Cubic & Higher Degree Polynomial Factorization

The sum and difference of cubes frequently finds its way on to standardized exams and competition tests in more specialized areas of mathematics. We could, of course, obtain the difference identity from equation 3 above. Notice that the signing is reversed on the right hand side of the equality when compared to the left hand side. You can verify these two identities for yourself.

$$a^3 \pm b^3 = (a \pm b)(a^2 \mp ab + b^2)$$

Factoring by grouping is a useful technique that you will come across with some frequency. I employed this to prove equation 3 above. Here, we will consider a short collection of examples.

---

---

---

The Rational Roots Theorem states that given a polynomial of any degree of the form $a_nx^n + a_{n-1}x^{n-1} + \cdots + a_1x + a_0 = 0$ then the roots of that polynomial which are rational (if they exist at all) are a factor of $a_0$ over a factor of $a_n$. It is possible that a polynomial has no rational roots.

Proof
Suppose \frac{p}{q} is a rational number expressed in lowest terms such that it solves the polynomial $f(x) = a_nx^n + a_{n-1}x^{n-1} + \cdots + a_1x + a_0$ Then we can write

$f(\frac{p}{q}) = 0$

$\sum_{k=0}^n a_k\left(\frac{p}{q}\right)^k = 0$

$\sum_{k=1}^n a_k\left(\frac{p}{q}\right)^k = -a_0$

$\sum_{k=1}^n a_kq^{n-k}p^k = -a_0q^n$

And now we see, every term on the lefthand side contains 1 or more factors of p. Since p does not divide q because \frac{p}{q} is a rational number expressed in lowest terms, this implies p must divide $a_0$. Now, we repeat the process, but moving over the leading term:

$f(\frac{p}{q}) = 0$

$\sum_{k=0}^n a_k\left(\frac{p}{q}\right)^k = 0$

$\sum_{k=0}^n a_k\left(\frac{p}{q}\right)^k = a_n\left(\frac{p}{q}\right)^n$

$\sum_{k=0}^n a_kq^{n-k}p^k = a_np^n$

Now we notice that every term on the lefthand side contains one or more factors of q. q does not divide p, so q must divide $a_n$.

The complex conjugate theorem states given a polynomial with real number coefficients that any complex solution necessarily comes with its conjugate.

The conjugate of a complex number $a + bi$ is $a – bi$. If you think about this carefully, you’ll note that what you’re observing is the difference of perfect squares in complex numbers. Consider a real quadratic with factor $(x – 1 + i)$. Its other factor must be $(x – 1 – i)$. Thus, the quadratic is $(x – 1 + i)(x – 1 – i) = x^2 – x – ix -x +1 +i +ix -i + 1 = x^2 – 2x + 2$.