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

Aion

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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 \(\displaystyle q(x) = a_n\) ?
 
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Subhotosh Khan

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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

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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...
 
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HallsofIvy

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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

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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 \(\displaystyle (x-c_1)^{k_1}(x-c_2)^{k_2}(x-c_3)^{k_3}... = x^n + x^{n-1} + x^{n-2}... \)?
 
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Jomo

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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?
 
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HallsofIvy

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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.
 
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