$

Therefore $\displaystyle q$ much have degree $\displaystyle 0$ hence be a constant. If we look at the $\displaystyle x^n$ terms on both sides we can see that $\displaystyle q(x) = a_n$ i.e. the highest coefficient in $\displaystyle p$, thus

$\displaystyle p(x) = a_n(x-c_1)^{k_1}...(x-c_m)^{k_m}$


My question is​:

Why/how does $\displaystyle q(x) = a_n$ ?

You wrote the polynomial $\displaystyle a_nx^n+ a_{n-1}x^{n-1}+ \cdot\cdot\cdot+ a_1x+ a_0$ and then factored it as $\displaystyle p(x) = (x-c_1)^{k_1} ... (x-c_m)^{k_m}q(x)$. As you say above q(x) must be a constant, not a function of x. Further, if you were to multiply that product of linear terms, $\displaystyle p(x) = (x-c_1)^{k_1} ... (x-c_m)^{k_m}$ it should be clear that the hist power term is just $\displaystyle x^n$ with no coefficient. In order to make that equal to $\displaystyle a_nx^n+ \cdot\cdot\cdot$, you have to multiply by $\displaystyle a_n$. That is the constant, q(x).

 

[Aion]

And so it follows that $\displaystyle q(x)$ is a constant.



You wrote the polynomial $\displaystyle a_nx^n+ a_{n-1}x^{n-1}+ \cdot\cdot\cdot+ a_1x+ a_0$ and then factored it as $\displaystyle p(x) = (x-c_1)^{k_1} ... (x-c_m)^{k_m}q(x)$. As you say above q(x) must be a constant, not a function of x. Further, if you were to multiply that product of linear terms, $\displaystyle p(x) = (x-c_1)^{k_1} ... (x-c_m)^{k_m}$ it should be clear that the hist power term is just $\displaystyle x^n$ with no coefficient. In order to make that equal to $\displaystyle a_nx^n+ \cdot\cdot\cdot$, you have to multiply by $\displaystyle a_n$. That is the constant, q(x).

So you're saying that $\displaystyle (x-c_1)^{k_1}(x-c_2)^{k_2}(x-c_3)^{k_3}... = x^n + x^{n-1} + x^{n-2}...$?

 

[Steven G]

So you're saying that $\displaystyle (x-c_1)^{k_1}(x-c_2)^{k_2}(x-c_3)^{k_3}... = x^n + x^{n-1} + x^{n-2}...$?

No, prof Halls is not saying that. All that is being said is the leading term will 1xⁿ.

Consider (x-c₁)(x-c₂) = 1x² - (c₁ + c₂)x + (c₁c₂). Notice that the non leading coefficients do not have be 1 while the leading coefficient is 1. To get the leading coefficient to be say 7 we need to multiply both sides by 7.

We will get 7(x-c₁)(x-c₂) = 7x² - 7(c₁ + c₂)x + 7(c₁c₂).

Does this help?

 

[HallsofIvy]

No, I am not saying that at all! I am saying that the leading coefficient, the coefficient of the highest power of x is 1. The other coefficients are not necessarily 1.

For example, the product (x- 1)(x- 2)(x- 3)= (x^2- 3x+ 2)(x- 1)= x^3- 4x^2+ 5x- 2. The coefficient of the highest power of x is 1, the other coefficients are not 1.

 

[Aion]

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Thanks, I think I get it now!