Finding Roots of Polynomials

For a linear polynomial and quadratic polynomial, we have no difficulty in finding the roots. But for cubic and higher degrees either the formula to find roots is too complex or it does not exist. Computers can utilize numerical methods to calculate the roots of polynomials with high precision. However, there are times when we might want to determine the roots or gain insight into their nature without resorting to numerical computations. For such purposes, various analytical methods are available. We will explore a selection of these techniques that enable us to understand the roots of polynomials and, in certain cases, find them explicitly without relying on numerical approximations.

Searching roots by hit and trial

Theorem: For a monic polynomial with integer coefficients, if there is an integer root, then it must be one of the factors of the coefficient with degree zero. Let \( p(x) = x^n + a_{n-1}x^{n-1} + a_{n-2}x^{n-2} + ... + a_n \). Then by Vieta’s Formulae, the product of all roots is equal to \( (-1)^n a_n \). Thus the integer roots are the factors of \( a_n \).

So by hit and trial, we try to search for the roots by putting factors of \( a_n \) into the polynomial and checking whether we are getting a zero or not.

e.g. \( p(x) = x^3 - 3x^2 + 4 \), then if there is an integer root it must be the factor of 4. So we try putting \( \pm1, \pm2, \pm4 \) (these are the possible factors of 4)

By hit and trial, we see both \( -1 \) and \( 2 \) satisfy the equation. So \( (x + 1)(x - 2) = x^2 - x - 2 \) divides \( p(x) \). Using long division, we divide \(p(x)\) by \(x^2-x-2\), we get \(x-2\) as the quotient. Thus, \(p(x)=(x-2)^2(x-1)\).

In general, we use the following method to find rational roots.

Rational roots of the polynomial

Theorem: If there exists a rational root \(\frac{p}{q}\) (where \(p, q \in \mathbb{Z}\) and are coprime to each other) of a polynomial \(P(x) = a_0x^n + a_1x^{n-1} + a_2x^{n-2} + ... + a_n\), where coefficients \(a_0, a_1, a_2, ..., a_n\) are real numbers, then \(p\) is a factor of \(a_n\), and \(q\) is a factor of \(a_0\).

Since, if \(\frac{p}{q}\) is a root of \(P(x)\), then \(P\left(\frac{p}{q}\right) = a_0\left(\frac{p}{q}\right)^n + a_1\left(\frac{p}{q}\right)^{n-1} + a_2\left(\frac{p}{q}\right)^{n-2} + ... + a_n = 0\)

\[ \Rightarrow a_0p^n + a_1pq^{n-1} + a_2pq^{n-2} + ... + a_nq^n = 0 \quad ...(1) \]

\[ \Rightarrow a_1p^{n-1}q + a_2p^{n-2}q^2 + ... + a_nq^n = -a_0p^n \]

\[ \Rightarrow a_1p^{n-1} + a_2p^{n-2}q + ... + a_nq^{n-1} = \frac{-a_0p^{n}}{q} \]

Since the left-hand side is an integer value, the right-hand side is also an integer. Clearly, \(q\) cannot divide \(p\), thus \(q\) divides \(a_0\). Similarly, if we rewrite equation (1) as:

\[ \Rightarrow a_0p^n + a_1pq^{n-1} + a_2pq^{n-2} + ... + a_{n-1}pq^{n-1} = -a_nq^n \]

\[ \Rightarrow a_0p^{n-1} + a_1pq^{n-2} + a_2pq^{n-3} + ... + a_{n-1}q^{n-1} = \frac{-a_nq^n}{p} \]

Again, the left-hand side is an integer value, the right-hand side is an integer. Since \(p\) cannot divide \(q\), thus \(p\) divides \(a_n\).

Let us understand by an example how we can implement the above theorem to find, if not all, but some of the rational roots if they exist.

Example: Given a polynomial \(6x^3 -11x^2 -4x + 4\). Search for its rational roots.

Solution: Possible factors of 6 are 1, 2, 3, and 6 and of 4 are 1, 2, and 4.

Thus possible rational roots are \( \pm1, \pm\frac{1}{2}, \pm2, \pm\frac{2}{3}, \pm4, \pm\frac{1}{3}, \pm\frac{1}{6} \) (observe that we have put \( \pm \) sign).

We will check one by one each possibility and stop as soon as we get the first root.

We find that only \(\frac{1}{2}, -\frac{2}{3}, \) and \( 2 \) satisfy the equation. We need not find all the roots. As soon as we had found the first root, we must have stopped and found the rest of the roots by using long division and finding the quadratic factor which can then be factorized.