Recover and factor polynomials

Count polynomial roots with multiplicity

Check the coefficient type before using conjugate pairs. Count repeated roots with their multiplicities, then use known roots to build factors and reduce the degree.

A polynomial of degree $n\ge1$, with complex coefficients and nonzero leading coefficient $a_n$, has exactly $n$ complex roots counting multiplicity.

Some roots can repeat; the statement does not promise $n$ distinct roots.

For a polynomial with real coefficients, a nonreal root $p$ is accompanied by its conjugate $p^*$.

A real root equals its own conjugate and does not require a second distinct root.

To see why the real-coefficient condition matters, start with

$$a_np^n+a_{n-1}p^{n-1}+a_{n-2}p^{n-2}+\cdots+a_1p+a_0=0.$$

Conjugate both sides and use the sum, product and power rules:

$$\begin{aligned}(a_np^n+a_{n-1}p^{n-1}+a_{n-2}p^{n-2}+\cdots+a_1p+a_0)^*&=0^*,\\(a_np^n)^*+(a_{n-1}p^{n-1})^*+(a_{n-2}p^{n-2})^*+\cdots+(a_1p)^*+a_0^*&=0,\\a_n^*(p^n)^*+a_{n-1}^*(p^{n-1})^*+a_{n-2}^*(p^{n-2})^*+\cdots+a_1^*p^*+a_0^*&=0. \end{aligned}$$

Each coefficient is real, so $a_k^*=a_k$. Therefore

$$a_n(p^*)^n+a_{n-1}(p^*)^{n-1}+a_{n-2}(p^*)^{n-2}+\cdots+a_1p^*+a_0=0.$$

Thus $p^*$ is a root of the same polynomial.

Check the coefficient condition before assuming a conjugate root

If any coefficient is nonreal, the proof cannot replace every $a_k^*$ by $a_k$; the conjugate may fail to be a root. The source counterexample is $3z^2+iz-2=0$. The quadratic formula gives

$$\begin{aligned}z&=\frac{-i\pm\sqrt{i^2-4(3)(-2)}}{2(3)}\\&=\frac{-i\pm\sqrt{23}}6\\&=\frac{\sqrt{23}}6-\frac i6\quad\text{or}\quad-\frac{\sqrt{23}}6-\frac i6. \end{aligned}$$

Both imaginary coefficients are $-1/6$, so these two roots are not a conjugate pair.

Combine conjugate factors into a real quadratic

For real $x,y$, conjugate roots $x+iy$ and $x-iy$ give

$$\begin{aligned}[z-(x+iy)][z-(x-iy)]&=z^2-z(x+iy)-z(x-iy)+(x+iy)(x-iy)\\&=z^2-2xz+(x^2+y^2). \end{aligned}$$

This quadratic has real coefficients.

Pairing all nonreal roots this way shows that every nonconstant real polynomial factors over the reals into real linear and quadratic factors, together with its nonzero leading constant.

A real cubic has either three real roots or one real root and a nonreal conjugate pair, counting multiplicity.

Example 10 uses the pair $2\pm i$ to form $z^2-4z+5$, then long division gives the remaining factor $z+1$.