polynomials cie centre a-level further pure maths
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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.
Polynomials AgainPolynomials Again
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
Polynomials AgainPolynomials Again
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.
Polynomials AgainPolynomials Again
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.
Polynomials AgainPolynomials Again
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
Polynomials AgainPolynomials Again
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…
Polynomials AgainPolynomials Again
3 23 6 10 0x x The roots of
are denoted as: , ,
Find the value of: 7 7 7
Investigate!
Polynomials AgainPolynomials Again
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
Polynomials AgainPolynomials Again
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
Polynomials AgainPolynomials Again
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:
Polynomials AgainPolynomials Again
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 ?
Polynomials AgainPolynomials Again
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!).
Polynomials AgainPolynomials Again
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.
Worked SolutionWorked Solution
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
Worked SolutionWorked Solution
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?
Polynomials – tips and tricksPolynomials – tips and tricks
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!):
Polynomials – tips and tricksPolynomials – tips and tricks
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