Cubics

Cubic Equations

To the uninitiated, cubics would appear to be a simple step up from quadratics. However, they are not. The general method for

finding a real solution for a cubic equation was discovered by the Italians in the 16th century when algebraic methods were spreading

throughout Europe. Since a cubic equation always has at least one real root, this method was satisfactory for a while. However when

the Fundamental Theorem of Algebra was confirmed in the 18th century it became necessary to establish complete solutions of cubic

equations that took into account all real and nonreal solutions.

The next two problems allow you to play with the sum and difference of two cubes formulas and cubing a binomial.

is a number which satisfies , find the value of .

If you can establish one root of a cubic for certain, you can use Synthetic Division to reduce the original equation to a linear factor

times a quadratic and then apply the Quadratic Formula to find the remaining solutions. The next two problems take advantage of this fact.

STEP 2007, Math I, #8: A curve is given by the equation

(*)

where

is a real number. Show that this curve touches the curve with equation

at . [Hint: a good place to start is by using an interactive drawing program like Mathematica to graph the function for

various values of a and to notice that the point (2,8) is always a point of intersection.]

Determine the coordinates of any other point of intersection of the two curves.

(i) Sketch on the same axes the curves (*) and (**) when

.

(ii) Sketch on the same axes the curves (*) and (**) when

.

(iii) Sketch on the same axes the curves (*) and (**) when

.

Rewriting a cubic by guessing its Final Form

The next problem involves a clever tactic frequently used when you know a little bit about a polynomial. The idea is to guess the form of the

factors and then work backwards. In this problem, we first remove the known factor of (x-3) then we guess that the two remaining factors are

linear in x and y and will take the form of (x + ay + b) and (x + cy + d). This is reasonable based on the shape of the cubic itself. Then we

multiply out these factors and solve some non-linear equations to find out what the a, b, c, and d are.

STEP 2004, Math I, #3: (i) Show that

is a factor of

STEP 2011, Math III, #3: Show that, provided the polynomial

can be written in the form

and are the roots of the quadratic equation in given by

and . That will enable you to completely rewrite the polynomial.]

Hence show that one solution of the equation is

can be written in the form , where

, and are real, and are both non-zero, and . Show that

and determine a similar expression involving . Hence show that

and that . Deduce that either or the two equations are identical.

STEP 2010, Math II, #7: (i) By considering the positions of its turning points, show that the curve with equation

and , crosses the x-axis once only.

(ii) Given that

. Hence find the real root of the cubic equation in the case , .

(iii) The quadratic equation

and . Show that

It is given that one of these roots is the square of the other. By considering the expression

find a relationship between

, and is real, determine the value of in terms of .

The next problem does not discuss how to solve a cubic but how to gain certain information about the nature of its real solutions

without actually finding them explicitly.

STEP 2013, Math II, #3: (i) Given that the cubic equation has three distinct

real roots and

, show with the help of sketches that either exactly one of the roots is positive or all three

of the roots are positive.

(ii) Given that the equation has three distinct real positive roots show that

, , (*)

[Hint: Consider the turning points.]

(iii) Given that the equation has three distinct real roots and that

, and ) that it is possible for the conditions

(*) to be satisfied even though the corresponding cubic equation has only one real root.