polynomials cie centre a-level further pure maths

PolynomialsPolynomials

CIE Centre A-level CIE Centre A-level Further Pure MathsFurther Pure Maths

Polynomials AgainPolynomials Again

Recall the basic structure of the cubic polynomial. If

3 2 0ax bx cx d and:

( )( )( ) 0x x x

Then: 3 2( ) ( ) 0x x x

Sum of the roots =

Product of the roots =

Pairwise product of the roots =

b

a

d

a

c

a

Note why we have divided through by a.Note also the signs (+ or -) for each term.

http://en.wikipedia.org/wiki/Image:Descartes.jpg


Let’s put it in a broader context:

1

( )( ) 0

( )( )( ) 0

( )( )( )( ) 0

...

( ) 0n

ii

x x

x x x

x x x x

x r

We can always set the leading coefficient to 1 by dividingthrough by a. Q: What is the pattern in the second coefficient (b)?Q: What is the pattern in the trailing coefficient?

2

3 2

4 3 2

11 0

0

0

0

...

... 0n nn n

x bx c

x bx cx d

x bx cx dx e

a x a x a



Answers:The coefficient of xn-1 is always -1 x the sum of the roots(assuming you have made the leading coefficient 1).The constant term, or trailing coefficient = (-1)n x the product of the roots.The other coefficients can also be expressed in terms of the roots, but their form is obviously more complicated.

Now let’s start to look at how we can manipulate polynomials to find interesting results.


3 22 9 4 0x x x The roots of

are denoted as: , ,

Write down a polynomial with roots: ( 2), ( 2), ( 2)

Think …

Let ' 2. We know that solves the given

polynomial. So write ' 2 and substitute this

into the equation. This will generate a new polynomial

which is solved by '; and the same for ', '.

Now write down the required polynomial.


3 2

3 2

3 2 2

3 2

2 9 4 0

( ' 2) 2( ' 2) 9( ' 2) 4 0

' 6 ' 12 ' 8 2 ' 8 ' 8 9 ' 18 4 0

' 4 ' 5 ' 14 0

x x x

3 2

Exactly the same calculation can be performed

for ', '. Let's write the new variable as y:

4 5 14 0

where 2.

y y y

y x


We solved this problem using substitution. The method can be applied where the new roots are related to the old roots through an invertible function:

1

' ( ), ' ( ), ' ( )

( ), ( )

f f f

y f x x f y

Now we will look at a more difficult probleminvolving the roots of a cubic…


3 23 6 10 0x x The roots of

are denoted as: , ,

Find the value of: 7 7 7

Investigate!



3 23 6 10 0x x

Factorization – not possible

Sketching the graph might help

From this we can only see that there isone real root > 2; but we need exact values.

This curious question can actually be answered bydeducing a recurrence relation for the sum of thepowers of the roots. Let’s see how it works.

-20

-15

-10

-5

0

5

10

15

20

-1.5 -1 -0.5 0 0.5 1 1.5 2 2.5 3



3 2 0ax bx cx d We know that:

for each root. So we can write three equations:

3 2

3 2

3 2

0

0

0

b c d

a a ab c d

a a ab c d

a a a

1

2

3

X

X

X

n

n

n



So by adding the equations together vertically we get:

3 3 3

2 2 2

1 1 1

0

n n n

n n n

n n n

n n n

b

ac

ad

a

The terms in {} all have the same structure. Let’s call this “sum of powers of roots” Sn, i.e.:

, n n nnS n

Rewriting the equation using Sn gives the more compact form onthe next slide:



3 2 1 0n n n n

b c dS S S S

a a a

Q: What is S0 ?

A: 3.

We therefore have a “recurrence relation” definingan inductive sequence. But to find S7 it is not enough to know S0!

Q: What are S1 and S2 ?



1

bS

a

S1 (and slightly less easily, S2) can be calculated fromthe previous equations:

2 2 22 ...?S

2 2 2 22

2

( ) 2( )

2

S

b c

a a

Thus we can find S1 and S2 just by examining the coefficients of the cubic equation. All other Sns can bededuced (although it might be quite tedious!).



You should now be able to do this problem…

3 23 6 10 0x x The roots of

are denoted as: , ,

Find the value of: 7 7 7


Worked SolutionWorked Solution

0 3S

1 sum of roots = 2b

Sa

2 2 2 22

21

( ) 2( )

2( / )

S

S c a

22 2 2 0 4S

To go further and find S7, you must write downthe recurrence relation for Sn.



3 23 6 10 0x x 3 23 6 10 0n n nS S S

Do you see how to getthis?

The remainder is trivial, if laborious, calculation:

3 2 0

133

Let 0;

3 6 10 0

6 4 10 3 18

n

S S S

S

4 3 1

134

Let 1;

3 6 10 0

1286 18 10 2

3

n

S S S

S

6 5 3

136

Let 3;

3 6 10 0

296 7726 10 18

3 3

n

S S S

S

5 4 2

135

Let 2;

3 6 10 0

128 2966 10 4

3 3

n

S S S

S



13

4

7

7 6

772 128 59126 10

Let 4;

3 3

1

9

3 6 0 0

n

S S S

S

…which finally gives the desired result:

This example is rather complex, but only a little moredifficult than the questions you will find on the exam.


ReviewReview

Here are the key points so far:

•Remember the relationship between the coefficients and the roots of a cubic polynomial

•Remember what you learnt in P3 about real/complex roots, curve sketching and factorizing using the factor theorem

•Realize that it is possible to find relationships between the sum of the powers of the roots (recurrence relationships, specifically)

•Realize that sometimes by making a substitution you can find further results about the roots


PROBLEMSPROBLEMS

1. A cubic polynomial has root +2. The product of its three roots is -9/2. The sum of its roots is 43/4. Write down the polynomial.

2. How many different cubic polynomials with positive integer roots have -98 as their constant term? (Assume the coefficient of x3 is 1).

3. If that was easy, try this one: how many different polynomials of degree 7, with positive integer roots, have -9800 as their constant term? Can you find a generalized algorithm to solve this kind of problem?Answers: 1) (x-2)(x-9)(x+1/4) 2) 5 3) 65

General formula is , where n is the number of prime factors.

1 1 122 1, n odd or 1 2 , n even

/ 2n n n

n


Polynomials – tips and tricksPolynomials – tips and tricks

The most useful ideas to remember:

22 1 2S S

Note that this is always true, for polynomials of any degree. For example, degree 4:

2 2 2 2 22 ( ) 2S

Can you see why this must be true?


For degree n:

1

(sum of the (n-1)th products of roots)

product of the rootsS

3 2

1

0

1 1 1 /

/

ax bx cx d

c aS

d a

To see this, consider degree 3 (note the reason for the negative sign!):


Having these two simple formulas for S(2) and S(-1), combined with the fact that S(1) and S(0) can be observed immediately, makes it easy to find any S(n).

1

(sum of the (n-1)th products of roots)

product of the rootsS

22 1 2S S 1 /S b a

1for ... 0n nax bx k

0S n

polynomials cie centre a-level further pure maths

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