# Cubic polynomials and their roots

- see how Descartes’ factor theorem applies to cubic functions
- review how to find zeroes of a cubic in special situations.

## Zeroes of a cubic polynomial

One simple way of cooking up a cubic polynomial is just to take a product of linear factors, for example \[\Large{y=(x-4)(x-1)(x+2).}\]We hope that you can all expand the right-hand side to get \[\Large{y=x^3-3x^2-6x+8.}\]The graph of this function is shown hereNote that it does pass through the points \(\normalsize{[4,0]}\), \(\normalsize{[1,0]}\) and \(\normalsize{[-2,0]}\), as it should by Descartes’ Factor theorem.## Factorising simple cubics

Here is a simple cubic polynomial that has been chosen to have a nice factorisation: \(\normalsize{f(x)=x^3-7x+6}.\)Let us note that the curve passes through the points \(\normalsize{[1,0]}\), \(\normalsize{[2,0]}\) and \(\normalsize{[-3,0]}\). This corresponds to the fact that \(\normalsize{f(1)=f(2)=f(-3)=0}\). We say that \(\normalsize{1}\), \(\normalsize{2}\) and \(\normalsize{-3}\) are the*zeroes*or

*roots*of \(\normalsize{f(x)}\).The zeroes of a polynomial \(\normalsize{f(x)}\) have a particular importance: from Descartes’ theorem if a polynomial \(\normalsize{p(x)}\) has a zero \(\normalsize{r}\), that is \(\normalsize{p(r)=0}\), then it follows that \(\normalsize{(x-r)}\) is a factor of \(\normalsize{p}\).In our example, since \(\normalsize{f}\) has zeroes \(\normalsize{1}\), \(\normalsize{2}\) and \(\normalsize{-3}\), we know that \(\normalsize{f(x)}\) has factors \(\normalsize{(x-1)}\), \(\normalsize{(x-2)}\) and \(\normalsize{(x+3)}\). And indeed \[\Large{f(x)=(x-1)(x-2)(x+3)}.\]So in practice trial and error may be used when looking for possible zeroes by hand. It is a good guess to try factors of the constant term if the polynomial is monic (that means that the highest degree coefficient is \(\normalsize{1}\)). Of course using a computer is much simpler: our electronic friends are particular adept at this kind of work!

Q1(M): Try to factor \(\normalsize{p(x)=x^3-3x^2+4x-70}\).

Q2(M): Find zeroes and factorisations of the following cubics:a) \(\normalsize{g(x)= x^3-4x}\)b) \(\normalsize{h(x)=x^3+4x^2-x-4}\)c) \(\normalsize{k(x)= 2x^3-7x^2-14x-5}\)

## Tartaglia, Cardano and factoring general cubics

*complex numbers*in an essential way, and also both square roots and cube roots. However it is not for the faint-hearted, and it is fair to say that these days it is rarely used.

^{Girolamo Cardano, By Wellcome Library, London, CC BY 4.0, via Wikimedia Commons}

## Answers

A1.Let’s look at this problem in some detail. To factor \(\normalsize{p(x)=x^3-3x^2+4x-70}\) one should look for zeroes which are factors of \(\normalsize{70}\), which are built from \(\normalsize{2,5}\) and \(\normalsize{7}\), and we should not forget about negative possible factors. Now with a bit of trial and error you can find that \(\normalsize{p(5)=0}\), so by Descartes’ theorem we know that \(\normalsize{(x-5)}\) is a factor.After that we want to actually divide \(\normalsize{p(x)}\) by \(\normalsize{(x-5)}\). This requires polynomial long division. If you have forgotten how to do long division, you will be happy to learn that polynomial long division is generally simpler than that with numbers.By looking at successive powers of x, we sequentially determine that the other factor must be \(\normalsize{x^2+2x+14}\). Please try it! The final factoring is\[\Large{p(x)=(x-5)(x^2+2x+14)}.\]A2.a) We have \(\normalsize{g(x)= x^3-4x=x(x-2)(x+2)}\).b) We have \(\normalsize{h(x)=x^3+4x^2-x-4=(x-1)(x+1)(x+4)}\).c) We have \(\normalsize{k(x)= 2x^3-7x^2-14x-5=(x-5)(x+1)(2x+1)}\).

#### Maths for Humans: Linear, Quadratic & Inverse Relations

## Our purpose is to transform access to education.

We offer a diverse selection of courses from leading universities and cultural institutions from around the world. These are delivered one step at a time, and are accessible on mobile, tablet and desktop, so you can fit learning around your life.

We believe learning should be an enjoyable, social experience, so our courses offer the opportunity to discuss what you’re learning with others as you go, helping you make fresh discoveries and form new ideas.

You can unlock new opportunities with unlimited access to hundreds of online short courses for a year by subscribing to our Unlimited package. Build your knowledge with top universities and organisations.

Learn more about how FutureLearn is transforming access to education