Polynomial Roots
The roots (sometimes called zeroes or solutions) of a polynomial \(P(x)\) are the values of \(x\) for which \(P(x)\) is equal to zero. Finding the roots of a polynomial is sometimes called solving the polynomial.
For example, if \(P(x)=x^2-5x+6\), then the roots of the polynomial \(P(x)\) are \(2\) and \(3\), since both \(P(2)\) and \(P(3)\) are equal to zero.
Contents
Introduction
A polynomial is a special kind of mathematical expression that looks like this:
\[a_n x^n+a_{n-1}x^{n-1}+a_{n-2}+x^{n-2}+\cdots+a_2x^2+a_1x+a_0=\displaystyle\sum_{i=0}^n a_i x^i.\]
If \(a_n\) is not equal to zero, then we say that the polynomial has degree \(n\). According to the fundamental theorem of algebra any polynomial with degree \(n\) has \(n\) complex roots, counted with multiplicity.
Finding Roots
Finding the root of a linear polynomial (a polynomial with degree one) \(ax+b\) is very straightforward. The formula for the root is \(-\frac{b}{a}\) (although calling this a formula is going a bit overboard).
The roots for a quadratic polynomial (a polynomial with degree two) \(ax^2+bx+c\) is given by the formula \[\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}.\]
The formula for the roots of a cubic polynomial (a polynomial with degree three) is a bit more complicated while the formula for the roots of a quartic polynomial (a polynomial with degree four) would fill two blackboards!
What about fifth-degree (quintic) polynomials? What about polynomials with higher degrees?
About \(170\) years ago, a young mathematician by the name of Henrik Abel proved that it is impossible to find a formula for the solutions of a quintic polynomial by adding, subtracting, multiplying, dividing and taking \(n^\text{th}\) roots. More formally speaking, a quintic polynomial is not solvable by radicals.
Henrik Abel
The same is true for polynomials with higher degrees.
Properties
\(1\). If the coefficients of a polynomial are real and if \(a+ib\) is a root of that polynomial, then so is \(a-ib\). See Complex Conjugate Root Theorem.
\(2\). If a polynomial with rational coefficients has \(a + \sqrt{b}\) as a root, where \(a, b\) are rational and \(\sqrt{b}\) is irrational, then \(a − \sqrt{b}\) is also a root.