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

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

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

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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