When we were playing around with partial fractions we appeared to make a bit of an assumption which was that the only forms that we had to deal with in the denominator of a fraction could always be written as a factor of either linear parts ($a+b x$) or quadratic parts which we could not factor into linear parts ($ax^2+bx+x$) where $b^2-4ac<0$, and of course multiple powers of these, for instance we could have terms like $(a+bx)^3$ in the denominator. How do we know that we can always split a polynomial up into these factors where the coefficients are real? Couldn’t it be for instance that if I gave you a cubic polynomial that all the roots were complex and so I couldn’t factor it in a way that every factor came out with real coefficients? It turns out that the answer is no, but we need a couple more ingredients to prove this.

We said that we were dealing with ratios of polynomials, eg. $\frac{P(x)}{Q(x)}$ but a polynomial where you only have linear and quadratic factors multiplied together doesn’t seem to be very general. How do we know that we can always write a polynomial in this way? Well, it turns out that using complex numbers we can prove this very powerful fact. The statement is:

Theorem:

Every polynomial with real coefficients factors into a product of factors which are either linear with real coefficients or quadratic irreducible with real coefficients.

Quadratic irreducible means that it can’t be split into a product of two linear factors with real coefficients. However, in order to prove this we need to look at a couple of important things. The first is another theorem which we will prove:

Theorem:

The non-real roots of a polynomial equation with real coefficients occur in complex conjugate pairs.

This says that if some complex number $q$ is a root of a polynomial, then so is $\bar{q}$. This sounds rather strange, but let’s prove it in a few simple lines.

Let our polynomial with real coefficients be:

$f(z)=a_nz^n+a_{n-1}z^{n-1}+...a_1z+a_0$

where by definition, the $a_i$ are all real. If $\alpha$ is a root then, by definition $f(\alpha)=0$ (this is what a root means in the context of a polynomial equation). Then we want to show that $f(\overline{\alpha})$ is also a root which simply means that the complex conjugate of a root is also a root. What is $f(\overline{\alpha})$? Well, we just plug it in:

$f(\overline{\alpha})=a_n\overline{\alpha}^n+a_{n-1}\overline{\alpha}^{n-1}+...a_1\overline{\alpha}+a_0$

but we know that the conjugate of a product of complex numbers is the same as the product of the conjugate (ie. $\overline{a}^n=\overline{a^n}$). See we can rewrite the above as:

$f(\overline{\alpha})=a_n\overline{\alpha^n}+a_{n-1}\overline{\alpha^{n-1}}+...a_1\overline{\alpha}+a_0$

but we also said that the coefficients were real, and the conjugate of a real coefficient is just that coefficient, so we can “expand” the conjugation:

$f(\overline{\alpha})=\overline{a_n\alpha^n}+\overline{a_{n-1}\alpha^{n-1}}+...\overline{a_1\alpha}+\overline{a_0}$

and furthermore, the sum of a set of conjugates is the conjugate of the sum:

$f(\overline{\alpha})=\overline{a_nz^n+a_{n-1}z^{n-1}+...a_1z+a_0}$

but this is just the same as:

$f(\overline{\alpha})=\overline{f(\alpha)}=\overline{0}=0$

So we have shown that if $\alpha$ is a root, then so is $\overline{\alpha}$. This means that any time you find a complex root of an equation, it must have another complex root which is the conjugate of the first root. In fact, for the case of a quadratic equation we can see this immediately because if $a$, $b$ and $c$ are real, then the only place that an imaginary number can come into the root is from the part which is $\pm\sqrt{b^2-4ac}$ which means that the two solutions, if $b^2-4ac<0$, are going to be conjugate of each other (the $+$ part and the $-$ part). We have just shown that this is true also for higher order polynomials. In fact this is true for any polynomial at all.

There’s a second crucial theorem that we need to show that we can factor any polynomial into a product of linear and irreducible quadratic factors. This is the Fundamental Theorem of Algebra:

Theorem: The fundamental theorem of algebra

Every polynomial (with real OR complex coefficients) which is of degree at least one, has a zero.

This says that there is always a solution to any polynomial (equaling zero) which is order more than 0. A zeroth order polynomial is just a constant and so it’s clear that this will not have a solution which is equal to zero, unless the constant is zero.

We will not prove the Fundamental Theorem of algebra but we will use it to prove our initial statement about factoring polynomials.

Let’s take a polynomial $P(z)$. by the Fundamental Theorem, it has at least one root, let’s call it $r_1$. That root can either be real or complex. If it’s real, then we know that the Polynomial can be written as $(z-r_1)\tilde{P}(z)$ where $\tilde{P}(z)=\frac{P(z)}{(Z-r_1)}$. We can always divide a polynomial by another to get another polynomial as long as the order of the polynomial on top is greater than that on the bottom.

The other option for our root is that it is complex. If it is complex then its conjugate is also a root. This means that $(z-r_1)(z-\bar{r}_1)$ is a factor of the polynomial, but this is purely real. We can see this by noting that if $r_1=a+bi$ then

$(z-r_1)(z-\bar{r}_1)=(z-a-bi)(z-a+bi)=((z-a)-bi)((z-a)+bi)=((z-a)^2+b^2)$

which is an irreducible quadratic (ie. we can’t write it as the product of linear factors). In this case we can write:

$P(z)=((z-a)^2+b^2)\tilde{P}(z)$.

Then we can take $\tilde{P}(z)$ and play exactly the same game. In fact we can do this until we are left with just a constant and this clearly can’t be written as the product of linear and quadratic factors – it’s as simple as it can be.

So we have proved our initial claim using both the fundamental theorem of algebra and the fact that roots always come in conjugate pairs. This fact allows us to play with partial fractions as we did before, so it’s really a very powerful statement.

Let’s just see where that gets us. Now if I give you the polynomial $2z^3-9z^2+14z-5$ and tell you that it has a zero at $z=2-i$, then you can immediately factor this cubic. The first thing you know for sure is that $z=2+i$ is also going to be a root. Now you can take this function and divide by $(z-(2-i))(z-(2+i))=((z-2)^2+1)$ This gives:

$\frac{2z^3-9z^2+14z-5}{((z-2)^2+1)}=(2z-1)$

and so:

$2z^3-9z^2+14z-5=((z-2)^2+1)(2z-1)$

We have factored the polynomial into a linear factor and an irreducible quadratic factor. Now we know the solutions to this equation.

 How clear is this post?