Skip to 0 minutes and 14 seconds In the linear and the quadratic situations, we saw that a general linear function, or a general quadratic function, could be obtained from y equals x, or y equals x squared, by scaling and translating appropriately. You might think the same is true for a cubic function, that we could take our basic cubic function, y equals x cubed, and just scale it and translate it, and we would get a general cubic. But it’s not quite so simple in the cubic world. A general cubic actually has quite a different shape from this special y equals x cubed cubic. This is more typical of the shape of a cubic. It’s still very pleasant and regular.

Skip to 0 minutes and 56 seconds It’s still actually very important in design issues, like with Bezier curves, but it’s given by a more general equation. Here’s the general form for a cubic function, y equals a x cubed plus b x squared plus cx plus d. So there’s four variables that can be introduced there, four degrees of freedom. When we’re studying a cubic function like this, one of the most important questions is, does it have 0’s and if so, where are they? If there are 0’s, as in this case, alpha, beta, gamma– that means x-intercepts– then by Descartes, we can rewrite the cubic, or try to at least, in the form a times x minus alpha, x minus beta, x minus gamma.

Skip to 1 minute and 39 seconds So each of these 0’s contributes a linear factor to the expression of the cubic. It’s important to realise though that the world of cubic functions actually extends beyond these particular forms. Now you may remember that when we were talking about quadratics, that quadratic functions geometrically always turned out to be parabolas. But in fact, second degree curves, more generally, also included ellipses and hyperbolas. So similarly, in the cubic situation, if we move to more general cubics that are given by arbitrary polynomials in x and y of degree 3, then we get a wide variety of other kinds of curves not just ones that look like this. So cubic analogues of ellipses and hyperbolas, whatever that might be.

Skip to 2 minutes and 33 seconds Newton and other 17th century mathematicians were very interested in these new kinds of cubics that they could study and explore using this Cartesian framework. For example, Fermat, one of the founders of the subject, studied x cubed plus y cubed equals 1, kind of a cubic catalogue of the unit circle. And he determined that it has a shape, something like this.

Skip to 3 minutes and 3 seconds That’s the Fermat curve, x cubed plus y cubed equals 1.

# Cubic curves II: cubic functions and their graphs

In this video we discuss cubic curves and cubic functions.

## General cubic functions

The general cubic function has the form \(\normalsize{y=ax^3+bx^2+cx+d}\) and has a somewhat different shape to the standard cubic \(\normalsize{y=ax^3}\). We discuss the general form of such functions, and the relation with any zeroes it might have: there are at most three zeroes, but a general cubic need not have all three zeroes, even approximately.

## More general cubic curves

Isaac Newton investigated more general cubic curves, given by arbitrary degree-three polynomials in \(\normalsize{x}\) and \(\normalsize{y}\). Earlier, both Fermat and Descartes had investigated special cubics. The theory of these kinds of curves is rich and full of surprises.

## Zeroes and coefficients

The zeroes of a polynomial, if they are known, and the coefficients of that polynomial are two different sets of numbers that have interesting relations. If we know the zeroes, then we can write down algebraic expressions for the coefficients. Going the other way is much harder and cannot be done in general. But in special cases, it can be done.

Q1(M): What is the factorisation of \(\normalsize{f(x)=x^3+2x^2-5x-6}\)?

One trick that is useful is that if we have a monic cubic (meaning that the coefficient of the highest order term is 1), such as

\[\Large{f(x)= x^3-5x^2-29x+105}\]and it does factor into integer factors, say

\[\Large{f(x)=(x-a)(x-b)(x-c)}\]then by expanding this out we must have that

\[\Large{a+b+c=5}\] \[\Large{ab+bc+ac=-29}\] \[\Large{abc=-105}.\]In particular if \(\normalsize{a}\), \(\normalsize{b}\), \(\normalsize{c}\) are integers and we want to solve for them, we should try various numbers that are factors of \(\normalsize{105}\), including negative integers. So we might try \(\normalsize{3,5,7,15,21,35}\) and their negatives. In fact it is the case that \(\normalsize{-105 = 3 \times (-5) \times 7}\), and

\[\Large{f(x)= x^3-5x^2-29x+105=(x-3)(x+5)(x-7)}.\]Hopefully this hint should make finding the factorisation in Q1 easier.

## Answers

\[\Large{f(x)=x^3+2x^2-5x-6=(x+1)(x-2)(x+3)}.\]

A1.The factorisation is

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