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

#### Aion

##### New member
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 $$\displaystyle q(x) = a_n$$ ?

Last edited:

#### Subhotosh Khan

##### Super Moderator
Staff member
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

##### New member
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...

Last edited:

#### HallsofIvy

##### Elite Member
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

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

Last edited:

#### Jomo

##### Elite Member
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?

Last edited:

#### HallsofIvy

##### Elite Member
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.

Last edited:

#### Aion

##### New member
Thanks, I think I get it now!

Last edited: