Further Pure | Topic 1: Polynomials and Rational Functions (Part 1)

Curriculum Objectives:

• recall and use the relations between the roots and coefficients of polynomial equations, for equations of degree 2, 3, 4 only;
• use a given simple substitution to obtain an equation whose roots are related in a simple way to those of the original equation;

Relation between roots of and coefficients of polynomial equation of degree 2

We begin by recalling that the general form of a polynomial equation of the second degree, or a quadratic equation is $p(x) = a{x^2} + bx + c$, where a, b, c are constants. The roots of this polynomial are then given by $x = \frac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$.

Let $\alpha $ and $\beta $ be the two roots of this polynomial equation, i.e.

$\alpha  = \frac{{ - b + \sqrt {{b^2} - 4ac} }}{{2a}}$

$\beta  = \frac{{ - b - \sqrt {{b^2} - 4ac} }}{{2a}}$

By then adding and multiplying these roots, we find an interesting general result for the quadratic equation, that can be easily extended to equations of any degree. We shall particularly see this for the degrees 3 and 4, later on.

First considering the sum of the roots:

$\alpha  + \beta  = \frac{{ - b + \sqrt {{b^2} - 4ac} }}{{2a}} + \frac{{ - b - \sqrt {{b^2} - 4ac} }}{{2a}}$

$ = \frac{{ - b + \sqrt {{b^2} - 4ac}  + ( - b) - \sqrt {{b^2} - 4ac} }}{{2a}}$

$ = \frac{{ - 2b}}{{2a}}$

$ = \frac{{ - b}}{a}$

Then considering the product of the roots:

$\alpha  \times \beta  = \frac{{ - b + \sqrt {{b^2} - 4ac} }}{{2a}} \times \frac{{ - b - \sqrt {{b^2} - 4ac} }}{{2a}}$

$ = \frac{{{{( - b)}^2} - {{\left( {\sqrt {{b^2} - 4ac} } \right)}^2}}}{{4{a^2}}}$

$ = \frac{{{b^2} - {b^2} + 4ac}}{{4{a^2}}}$

$ = \frac{{4ac}}{{4{a^2}}}$

$ = \frac{c}{a}$

So to summarise, for the polynomial $p(x) = a{x^2} + bx + c$ with the roots $\alpha $ and $\beta $:

Sum of the roots $ = \frac{{ - b}}{a}$

Product of the roots $ = \frac{{ c}}{a}$

I was unable to find a question from the past papers for the topic (they are generally on equations of 3 or 4 order), hence the following example shall have to suffice:

Ex: Given that the equation $4{x^2} + 7x - 5 = 0$ has the roots $\alpha $ and $\beta $, find the equation such that its roots are ${\alpha ^2}$ and ${\beta ^2}$.

Sol:

Sum of the roots $ = \frac{{ - b}}{a}$ $ = \frac{{ - 7}}{4}$

Product of the roots $ = \frac{{ c}}{a}$ $ = \frac{{ -5}}{4}$

To find ${\alpha ^2} + {\beta ^2}$ we may use the relationship ${\left( {\alpha  + \beta } \right)^2} = {\alpha ^2} + 2\alpha \beta  + {\beta ^2}$.

Hence  ${\left( {\alpha  + \beta } \right)^2} - 2\alpha \beta  = {\alpha ^2} + {\beta ^2}$

We know that the sum of the roots and the product.

${\left( {\frac{{ - 7}}{4}} \right)^2} - 2\left( {\frac{{ - 5}}{4}} \right) = {\alpha ^2} + {\beta ^2}$

${\alpha ^2} + {\beta ^2} = \left( {\frac{{49}}{{16}}} \right) + \left( {\frac{{10}}{4}} \right)$

${\alpha ^2} + {\beta ^2} = \frac{{89}}{{16}}$

For ${\alpha ^2}{\beta ^2}$, we have

${\alpha ^2}{\beta ^2} = {(\alpha \beta )^2} = {\left( {\frac{{ - 5}}{4}} \right)^2} = \frac{{25}}{{16}}$

Hence the equation with the roots ${\alpha ^2}$ and ${\beta ^2}$ is:

${x^2} - \frac{{89}}{{16}}x + \frac{{25}}{{16}} = 0$

$16{x^2} - 89x + 25 = 0$

Relation between roots of and coefficients of polynomial equation of degree 3 and 4

Consider the following cubic equation (a polynomial equation of the third degree):

$p(x) = a{x^3} + b{x^2} + cx + d$, further assume that the roots of the equation are $\alpha $, $\beta $, and $\delta $. It is evident then:

$a{x^3} + b{x^2} + cx + d \equiv k(x - \alpha )(x - \beta )(x - \delta )$

Expanding the right hand side of the identity, and simplifying yields the following:

$a{x^3} + b{x^2} + cx + d \equiv k{x^3} - k(\alpha  + \beta  + \delta ){x^2} + k(\alpha \beta  + \beta \delta  + \alpha \delta )x - k\alpha \beta \delta $

As both sides are identical we can compare the coefficients, and derive the general result for cubic equations. I use the following notation:

$\sum \alpha  $ for the sum of the roots.

$\sum {\alpha \beta } $ for the sum of the product of the roots two at a time.

Thence we can generalise:

$\sum \alpha   = \frac{{ - b}}{a}$

$\sum {\alpha \beta  = \frac{c}{a}} $

$\alpha \beta \delta  = \frac{{ - d}}{a}$

Furthermore, in the same vein it can be proved the roots of the quartic polynomial($\alpha $, $\beta $, $\delta $, and $\gamma$) (a polynomial equation of the fourth degree): $a{x^4} + b{x^3} + c{x^2} + dx + e = 0$ satisfy the following relationships

$\sum \alpha   = \frac{{ - b}}{a}$

$\sum {\alpha \beta  = \frac{c}{a}} $

$\sum \alpha  \beta \delta  = \frac{{ - d}}{a}$

$\alpha \beta \delta \gamma  = \frac{e}{a}$

An Important Result

The following result is an indispensable one; one that you will most likely encounter in the past papers.

${\left( {\sum \alpha  } \right)^2} = \sum {{\alpha ^2} + 2\sum {\alpha \beta } } $

We used this result when solving the example above where only two roots were involved, but this result will hold for three and four roots as well.

Using substitutions to find equations of roots related to the roots of an equation already given.

The general method is outlined in the following example:

Ex: Given the equation ${x^3} - 3{x^2} + 4 = 0$, find the equation such that its roots are twice that of the equation given.

Sol:

Let x be a root of the equation given, and y the root of the equation to be found. They are obviously related thusly:

$y = 2x$

$\frac{y}{2} = x$

Substituting this into the given equation would then yield the equation required:

${\left( {\frac{y}{2}} \right)^3} - 3{\left( {\frac{y}{2}} \right)^2} + 4 = 0$

Which can then be simplified to give:

$y - 6y + 32 = 0$

 

You may use this method to check the answer obtained in the first example at the start of the chapter. The substitution to be used there would be $x^2 = y$ => $x = {y^{\frac{1}{2}}}$
 

An important strategy to deal with such questions:

The roots of an equation satisfy a recurrence relationship defined by their own equation, which in the crudest of terms means that one result can be used to derive another, which in turn maybe used to find another. Perhaps it is better understood directly with an example.

The following example has been taken from the past paper: October/November 2013 Paper 12 Q2:

Ex: The cubic equation ${x^3} - px - q = 0$, where p and q are constants, has the roots $\alpha $, $\beta $ and $\gamma $. Show that:
i) ${\alpha ^2} + {\beta ^2} + {\gamma ^2} = 2p$
ii) ${\alpha ^3} + {\beta ^3} + {\gamma ^3} = 3q$
iii) $6({\alpha ^5} + {\beta ^5} + {\gamma ^5}) = 5({\alpha ^3} + {\beta ^3} + {\gamma ^3})({\alpha ^2} + {\beta ^2} + {\gamma ^2})$

Sol:

i) The result stated above will be used here:

From the equation we see that the coefficient of $x^2$ is zero therefore:

$\sum {\alpha  = 0} $
$\sum {\alpha \beta  =  - p} $

We have that: ${\left( {\sum \alpha  } \right)^2} = \sum {\alpha  + 2\sum {\alpha \beta } } $
Putting in the data the first part is easily proven.
Here you can see that the substitution strategy fails. While it is the easier one and perhaps less prone to errors, you will find it may not work all the time.

ii) For this we have many method at our disposal. Some relevant identity that links the roots to their cube may be used. The method of substitution maybe used here with the relevant substitution being $x^3 = y$
I however would like to use the recurrence method mentioned. Consider the original equation. The root of the equation is such that once substituted into the equation yield a zero. Which is to say that:

${\alpha ^3} - p\alpha  - q = 0$

We can make similiar equations for the other roots. We can then add them to yield the following equation:
${\alpha ^3} + {\beta ^3} + {\gamma ^3} - p\alpha  - p\beta  - p\gamma  - q - q - q = 0$
But this perhaps better written as:
$\sum {{\alpha ^3} - p\sum \alpha   - 3q = 0} $

Putting in the relevant values(the sum of the roots is zero) we can prove the second part.

3) For this recurrence method is used once again.

If make $x^3$ the subject of the equation and then multiplying through out by $x^2$, we can make the following equation.

${x^5} = p{x^3} + q{x^2}$

Following the reasoning presented in the previous part, this means:

$\sum {{\alpha ^5} = p\sum {{\alpha ^3} + q\sum {{\alpha ^2}} } } $

Putting in relevant values I get that

$\sum {{\alpha ^5} = 5pq} $

Having found this the required result is easily proved by putting in the relevant values.

The next question is from the May/June 2013 session, paper 12

Q. The cubic equation $x^3 - 2x^2 - 3x + 4 = 0$ has the roots $\alpha $, $\beta $ and $\gamma $. Find the sum of the roots, denoted by c.
Sol:  2, from the equation
i) Use the substitution y = c - x to find an equation with the roots $\alpha  + \beta ,\alpha  + \gamma ,\beta  + \gamma $
Sol: $y = c - x$ => $x = c - y$ => $x = 2 - y$

Making this substitution:
${(2 - y)^3} - 2{(2 - y)^2} - 3(2 - y) + 4 = 0$

Which can be simplified to give:
$y^3 - 4y^2 + y + 2 = 0$

ii) Find the equation with the roots $\frac{1}{{\alpha  + \beta }},\frac{1}{{\beta  + \gamma }},\frac{1}{{\alpha  + \gamma }}$
Sol: Using the substitution $z = \frac{1}{y}$ => $y = \frac{1}{z}$ the result can be easily found

iii) Find the value of the sum: $\frac{1}{{{{\left( {\alpha  + \beta } \right)}^2}}} + \frac{1}{{{{\left( {\beta  + \gamma } \right)}^2}}} + \frac{1}{{{{\left( {\alpha  + \gamma } \right)}^2}}}$

Sol: the result involving the squares of the sum of roots, and the sum of the squares of the roots, given above is to be used here.